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Mathematical Tools & Measurement

  1. 01Basic Essential Mathematics
  2. 02Vectors
  3. 03Differentiation
  4. 04Applications of Differentiation
  5. 05Integration
  6. 06Applications of Integration
  7. 07Physical Quantities and Units
  8. 08Dimensional Formula
  9. 09Dimensional Analysis and Its Applications
  10. 10Experimental Skills

Kinematics

  1. 01Motion in One Dimension
  2. 02Motion in Multiple Dimensions
  3. 03Relative Velocity
  4. 04Circular Motion Kinematics
  5. 05Circular Motion Dynamics

Dynamics

  1. 01Forces and Laws of Motion
  2. 02Laws of Motion
  3. 03Friction
  4. 04Force and Potential Energy
  5. 05Fundamentals of Force
  6. 06Newton's Laws and Free Body Diagrams
  7. 07Applications: Objects in Equilibrium
  8. 08Applications: Objects in Motion
  9. 09Constraint Relations
  10. 10Inertial and Non-Inertial Frames
  11. 11Basics of Friction
  12. 12Applications of Friction

Work, Energy, and Power

  1. 01Conservation of Mechanical Energy
  2. 02Work and the Work-Energy Theorem
  3. 03Work and Kinetic Energy Theorem
  4. 04Energy and its Conservation
Theory/Mathematical Tools & Measurement

Mathematical Tools & Measurement · Chapter 02

Vectors

Comprehensive study guide on Vectors, Vector Algebra (Addition and Subtraction), and Vector Multiplication.

218 min read · 16 topics

01

Introduction to Vectors

Why Numbers Alone Are Not Enough

Consider a simple question: if a force of 10 N and a force of 8 N act on a block, what is the total resulting force?

Treated as ordinary numbers, the two forces seem to add to 18 N. In fact, that is only one of many possibilities. What if the 10 N force is pushing the block forward, and the 8 N force is pushing it backward? The net effect would be a mere 2 N forward. What if they pull at right angles, or at some other diagonal angle to each other? The answer changes completely. You simply cannot calculate the total effect without knowing their directions.

Numbers alone, therefore, are not always enough to describe nature. Some quantities require both a magnitude (how much) and a direction (which way) to make sense. This chapter develops the tools used to describe and manipulate such quantities.

02

Physical Quantities: Scalars and Vectors

Classifying Our Measurements

In physics, we describe the universe by measuring its properties. Any property of an object or system that can be measured and quantified is called a physical quantity. For instance, we can measure how heavy a stone is (mass), how hot a cup of tea is (temperature), or how fast a runner is moving (speed).

To handle these measurements mathematically, we group physical quantities into two fundamental classes based on whether they care about spatial direction: scalars and vectors.

Scalar Quantities

A scalar quantity is one that is completely described by a single numerical value—its magnitude—accompanied by its appropriate physical unit. Scalars are simple: they tell us "how much" or "how many," and they do not have any direction in space.

Consider these everyday examples of scalars:

  • Mass: An apple has a mass of 150 g150 \text{ g}150 g. It makes no sense to ask "in what direction" the mass points.
  • Temperature: A bowl of soup is at 60∘C60^\circ\text{C}60∘C. It is hot in all directions equally.
  • Distance: A walk along a winding path may cover a distance of 5 km5 \text{ km}5 km. This is the total length of the path you walked, regardless of where it led.
  • Speed: A car's speedometer reads 60 km/h60 \text{ km/h}60 km/h. It tells you how fast the car is moving, but not where it is heading.
  • Energy: A battery stores 10 Wh10 \text{ Wh}10 Wh of energy. Energy is a capacity, not a path.

Because scalars are just numbers, you can add, subtract, multiply, and divide them using the ordinary rules of algebra.

Which of the following physical quantities is completely described by its magnitude and unit alone, requiring no directional information?

Vector Quantities

A vector quantity is one that requires both a magnitude and a direction in space for its complete description. If you omit the direction, the description is physically incomplete.

The following examples show how vectors differ from scalars:

  • Displacement: Instead of just distance, displacement tells you how your position changed. If you walk 5 km5 \text{ km}5 km East, you have specified a vector.
  • Velocity: Velocity is speed with a direction. A car traveling at 60 km/h60 \text{ km/h}60 km/h North has a velocity vector.
  • Force: Pushing a box with 20 N20 \text{ N}20 N of force downwards has a very different physical outcome than pushing it 20 N20 \text{ N}20 N upwards.
  • Acceleration: Gravity accelerates falling objects at 9.8 m/s29.8 \text{ m/s}^29.8 m/s2 straight down.

Because vectors carry spatial direction, we cannot add them using standard arithmetic. A displacement of 3 m3 \text{ m}3 m plus a displacement of 4 m4 \text{ m}4 m does not necessarily equal 7 m7 \text{ m}7 m—it depends entirely on how the two movements align. The rules for combining vectors are developed in the sections that follow.

A speed of 60 km/h is a scalar: one number on a dial. The same 60 km/h given a heading, drawn as an arrow pointing North, is a velocity vector.
A speed of 60 km/h is a scalar: one number on a dial. The same 60 km/h given a heading, drawn as an arrow pointing North, is a velocity vector.

03

Representing Vectors

Two Ways to Represent a Vector

To work with vectors, we need a clear system to talk about them (notation) and visualize them (diagrams). In physics, we rely on two main methods of representation: a graphical representation (drawing arrows) and an algebraic representation (using symbols and numbers). We switch between these two modes constantly, using diagrams to build intuition and algebra to perform exact calculations.

Graphical Representation

The most intuitive way to represent a vector is diagrammatically as an arrow in space. This arrow is a literal model of the vector's two properties:

  • The length of the arrow is drawn to scale to represent the magnitude of the vector. A longer arrow means a larger quantity (e.g., a stronger force or faster speed).
  • The direction the arrowhead points is the direction of the vector.

The starting point of the arrow is called the tail, and the pointed end is the head. In diagrams, the angle of the arrow is typically measured from a reference direction, such as the positive x-axis or North.

A vector drawn as an arrow: the tail is its starting point, the head its end, the length of the arrow represents the magnitude, and the angle from the x-axis gives its direction.
A vector drawn as an arrow: the tail is its starting point, the head its end, the length of the arrow represents the magnitude, and the angle from the x-axis gives its direction.

Symbolic Notation

When writing equations, how do we tell the reader that a letter represents a vector rather than a normal number? We use special typographical symbols:

  • The Arrow Notation: We write a small arrow above the variable, such as A⃗\vec{A}A or F⃗\vec{F}F. The arrow marks the symbol as a vector: a reminder that it cannot be combined with other terms by ordinary arithmetic.
  • Boldface Notation: In printed textbooks, vectors are often printed in bold, like A or F. Because boldface is difficult to write by hand, arrow notation is preferred for handwritten work.

The magnitude of a vector is a scalar—it is simply a positive number with a unit. We write the magnitude of vector A⃗\vec{A}A using absolute value bars as ∣A⃗∣|\vec{A}|∣A∣, or simply as the letter in standard italics, AAA.

For instance, if v⃗\vec{v}v represents a velocity vector of 10 m/s10 \text{ m/s}10 m/s East, its magnitude is v=∣v⃗∣=10 m/sv = |\vec{v}| = 10 \text{ m/s}v=∣v∣=10 m/s.

If p⃗\vec{p}p​ represents the momentum vector of an object, which of the following expressions correctly represents its scalar magnitude?

Analytical Representation: Component Form

Drawing arrows helps us visualize physical situations, but it is not convenient for precise mathematics. Instead of drawing, we can represent a vector analytically by breaking it down into components along perpendicular coordinate axes (usually the horizontal x-axis and vertical y-axis).

You can think of a component as the "projection" or shadow of the vector along a specific direction. To write vectors in this component form, we introduce unit vectors, which are vectors of magnitude 1 pointing along the coordinate axes:

  • i^\hat{i}i^ (pronounced "i-hat") points in the positive x-direction.
  • j^\hat{j}j^​ (pronounced "j-hat") points in the positive y-direction.
  • k^\hat{k}k^ (pronounced "k-hat") points in the positive z-direction (for three dimensions).

Using these unit vectors as direction markers, we write any vector A⃗\vec{A}A as a sum of its horizontal component AxA_xAx​ and its vertical component AyA_yAy​:

A⃗=Axi^+Ayj^\vec{A} = A_x\hat{i} + A_y\hat{j}A=Ax​i^+Ay​j^​

For example, if a force pulls 3 Newtons to the right and 4 Newtons upward, we write it as F⃗=3i^+4j^ N\vec{F} = 3\hat{i} + 4\hat{j} \text{ N}F=3i^+4j^​ N. This is the analytical component form of the vector.

Component Form in Use: Two Walks

Take i^\hat{i}i^ to point East, j^\hat{j}j^​ North, and k^\hat{k}k^ vertically up.

A walk in the plane. A person walks 30 m30 \text{ m}30 m East and then 40 m40 \text{ m}40 m North. The two legs are 30i^30\hat{i}30i^ and 40j^40\hat{j}40j^​, so the total displacement is

d⃗=30i^+40j^ m,d=302+402=50 m\vec{d} = 30\hat{i} + 40\hat{j} \text{ m}, \qquad d = \sqrt{30^2 + 40^2} = 50 \text{ m}d=30i^+40j^​ m,d=302+402​=50 m

A walk that leaves the plane. Another person walks 20 m20 \text{ m}20 m West, then 10 m10 \text{ m}10 m North, then enters a lift and rises 15 m15 \text{ m}15 m. West is the negative i^\hat{i}i^ direction and the lift ride adds a k^\hat{k}k^ component:

d⃗=−20i^+10j^+15k^ m,d=(−20)2+102+152=725≈26.9 m\vec{d} = -20\hat{i} + 10\hat{j} + 15\hat{k} \text{ m}, \qquad d = \sqrt{(-20)^2 + 10^2 + 15^2} = \sqrt{725} \approx 26.9 \text{ m}d=−20i^+10j^​+15k^ m,d=(−20)2+102+152​=725​≈26.9 m

Each movement simply contributes its own component, and the components never mix — that is what makes the component form the working language of vector calculations.

The Position Vector

One particular vector deserves its own name because it appears everywhere in mechanics: the position vector. The position vector of a point is the displacement vector starting from the origin and ending at that point.

Because it is measured from the origin, a position vector depends entirely on the choice of origin and the orientation of the coordinate axes you are using — change either one, and the position vector of the same physical point changes too. This is different from a general vector like a force or a velocity, whose value doesn't care where you place the origin.

The origin of the coordinate system is moved to a new location. Which of the following quantities changes?

The Cartesian System of Vectors in 3D

So far we have written vectors using two perpendicular unit vectors, i^\hat{i}i^ and j^\hat{j}j^​, for problems confined to a flat plane. But most of the real world is three-dimensional, so we extend the same idea by introducing a third unit vector, k^\hat{k}k^, pointing along the z-axis. Together, i^\hat{i}i^, j^\hat{j}j^​, and k^\hat{k}k^ are three mutually perpendicular unit vectors along the x, y, and z axes of a right-handed coordinate system.

The three mutually perpendicular unit vectors i-hat, j-hat, and k-hat along the x, y, and z axes of a right-handed coordinate system; drag the scene to turn it, and zoom with the buttons

The Position Vector in 3D

Consider a point P(a,b,c)P(a, b, c)P(a,b,c) in three-dimensional space. The position vector r⃗\vec{r}r of PPP, drawn from the origin OOO, is written as the sum of its three components along the axes:

OP→=r⃗=ai^+bj^+ck^\overrightarrow{OP} = \vec{r} = a\hat{i} + b\hat{j} + c\hat{k}OP=r=ai^+bj^​+ck^

The position vector r of point P(a,b,c) built from its components: a along the x axis, then b parallel to the y axis, then c straight up. Drag the scene to turn it, and zoom with the buttons.

The magnitude of this position vector follows directly from the three-dimensional Pythagorean theorem:

r=a2+b2+c2r = \sqrt{a^2 + b^2 + c^2}r=a2+b2+c2​

Equality of Vectors in Component Form

Writing vectors in Cartesian form also gives us a clean, purely algebraic test for equality: two vectors expressed in i^,j^,k^\hat{i}, \hat{j}, \hat{k}i^,j^​,k^ form are equal only when their corresponding components are separately equal. That is, if A⃗=Axi^+Ayj^+Azk^\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}A=Ax​i^+Ay​j^​+Az​k^ and B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}B=Bx​i^+By​j^​+Bz​k^, then A⃗=B⃗\vec{A} = \vec{B}A=B if and only if Ax=BxA_x = B_xAx​=Bx​, Ay=ByA_y = B_yAy​=By​, and Az=BzA_z = B_zAz​=Bz​ all hold simultaneously.

Example 14: Position Vector and its Magnitude

Problem: For a point P(2,4,−5)P(2, 4, -5)P(2,4,−5) in a three-dimensional coordinate system, write the position vector r⃗\vec{r}r of PPP and find its magnitude ∣r⃗∣|\vec{r}|∣r∣.

Solution: Reading the coordinates directly as the components of the position vector,

r⃗=2i^+4j^−5k^\vec{r} = 2\hat{i} + 4\hat{j} - 5\hat{k}r=2i^+4j^​−5k^

and its magnitude follows from the three-dimensional Pythagorean theorem:

∣r⃗∣=rx2+ry2+rz2=22+42+(−5)2=45=35|\vec{r}| = \sqrt{r_x^2 + r_y^2 + r_z^2} = \sqrt{2^2 + 4^2 + (-5)^2} = \sqrt{45} = 3\sqrt{5}∣r∣=rx2​+ry2​+rz2​​=22+42+(−5)2​=45​=35​

Now Solve: Find the magnitude of the position vector of the point Q(1,−2,2)Q(1, -2, 2)Q(1,−2,2).
Answer: 3.

04

Properties of Vectors

Properties of Vectors

Just as numbers have rules and properties (like the identity property x+0=xx + 0 = xx+0=x), vectors have their own unique properties that govern how they behave. Before we perform algebra on vectors, we must establish these fundamental properties.

1. Equality of Vectors

Two vectors, A⃗\vec{A}A and B⃗\vec{B}B, are defined as equal if, and only if, they have the same magnitude and the same direction. We write this as A⃗=B⃗\vec{A} = \vec{B}A=B.

Notice what is missing from this definition: position. It does not matter where a vector starts or ends in space. As long as its length is the same and it points in the same direction, it is the same vector. This means vectors are portable—you can slide them anywhere in space without altering their value, which is a powerful shortcut when adding vectors graphically.

Two arrows with the same length and the same direction are the same vector, no matter where they start: A = B.
Two arrows with the same length and the same direction are the same vector, no matter where they start: A = B.

2. The Unit Vector

A unit vector is a special vector with a magnitude of exactly 1. It carries no physical units of its own; its sole purpose is to point, establishing a direction in space. It is a mathematical signpost.

If you have any vector A⃗\vec{A}A, you can find its corresponding unit vector, written with a caret or "hat" as A^\hat{A}A^ (pronounced "A-hat"), by dividing the vector by its own magnitude:

A^=A⃗∣A⃗∣\hat{A} = \frac{\vec{A}}{|\vec{A}|}A^=∣A∣A​

This operation scales the vector's length down to exactly 1 while preserving its direction. Consequently, any vector can be written as the product of its size (magnitude) and its direction (unit vector):

A⃗=AA^\vec{A} = A \hat{A}A=AA^

A velocity vector is given in component form as v⃗=3i^+4j^ m/s\vec{v} = 3\hat{i} + 4\hat{j} \text{ m/s}v=3i^+4j^​ m/s. What is the unit vector v^\hat{v}v^ pointing in the direction of this velocity?

3. The Negative of a Vector

The negative of a vector A⃗\vec{A}A, written as −A⃗-\vec{A}−A, is a vector with the same magnitude (length) as A⃗\vec{A}A, but pointing in the exact opposite direction (a 180∘180^\circ180∘ rotation).

A vector and its negative: equal lengths, marked by matching ticks, but exactly opposite directions.
A vector and its negative: equal lengths, marked by matching ticks, but exactly opposite directions.

In vector algebra, the negative sign does not mean "less than zero"—magnitude can never be negative. Instead, the negative sign is a directional instruction to flip the arrow by 180∘180^\circ180∘.

4. The Zero (Null) Vector

The zero vector, denoted as 0⃗\vec{0}0, is a vector with a magnitude of exactly zero. Because it has no length, its direction is undefined. It acts as the identity element for vector addition, meaning A⃗+0⃗=A⃗\vec{A} + \vec{0} = \vec{A}A+0=A.

A zero vector arises naturally in physics when physical effects cancel out (e.g., when equal forces pull in opposite directions) or when an object is at rest, resulting in a velocity of 0⃗\vec{0}0.

5. Parallel and Antiparallel Vectors

Vectors are described as parallel if they point in the exact same direction, regardless of whether their magnitudes are equal. The angle between parallel vectors is 0∘0^\circ0∘.

Vectors are described as antiparallel if they point in exact opposite directions. The angle between antiparallel vectors is 180∘180^\circ180∘.

Parallel vectors point the same way and antiparallel vectors point opposite ways; their magnitudes need not be equal.
Parallel vectors point the same way and antiparallel vectors point opposite ways; their magnitudes need not be equal.

6. The Angle Between Two Vectors

We often need to speak precisely about how two vectors are oriented relative to each other. The angle between two vectors A⃗\vec{A}A and B⃗\vec{B}B is defined as the angle through which one vector must be rotated about their common tail so that its direction coincides with the direction of the other vector.

To measure this angle correctly, always slide the two vectors (recall that vectors are portable) so that their tails meet at a common point, and then read off the angle between them at that point — never the angle measured between their arrowheads or along some other path.

Example 5: Reading the Angle Between Two Vectors

Problem: For each of the three configurations below, find the angle between a⃗\vec{a}a and b⃗\vec{b}b.

Case (i): The vectors are drawn tail-to-tail with an included angle of 60∘60^\circ60∘ directly visible between them.
Solution: Since the vectors already share a common tail and the angle is read directly between them, the angle between a⃗\vec{a}a and b⃗\vec{b}b is simply 60∘60^\circ60∘.

Case (ii): Vector b⃗\vec{b}b is drawn making an angle of 60∘60^\circ60∘ with the ray opposite to a⃗\vec{a}a (i.e., with the extension of a⃗\vec{a}a through the common tail).
Solution: The angle between a⃗\vec{a}a and b⃗\vec{b}b must be measured from a⃗\vec{a}a itself, not its extension. Since b⃗\vec{b}b makes 60∘60^\circ60∘ with the opposite ray of a⃗\vec{a}a, it makes 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ180∘−60∘=120∘ with a⃗\vec{a}a itself. The angle between them is 120∘120^\circ120∘.

Case (iii): The vectors are drawn tail-to-tail, nearly opposing each other, with the reflex/adjacent construction showing an angle of 120∘120^\circ120∘ on one side.
Solution: The actual angle between a⃗\vec{a}a and b⃗\vec{b}b, taken as the angle you would rotate a⃗\vec{a}a through to align with b⃗\vec{b}b (always the value between 0∘0^\circ0∘ and 180∘180^\circ180∘), works out to 60∘60^\circ60∘ here.

Now Solve: What is the angle between i^\hat{i}i^ and i^+j^\hat{i} + \hat{j}i^+j^​?
Answer: 45∘45^\circ45∘, since i^+j^\hat{i}+\hat{j}i^+j^​ points exactly along the bisector between the x-axis and y-axis.

05

Resolving Vectors into Components

Two Descriptions of the Same Vector

We have seen that a vector can be described in two ways: by its magnitude and direction (like a velocity of 10 m/s10 \text{ m/s}10 m/s at 30∘30^\circ30∘ above the horizontal) or by its components (like 8.66i^+5j^ m/s8.66\hat{i} + 5\hat{j} \text{ m/s}8.66i^+5j^​ m/s).

To solve physics problems, we must translate between these two descriptions. This process is called vector resolution. In this section, we will learn how to break a diagonal vector down into its perpendicular parts, and conversely, how to combine those parts back into a single magnitude and direction.

Resolving a Vector into Components

Consider a vector A⃗\vec{A}A making an angle θ\thetaθ with the positive x-axis. By dropping perpendicular lines from the head of the vector to the axes, we form a right-angled triangle where the vector is the hypotenuse and the components are the legs.

A vector at angle theta to the x-axis resolved into a horizontal component A cos theta and a vertical component A sin theta.
A vector at angle theta to the x-axis resolved into a horizontal component A cos theta and a vertical component A sin theta.

Using basic trigonometry (sine and cosine), we can calculate the components immediately:

  • Horizontal Component (adjacent to the angle):
    Ax=Acos⁡θA_x = A \cos\thetaAx​=Acosθ
  • Vertical Component (opposite to the angle):
    Ay=Asin⁡θA_y = A \sin\thetaAy​=Asinθ

These components tell you exactly how much of the vector's influence points along the x-direction and how much points along the y-direction.

A force of 10 N10 \text{ N}10 N is applied to a block at an angle of 60∘60^\circ60∘ above the horizontal. What is the magnitude of the horizontal component of this force pulling the block forward? (Given cos⁡(60∘)=0.5\cos(60^\circ) = 0.5cos(60∘)=0.5 and sin⁡(60∘)≈0.866\sin(60^\circ) \approx 0.866sin(60∘)≈0.866)

From Components Back to Magnitude and Direction

What if you know the components AxA_xAx​ and AyA_yAy​ and want to find the overall vector's length and heading? We can reconstruct the vector using the same right triangle.

1. Calculating the Magnitude

Because the components AxA_xAx​ and AyA_yAy​ form the perpendicular legs of a right triangle, the magnitude AAA (the hypotenuse) is found using the Pythagorean theorem:

A=Ax2+Ay2A = \sqrt{A_x^2 + A_y^2}A=Ax2​+Ay2​​

Since squaring negative numbers yields positive results, the magnitude is always a positive number (or zero), representing the absolute length of the vector.

2. Calculating the Direction (with Quadrant Correction)

The direction angle θ\thetaθ is related to the components by the tangent function, which is the ratio of the opposite leg to the adjacent leg:

tan⁡θ=AyAx\tan\theta = \frac{A_y}{A_x}tanθ=Ax​Ay​​

To solve for θ\thetaθ, we use the inverse tangent function:

θ=tan⁡−1(AyAx)\theta = \tan^{-1}\left( \frac{A_y}{A_x} \right)θ=tan−1(Ax​Ay​​)

CRITICAL WARNING: The Quadrant Trap
Standard calculators and mathematical definitions of tan⁡−1\tan^{-1}tan−1 only return angles between −90∘-90^\circ−90∘ and 90∘90^\circ90∘. To find the true physical angle from the positive x-axis, you must check the signs of the components and adjust as follows:
  • Quadrant I (+Ax,+Ay+A_x, +A_y+Ax​,+Ay​): θ=θref\theta = \theta_{\text{ref}}θ=θref​
  • Quadrant II (−Ax,+Ay-A_x, +A_y−Ax​,+Ay​): θ=180∘−θref\theta = 180^\circ - \theta_{\text{ref}}θ=180∘−θref​
  • Quadrant III (−Ax,−Ay-A_x, -A_y−Ax​,−Ay​): θ=180∘+θref\theta = 180^\circ + \theta_{\text{ref}}θ=180∘+θref​
  • Quadrant IV (+Ax,−Ay+A_x, -A_y+Ax​,−Ay​): θ=360∘−θref\theta = 360^\circ - \theta_{\text{ref}}θ=360∘−θref​

Where θref=tan⁡−1∣AyAx∣\theta_{\text{ref}} = \tan^{-1}\left| \frac{A_y}{A_x} \right|θref​=tan−1​Ax​Ay​​​ is the positive reference angle.

A displacement vector has components dx=−3 md_x = -3 \text{ m}dx​=−3 m and dy=3 md_y = 3 \text{ m}dy​=3 m. What is the correct direction angle θ\thetaθ of this vector measured counter-clockwise from the positive x-axis? (Given tan⁡−1(1)=45∘\tan^{-1}(1) = 45^\circtan−1(1)=45∘)

06

Introduction to Vector Addition

Combining Vectors: The Net Effect

Suppose two persons push a box, each with a force of 10 N10 \text{ N}10 N. What is the total force on the box? The question cannot be answered as it stands. If both push the same way, the total is 20 N20 \text{ N}20 N. If they push in opposite directions, the pushes cancel and the total is zero. If they push at right angles, the total is 102+102≈14.1 N\sqrt{10^2 + 10^2} \approx 14.1 \text{ N}102+102​≈14.1 N, directed between them. Without the directions, the sum of two forces is simply not determined.

Displacements show the same behaviour, and are the easiest vectors to picture. Walk 3 m3 \text{ m}3 m and then 4 m4 \text{ m}4 m: you end 7 m7 \text{ m}7 m from the start only if both walks point the same way. Walk them at right angles and you end 32+42=5 m\sqrt{3^2 + 4^2} = 5 \text{ m}32+42​=5 m from the start; walk the second leg straight back and you end only 1 m1 \text{ m}1 m away. The order and directions of the two legs decide where you finish.

Adding vectors therefore needs its own rules. The sections that follow build those rules: the Triangle Law and the Parallelogram Law add vectors by drawing, and the component method adds them by calculation.

The Resultant Vector

The sum of two or more vectors is called the resultant vector (often denoted as R⃗\vec{R}R). The resultant is a single vector that produces the same net effect as all the individual vectors combined. For example, the resultant force on an object determines its acceleration.

R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B

Example: A Tug-of-War

Consider a tug-of-war. Team A pulls left with 100 N (F⃗A\vec{F}_AFA​) and Team B pulls right with 80 N (F⃗B\vec{F}_BFB​). The resultant force is not 180 N. Since they are in opposite directions, the net effect is a force of 20 N to the left. So, R⃗=F⃗A+F⃗B=20\vec{R} = \vec{F}_A + \vec{F}_B = 20R=FA​+FB​=20 N to the left.

Range of Possible Resultants

A fascinating aspect of vector addition is that two vectors with fixed magnitudes can produce a resultant with a wide range of possible magnitudes, depending on the angle between them.

Suppose we have two vectors, A⃗\vec{A}A with magnitude AAA and B⃗\vec{B}B with magnitude BBB. The magnitude of their resultant, RRR, will be:

  • Maximum when A⃗\vec{A}A and B⃗\vec{B}B are parallel (angle = 0°). Then, Rmax=A+BR_{max} = A + BRmax​=A+B.
  • Minimum when A⃗\vec{A}A and B⃗\vec{B}B are antiparallel (angle = 180°). Then, Rmin=∣A−B∣R_{min} = |A - B|Rmin​=∣A−B∣.

For any other angle between them, the resultant's magnitude RRR will be somewhere in between these two extremes: ∣A−B∣≤R≤A+B|A - B| \le R \le A + B∣A−B∣≤R≤A+B.

The same two magnitudes give different resultants: at 0 degrees the lengths add, at 90 degrees they form a right triangle, and at 180 degrees they partly cancel.
The same two magnitudes give different resultants: at 0 degrees the lengths add, at 90 degrees they form a right triangle, and at 180 degrees they partly cancel.

A boat can travel at 8 m/s in still water. It is crossing a river that flows at 6 m/s. What is the maximum possible speed (magnitude of the resultant velocity) of the boat relative to the ground?

Methods of Addition

So, how do we actually perform this addition, especially when the vectors are not parallel or antiparallel? We will learn two primary approaches:

  1. Graphical Methods: Using geometric constructions like the 'Triangle Law' and 'Parallelogram Law' to visualize the resultant.
  2. Analytical Method: Using vector components and trigonometry for precise calculations.

We will explore these methods in detail in the following sections.

07

Triangle Law of Vector Addition

The Head-to-Tail Method

The Triangle Law provides our first graphical method for adding two vectors. It's an intuitive approach based on a 'journey' metaphor. If you walk along vector A⃗\vec{A}A and then immediately walk along vector B⃗\vec{B}B, where do you end up? The resultant vector, R⃗\vec{R}R, is the single vector that takes you directly from your starting point to your final destination.

This method is formally called the head-to-tail method because of how we arrange the vectors.

Statement of the Triangle Law

If two vectors are represented in magnitude and direction by two sides of a triangle taken in the same order (head-to-tail), then their resultant is represented in magnitude and direction by the third side of the triangle, taken in the opposite order (from the tail of the first to the head of the second).

Step-by-Step Graphical Construction:

  1. Draw the First Vector: Draw vector A⃗\vec{A}A to a chosen scale and in its specified direction.
  2. Draw the Second Vector: Place the tail of vector B⃗\vec{B}B at the head of vector A⃗\vec{A}A. Draw B⃗\vec{B}B to the same scale and in its correct direction.
  3. Draw the Resultant: The resultant vector R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B is the vector drawn from the tail of A⃗\vec{A}A to the head of B⃗\vec{B}B.
  4. Measure the Resultant: Measure the length of R⃗\vec{R}R and use the scale to find its magnitude. Measure its angle to find its direction.

The triangle law step by step: draw A, draw B starting from the head of A, then draw the resultant R from the tail of A to the head of B.
The triangle law step by step: draw A, draw B starting from the head of A, then draw the resultant R from the tail of A to the head of B.

An activity: drag the scattered vector B until its tail sits on the head of A. The triangle closes and the resultant R appears, drawn from the tail of A to the head of B.
An activity: drag the scattered vector B until its tail sits on the head of A. The triangle closes and the resultant R appears, drawn from the tail of A to the head of B.

Vector Addition is Commutative

The order in which two vectors are added does not change the result. Suppose we add B⃗+A⃗\vec{B} + \vec{A}B+A? We would first draw B⃗\vec{B}B, then draw A⃗\vec{A}A starting from the head of B⃗\vec{B}B. As you can see from the diagram, we arrive at the exact same resultant vector R⃗\vec{R}R.

A⃗+B⃗=B⃗+A⃗\vec{A} + \vec{B} = \vec{B} + \vec{A}A+B=B+A

This shows that vector addition is commutative, just like the addition of scalar numbers.

Vector Addition is Associative

When we add three or more vectors, does the way we group them matter? Suppose we want A⃗+B⃗+C⃗\vec{A} + \vec{B} + \vec{C}A+B+C. We could first add A⃗\vec{A}A and B⃗\vec{B}B, then add C⃗\vec{C}C to that sum; or we could first add B⃗\vec{B}B and C⃗\vec{C}C, then add A⃗\vec{A}A to that sum. If you draw both groupings head-to-tail, you'll find they produce the exact same final resultant vector, closing the same overall polygon. This confirms that vector addition is associative:

(A⃗+B⃗)+C⃗=A⃗+(B⃗+C⃗)(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})(A+B)+C=A+(B+C)

This means that when adding several vectors, you're free to group and reorder them however is most convenient, and the final resultant will never change.

Magnitude and Direction of the Resultant

The triangle construction also gives an exact formula for the resultant. Let B⃗\vec{B}B be drawn from the head of A⃗\vec{A}A, leaving it at the angle α\alphaα to the direction of A⃗\vec{A}A. Extend the line of A⃗\vec{A}A and drop a perpendicular from the head of B⃗\vec{B}B onto it. The perpendicular has length Bsin⁡αB\sin\alphaBsinα, and it meets the extension at a distance Bcos⁡αB\cos\alphaBcosα past the head of A⃗\vec{A}A.

The head-to-tail triangle construction with the line of A extended: the dashed right triangle has base A + B cos alpha and height B sin alpha, and R is its hypotenuse.
The head-to-tail triangle construction with the line of A extended: the dashed right triangle has base A + B cos alpha and height B sin alpha, and R is its hypotenuse.

The resultant R⃗\vec{R}R is then the hypotenuse of a right triangle with base A+Bcos⁡αA + B\cos\alphaA+Bcosα and height Bsin⁡αB\sin\alphaBsinα. Applying the Pythagorean theorem and using cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1cos2α+sin2α=1:

R2=(A+Bcos⁡α)2+(Bsin⁡α)2⇒R=A2+B2+2ABcos⁡αR^2 = (A + B\cos\alpha)^2 + (B\sin\alpha)^2 \quad \Rightarrow \quad R = \sqrt{A^2 + B^2 + 2AB\cos\alpha}R2=(A+Bcosα)2+(Bsinα)2⇒R=A2+B2+2ABcosα​

From the same right triangle, the angle θ\thetaθ that R⃗\vec{R}R makes with A⃗\vec{A}A is

tan⁡θ=Bsin⁡αA+Bcos⁡α\tan\theta = \frac{B\sin\alpha}{A + B\cos\alpha}tanθ=A+BcosαBsinα​

Special Cases

Putting particular angles into R=A2+B2+2ABcos⁡αR = \sqrt{A^2 + B^2 + 2AB\cos\alpha}R=A2+B2+2ABcosα​ covers the situations met most often:

  • α=0∘\alpha = 0^\circα=0∘ (same direction): cos⁡0∘=1\cos 0^\circ = 1cos0∘=1, so R=(A+B)2=A+BR = \sqrt{(A+B)^2} = A + BR=(A+B)2​=A+B, and θ=0\theta = 0θ=0. The magnitudes simply add. This is the largest possible resultant.
  • α=90∘\alpha = 90^\circα=90∘ (perpendicular): cos⁡90∘=0\cos 90^\circ = 0cos90∘=0, so R=A2+B2R = \sqrt{A^2 + B^2}R=A2+B2​ and tan⁡θ=B/A\tan\theta = B/Atanθ=B/A.
  • α=180∘\alpha = 180^\circα=180∘ (opposite directions): cos⁡180∘=−1\cos 180^\circ = -1cos180∘=−1, so R=(A−B)2=∣A−B∣R = \sqrt{(A-B)^2} = |A - B|R=(A−B)2​=∣A−B∣, directed along the larger vector. This is the smallest possible resultant.

The three special cases side by side: the magnitudes add at 0 degrees, combine by the Pythagorean theorem at 90 degrees, and subtract at 180 degrees.
The three special cases side by side: the magnitudes add at 0 degrees, combine by the Pythagorean theorem at 90 degrees, and subtract at 180 degrees.

For any angle between these extremes, the magnitude of the resultant lies between ∣A−B∣|A-B|∣A−B∣ and A+BA+BA+B.

Extending to Many Vectors: The Polygon Law

The head-to-tail idea behind the Triangle Law extends naturally to more than two vectors. To add several vectors, draw them one after another in order, always starting the next vector from the head of the previous one. The resultant is then the single vector drawn from the tail of the very first vector to the head of the very last vector — closing the polygon formed by all the vectors. This generalization is called the Polygon Law of Vector Addition.

A Useful Corollary: Closed Polygons Sum to Zero

An important special case falls directly out of the Polygon Law: if a set of nnn vectors, added head-to-tail in this way, forms a closed polygon — meaning the head of the last vector lands exactly back on the tail of the first — then their resultant is the zero vector. There is no 'gap' left to close, so there is nothing left for the resultant to represent.

Five vectors drawn head-to-tail form a closed pentagon: the head of the last vector touches the tail of the first. What is their resultant?

The diagram shows two vectors, P and Q, being added. Which vector correctly represents the resultant R = P + Q? ___img[m2_triangle_law_quiz.svg, 1.0, Vectors P and Q drawn head-to-tail, with three dashed candidate arrows numbered 1, 2 and 3.]

Example 6: Constructing the Vector Diagram for a Resultant

Problem: For each configuration below, draw the vector diagram (using the Triangle Law) to find the resultant R⃗=a⃗+b⃗\vec{R} = \vec{a} + \vec{b}R=a+b.

(i) a⃗\vec{a}a and b⃗\vec{b}b are inclined to each other at some angle θ\thetaθ.
Construction: Draw a⃗\vec{a}a first. From the head of a⃗\vec{a}a, draw b⃗\vec{b}b at the given angle θ\thetaθ to a⃗\vec{a}a, preserving its own magnitude and direction. The resultant R⃗\vec{R}R is the vector from the tail of a⃗\vec{a}a to the head of b⃗\vec{b}b.

(ii) a⃗\vec{a}a and b⃗\vec{b}b are perpendicular to each other.
Construction: Draw a⃗\vec{a}a, then draw b⃗\vec{b}b starting from the head of a⃗\vec{a}a at exactly 90∘90^\circ90∘ to it. The resultant R⃗\vec{R}R, from the tail of a⃗\vec{a}a to the head of b⃗\vec{b}b, is now the hypotenuse of a right triangle, so R=a2+b2R = \sqrt{a^2+b^2}R=a2+b2​.

(iii) a⃗\vec{a}a and b⃗\vec{b}b are inclined at the same angle θ\thetaθ, but with b⃗\vec{b}b drawn on the opposite side compared to case (i).
Construction: The construction proceeds exactly as before — draw a⃗\vec{a}a, then b⃗\vec{b}b from its head at angle θ\thetaθ, now measured on the other side. The resultant is still the vector from the tail of a⃗\vec{a}a to the head of b⃗\vec{b}b, but its direction in space is now mirrored relative to case (i), even though its magnitude R=a2+b2+2abcos⁡θR=\sqrt{a^2+b^2+2ab\cos\theta}R=a2+b2+2abcosθ​ is unchanged.

This exercise is a reminder that the Triangle Law is purely a geometric construction rule — the magnitude and direction of the resultant depend only on the magnitudes of a⃗\vec{a}a, b⃗\vec{b}b, and the angle between them, never on which side of a⃗\vec{a}a you happen to draw b⃗\vec{b}b.

The three constructions drawn head-to-tail: b at a general angle above a, b perpendicular to a, and b at the same angle on the opposite side, with the resultant R closing each triangle.
The three constructions drawn head-to-tail: b at a general angle above a, b perpendicular to a, and b at the same angle on the opposite side, with the resultant R closing each triangle.

08

Parallelogram Law of Vector Addition

The Tail-to-Tail Method

The Parallelogram Law is another graphical method for adding two vectors. While the Triangle Law is a 'head-to-tail' method, the Parallelogram Law is a 'tail-to-tail' method. It is particularly useful for visualizing problems where two vectors act on the same point, such as two forces pulling on an object.

Statement of the Parallelogram Law

If two vectors, acting simultaneously at a point, are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point.

Step-by-Step Graphical Construction:

  1. Draw Vectors Tail-to-Tail: Draw vector A⃗\vec{A}A and vector B⃗\vec{B}B starting from the same origin point.
  2. Complete the Parallelogram: Draw a line parallel to A⃗\vec{A}A from the head of B⃗\vec{B}B, and another line parallel to B⃗\vec{B}B from the head of A⃗\vec{A}A. These lines will intersect to form a parallelogram.
  3. Draw the Resultant: The resultant vector R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B is the diagonal of the parallelogram that starts from the common origin point.

The parallelogram law step by step: draw A and B tail-to-tail, complete the parallelogram with dashed sides, then draw R along the diagonal from the common tail.
The parallelogram law step by step: draw A and B tail-to-tail, complete the parallelogram with dashed sides, then draw R along the diagonal from the common tail.

Connection to the Triangle Law

The Parallelogram Law is just a different way to visualize the Triangle Law. Notice that the side of the parallelogram opposite to B⃗\vec{B}B is equal to B⃗\vec{B}B. If you hide the original B⃗\vec{B}B and the side parallel to A⃗\vec{A}A, you are left with a perfect 'head-to-tail' addition of A⃗\vec{A}A and B⃗\vec{B}B forming a triangle with the resultant R⃗\vec{R}R.

The Same Resultant, Seen on the Parallelogram

The formula for the resultant was derived in the previous section from the triangle construction:

R=A2+B2+2ABcos⁡αtan⁡θ=Bsin⁡αA+Bcos⁡αR = \sqrt{A^2 + B^2 + 2AB\cos\alpha} \qquad \tan\theta = \frac{B\sin\alpha}{A + B\cos\alpha}R=A2+B2+2ABcosα​tanθ=A+BcosαBsinα​

The parallelogram shows exactly the same right triangle. Draw A⃗\vec{A}A and B⃗\vec{B}B tail-to-tail with the angle α\alphaα between them and complete the parallelogram; the diagonal from the common tail is R⃗\vec{R}R. Dropping a perpendicular from the head of R⃗\vec{R}R onto the extension of A⃗\vec{A}A reproduces the base A+Bcos⁡αA + B\cos\alphaA+Bcosα and the height Bsin⁡αB\sin\alphaBsinα, because the far side of the parallelogram is just B⃗\vec{B}B shifted to the head of A⃗\vec{A}A.

The parallelogram on A and B with the diagonal R: the dashed perpendicular from the head of R reproduces the same right triangle with base A + B cos alpha and height B sin alpha.
The parallelogram on A and B with the diagonal R: the dashed perpendicular from the head of R reproduces the same right triangle with base A + B cos alpha and height B sin alpha.

By the symmetric argument — resolving A⃗\vec{A}A along and perpendicular to B⃗\vec{B}B — the angle θ′\theta'θ′ that R⃗\vec{R}R makes with B⃗\vec{B}B satisfies tan⁡θ′=Asin⁡αB+Acos⁡α\tan\theta' = \dfrac{A\sin\alpha}{B + A\cos\alpha}tanθ′=B+AcosαAsinα​, and θ+θ′=α\theta + \theta' = \alphaθ+θ′=α.

Example 7: Applying the Resultant Formula

Problem: Use the formula R=A2+B2+2ABcos⁡αR = \sqrt{A^2+B^2+2AB\cos\alpha}R=A2+B2+2ABcosα​ to find the magnitude of the resultant of a⃗\vec{a}a and b⃗\vec{b}b in each case.

(i) ∣a⃗∣=5,∣b⃗∣=4,θ=60∘|\vec{a}| = 5, |\vec{b}| = 4, \theta = 60^\circ∣a∣=5,∣b∣=4,θ=60∘

∣R⃗∣=52+42+2(5)(4)cos⁡60∘=61|\vec{R}| = \sqrt{5^2+4^2+2(5)(4)\cos60^\circ} = \sqrt{61}∣R∣=52+42+2(5)(4)cos60∘​=61​

(ii) ∣a⃗∣=2,∣b⃗∣=3,θ=135∘|\vec{a}| = \sqrt{2}, |\vec{b}| = 3, \theta = 135^\circ∣a∣=2​,∣b∣=3,θ=135∘

∣R⃗∣=(2)2+32+22(3)cos⁡135∘=2+9−6=5|\vec{R}| = \sqrt{(\sqrt2)^2+3^2+2\sqrt2(3)\cos135^\circ} = \sqrt{2+9-6} = \sqrt{5}∣R∣=(2​)2+32+22​(3)cos135∘​=2+9−6​=5​

(iii) ∣a⃗∣=3,∣b⃗∣=4,θ=60∘|\vec{a}| = 3, |\vec{b}| = 4, \theta = 60^\circ∣a∣=3,∣b∣=4,θ=60∘

∣R⃗∣=32+42+2(3)(4)cos⁡60∘=37|\vec{R}| = \sqrt{3^2+4^2+2(3)(4)\cos60^\circ} = \sqrt{37}∣R∣=32+42+2(3)(4)cos60∘​=37​

Now Solve: Find the resultant of a 10 N force acting East and a 30230\sqrt{2}302​ N force acting North-East.
Answer: 50 N.

Example 8: Combining Two Forces into a Single Pull

Problem: Two forces of 60 N60 \text{ N}60 N and 80 N80 \text{ N}80 N act at an angle of 60∘60^\circ60∘ with each other on an object. What single pull would replace the given forces?

Solution: The two forces are drawn from a common origin OOO, making an angle of 60∘60^\circ60∘ with each other. OAOAOA and OCOCOC represent the forces of 60 N60 \text{ N}60 N and 80 N80 \text{ N}80 N respectively. Since a single force replacing both must produce the same combined effect, that single force is simply their resultant — the diagonal OBOBOB of the parallelogram completed on OAOAOA and OCOCOC.

Two forces of 60 N and 80 N at 60 degrees to each other, combined into a single resultant by the parallelogram law
Two forces of 60 N and 80 N at 60 degrees to each other, combined into a single resultant by the parallelogram law

R2=602+802+2(60)(80)cos⁡60∘=3600+6400+4800=14800R^2 = 60^2 + 80^2 + 2(60)(80)\cos60^\circ = 3600 + 6400 + 4800 = 14800R2=602+802+2(60)(80)cos60∘=3600+6400+4800=14800

⇒R=121.7 N\Rightarrow \quad R = 121.7 \text{ N}⇒R=121.7 N

The angle ϕ\phiϕ that this resultant makes with the 60 N60 \text{ N}60 N force is given by

tan⁡ϕ=80sin⁡60∘60+80cos⁡60∘\tan\phi = \frac{80\sin60^\circ}{60+80\cos60^\circ}tanϕ=60+80cos60∘80sin60∘​

which works out to ϕ=34.7∘\phi = 34.7^\circϕ=34.7∘.

Now Solve: Repeat the problem with both forces equal to 60 N, still at 60∘60^\circ60∘ to each other.
Answer: 603 N60\sqrt{3} \text{ N}603​ N at 30∘30^\circ30∘. Notice how the angle the resultant makes with the first force shrinks from 34.7∘34.7^\circ34.7∘ to 30∘30^\circ30∘ simply because the second force became smaller (from 80 N to 60 N) — a smaller partner pulls the resultant less far from the first force's own direction.

Example 9: Proving Perpendicularity When a Force is Doubled

Problem: The resultant of two non-zero forces P⃗\vec{P}P and Q⃗\vec{Q}Q​ has magnitude PPP. Prove that if P⃗\vec{P}P is doubled (keeping Q⃗\vec{Q}Q​ and the angle between them fixed), the new resultant becomes perpendicular to Q⃗\vec{Q}Q​.

Solution: Let θ\thetaθ be the angle between P⃗\vec{P}P and Q⃗\vec{Q}Q​. Since the original resultant has magnitude PPP,

P2=P2+Q2+2PQcos⁡θ⇒Q+2Pcos⁡θ=0P^2 = P^2 + Q^2 + 2PQ\cos\theta \quad \Rightarrow \quad Q + 2P\cos\theta = 0P2=P2+Q2+2PQcosθ⇒Q+2Pcosθ=0

Now double P⃗\vec{P}P to 2P⃗2\vec{P}2P, and let α\alphaα be the angle the new resultant makes with Q⃗\vec{Q}Q​:

tan⁡α=2Psin⁡θQ+2Pcos⁡θ\tan\alpha = \frac{2P\sin\theta}{Q + 2P\cos\theta}tanα=Q+2Pcosθ2Psinθ​

But we already showed Q+2Pcos⁡θ=0Q + 2P\cos\theta = 0Q+2Pcosθ=0, so the denominator vanishes:

tan⁡α=2Psin⁡θ0→∞⇒α=90∘\tan\alpha = \frac{2P\sin\theta}{0} \rightarrow \infty \quad \Rightarrow \quad \alpha = 90^\circtanα=02Psinθ​→∞⇒α=90∘

The new resultant is therefore perpendicular to Q⃗\vec{Q}Q​, exactly as claimed.

Now Solve: Solve the same problem using geometry instead of algebra — use the property that the angle subtended in a semicircle is always a right angle to prove the same result.

Two forces, F1 and F2, are pulling on a hook as shown in the diagram. Which vector best represents the resultant force? ___img[m2_parallelogram_quiz.svg, 1.0, Forces F1 and F2 pulling on a hook at point O, with the parallelogram completed in dashed lines and three dashed candidate arrows lettered A, B and C.]

09

Vector Subtraction

The Meaning of Vector Subtraction

Vector subtraction is not a new operation, but rather a special case of addition. The key insight is this: to subtract a vector, you add its negative.

So, the operation A⃗−B⃗\vec{A} - \vec{B}A−B is defined as:

R⃗=A⃗−B⃗=A⃗+(−B⃗)\vec{R} = \vec{A} - \vec{B} = \vec{A} + (-\vec{B})R=A−B=A+(−B)

This simple definition allows us to use all the tools we've already learned for vector addition (Triangle Law, Parallelogram Law, Component Method) to perform subtraction.

Graphical Subtraction

To subtract B⃗\vec{B}B from A⃗\vec{A}A graphically, we first need to find the negative of B⃗\vec{B}B, which we call −B⃗-\vec{B}−B. This is a vector with the same length as B⃗\vec{B}B but pointing in the exact opposite direction. Then, we add A⃗\vec{A}A and −B⃗-\vec{B}−B using the head-to-tail method.

To subtract B from A, reverse B to get -B, then add it to A head-to-tail: the orange arrow is A - B.
To subtract B from A, reverse B to get -B, then add it to A head-to-tail: the orange arrow is A - B.

A Graphical Shortcut

If you draw A⃗\vec{A}A and B⃗\vec{B}B tail-to-tail, the vector that points from the head of B⃗\vec{B}B to the head of A⃗\vec{A}A is the resultant R⃗=A⃗−B⃗\vec{R} = \vec{A} - \vec{B}R=A−B.

The shortcut: with A and B tail-to-tail, the difference A - B runs from the head of B to the head of A.
The shortcut: with A and B tail-to-tail, the difference A - B runs from the head of B to the head of A.

An activity: drag the reversed vector -B until its tail sits on the head of A. The orange arrow that closes the triangle is the difference A - B.
An activity: drag the reversed vector -B until its tail sits on the head of A. The orange arrow that closes the triangle is the difference A - B.

Subtraction by Components

The component method makes subtraction incredibly straightforward. To find the components of the resultant R⃗=A⃗−B⃗\vec{R} = \vec{A} - \vec{B}R=A−B, you simply subtract the components of B⃗\vec{B}B from the components of A⃗\vec{A}A:

Rx=Ax−BxR_x = A_x - B_xRx​=Ax​−Bx​
Ry=Ay−ByR_y = A_y - B_yRy​=Ay​−By​

Once you have RxR_xRx​ and RyR_yRy​, you can find the magnitude and direction of the resultant vector R⃗\vec{R}R just as you did with addition.

Example: Given A⃗=5i^+2j^\vec{A} = 5\hat{i} + 2\hat{j}A=5i^+2j^​ and B⃗=3i^+4j^\vec{B} = 3\hat{i} + 4\hat{j}B=3i^+4j^​. Find A⃗−B⃗\vec{A} - \vec{B}A−B.

Rx=5−3=2R_x = 5 - 3 = 2Rx​=5−3=2
Ry=2−4=−2R_y = 2 - 4 = -2Ry​=2−4=−2

So, R⃗=2i^−2j^\vec{R} = 2\hat{i} - 2\hat{j}R=2i^−2j^​.

Vector Subtraction is Not Commutative

We know vector addition is commutative, but subtraction is a different story. Consider A⃗−B⃗\vec{A} - \vec{B}A−B and B⃗−A⃗\vec{B} - \vec{A}B−A. Since B⃗−A⃗=−(A⃗−B⃗)\vec{B} - \vec{A} = -(\vec{A} - \vec{B})B−A=−(A−B), these two vectors are negatives of each other — same length, opposite direction. This means their magnitudes are equal:

∣A⃗−B⃗∣=∣B⃗−A⃗∣|\vec{A} - \vec{B}| = |\vec{B} - \vec{A}|∣A−B∣=∣B−A∣

but the vectors themselves are certainly not equal, since a vector and its negative point in opposite directions:

A⃗−B⃗=−(B⃗−A⃗)⇒A⃗−B⃗≠B⃗−A⃗\vec{A} - \vec{B} = -(\vec{B} - \vec{A}) \quad \Rightarrow \quad \vec{A} - \vec{B} \neq \vec{B} - \vec{A}A−B=−(B−A)⇒A−B=B−A

So vector subtraction is non-commutative: swapping the order flips the direction of the result even though it leaves the size unchanged. It's easy to mistake the equal-magnitude fact for full equality — don't make that error.

If A⃗−B⃗=D⃗\vec{A}-\vec{B} = \vec{D}A−B=D, what is B⃗−A⃗\vec{B}-\vec{A}B−A?

Subtraction as a Diagonal of the Parallelogram

Recall that when A⃗\vec{A}A and B⃗\vec{B}B are drawn tail-to-tail, the parallelogram they form has two diagonals. The diagonal starting from the common tail represents the sum A⃗+B⃗\vec{A} + \vec{B}A+B (the Parallelogram Law of addition). The other diagonal — the one connecting the two heads, which does not pass through the common tail — represents the subtraction: it is A⃗−B⃗\vec{A} - \vec{B}A−B when drawn pointing from the head of B⃗\vec{B}B toward the head of A⃗\vec{A}A, or B⃗−A⃗\vec{B} - \vec{A}B−A when drawn in the opposite sense, from the head of A⃗\vec{A}A toward the head of B⃗\vec{B}B. This single diagonal, read in either direction, geometrically encodes both non-commutative results at once.

The Physical Meaning of Subtraction

Vector subtraction is not just a mathematical curiosity; it's essential for describing important physical concepts. The most common use is to find the change in a vector quantity.

For example, change in velocity (Δv⃗\Delta \vec{v}Δv) is defined as the final velocity minus the initial velocity:

Δv⃗=v⃗final−v⃗initial\Delta \vec{v} = \vec{v}_{final} - \vec{v}_{initial}Δv=vfinal​−vinitial​

This Δv⃗\Delta \vec{v}Δv vector is crucial for understanding acceleration. A car turning a corner at a constant speed still has a change in velocity (and thus accelerates) because its direction vector has changed, and v⃗final−v⃗initial\vec{v}_{final} - \vec{v}_{initial}vfinal​−vinitial​ is not zero.

An object's velocity changes from v⃗i=10\vec{v}_i = 10vi​=10 m/s East to v⃗f=10\vec{v}_f = 10vf​=10 m/s North. What is the direction of the change in velocity, Δv⃗\Delta \vec{v}Δv?

10

Component Method of Vector Addition

The Analytical Method

Graphical methods like the Triangle and Parallelogram laws are excellent for visualizing vector addition. However, they rely on precise drawing and measurement, which can be cumbersome and inaccurate. For precise and complex problems (especially with more than two vectors), we need a more robust technique: the analytical method of addition by components.

The core idea is simple but powerful: instead of struggling with adding slanted arrows, we break each vector down into its simple, perpendicular components (x and y). We then add all the x-components together and all the y-components together, which is easy because they lie on the same line. Finally, we reassemble the resultant vector from its summed components.

A stepped activity: press next step to resolve the two vectors into dashed components, stack the components head to tail along each axis, and finally draw the resultant R from the origin.
A stepped activity: press next step to resolve the two vectors into dashed components, stack the components head to tail along each axis, and finally draw the resultant R from the origin.

The Four-Step Process

Adding vectors A⃗\vec{A}A, B⃗\vec{B}B, C⃗\vec{C}C, ... to find the resultant R⃗=A⃗+B⃗+C⃗+...\vec{R} = \vec{A} + \vec{B} + \vec{C} + ...R=A+B+C+... involves four clear steps.

One panel per step: resolve each vector, sum the x components along one line, sum the y components along another, then reassemble the resultant R from Rx and Ry.
One panel per step: resolve each vector, sum the x components along one line, sum the y components along another, then reassemble the resultant R from Rx and Ry.

  1. Step 1: Resolve All Vectors into Components.
    For each vector, find its x and y components using trigonometry. Be careful with signs based on the quadrant.
    Ax=Acos⁡θAA_x = A \cos\theta_AAx​=AcosθA​, Ay=Asin⁡θAA_y = A \sin\theta_AAy​=AsinθA​
    Bx=Bcos⁡θBB_x = B \cos\theta_BBx​=BcosθB​, By=Bsin⁡θBB_y = B \sin\theta_BBy​=BsinθB​
    ... and so on.
  2. Step 2: Sum the Components.
    Add all the x-components to get the resultant's x-component, RxR_xRx​. Add all the y-components to get the resultant's y-component, RyR_yRy​.
    Rx=Ax+Bx+Cx+...R_x = A_x + B_x + C_x + ...Rx​=Ax​+Bx​+Cx​+...
    Ry=Ay+By+Cy+...R_y = A_y + B_y + C_y + ...Ry​=Ay​+By​+Cy​+...
  3. Step 3: Calculate the Resultant's Magnitude.
    Use the Pythagorean theorem with the resultant components to find the magnitude of the resultant vector.
    R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}R=Rx2​+Ry2​​
  4. Step 4: Calculate the Resultant's Direction.
    Use the inverse tangent to find the direction angle θR\theta_RθR​ of the resultant. Remember to check the quadrant based on the signs of RxR_xRx​ and RyR_yRy​.
    θR=tan⁡−1(RyRx)\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)θR​=tan−1(Rx​Ry​​)

The component method in three steps: the vectors, each one resolved into components, and the components summed to build the resultant
The component method in three steps: the vectors, each one resolved into components, and the components summed to build the resultant

When Is the Resultant Exactly Zero?

Because R=Rx2+Ry2R = \sqrt{R_x^2+R_y^2}R=Rx2​+Ry2​​, the resultant magnitude RRR can only be zero if both terms under the square root vanish. In other words, for the resultant of several vectors to be zero, the x-component sum RxR_xRx​ and the y-component sum RyR_yRy​ must both separately be zero — it is not enough for one to cancel while the other doesn't, and it's not enough for the sum Rx+RyR_x+R_yRx​+Ry​ to be zero either.

Worked Example

We add two vectors: A⃗\vec{A}A = 5.0 m at 37° and B⃗\vec{B}B = 3.0 m at 150°.

Step 1: Components
Ax=5.0cos⁡(37°)=5.0×0.8=4.0A_x = 5.0 \cos(37°) = 5.0 \times 0.8 = 4.0Ax​=5.0cos(37°)=5.0×0.8=4.0 m
Ay=5.0sin⁡(37°)=5.0×0.6=3.0A_y = 5.0 \sin(37°) = 5.0 \times 0.6 = 3.0Ay​=5.0sin(37°)=5.0×0.6=3.0 m
Bx=3.0cos⁡(150°)=3.0×(−0.866)=−2.6B_x = 3.0 \cos(150°) = 3.0 \times (-0.866) = -2.6Bx​=3.0cos(150°)=3.0×(−0.866)=−2.6 m
By=3.0sin⁡(150°)=3.0×0.5=1.5B_y = 3.0 \sin(150°) = 3.0 \times 0.5 = 1.5By​=3.0sin(150°)=3.0×0.5=1.5 m

Step 2: Sum
Rx=4.0+(−2.6)=1.4R_x = 4.0 + (-2.6) = 1.4Rx​=4.0+(−2.6)=1.4 m
Ry=3.0+1.5=4.5R_y = 3.0 + 1.5 = 4.5Ry​=3.0+1.5=4.5 m

Step 3: Magnitude
R=(1.4)2+(4.5)2=1.96+20.25=22.21≈4.7R = \sqrt{(1.4)^2 + (4.5)^2} = \sqrt{1.96 + 20.25} = \sqrt{22.21} \approx 4.7R=(1.4)2+(4.5)2​=1.96+20.25​=22.21​≈4.7 m

Step 4: Direction
θR=tan⁡−1(4.51.4)=tan⁡−1(3.21)≈72.7°\theta_R = \tan^{-1}\left(\frac{4.5}{1.4}\right) = \tan^{-1}(3.21) \approx 72.7°θR​=tan−1(1.44.5​)=tan−1(3.21)≈72.7°
(Since RxR_xRx​ and RyR_yRy​ are both positive, the resultant is in the first quadrant and the angle is correct.)

When adding multiple vectors using the component method, why is it essential to find the components of each vector before summing them?

Example 10: Net Displacement from Three Successive Displacements

Problem: Find the net displacement of a particle from its starting point if it undergoes three successive displacements: S⃗1=20 m,45∘\vec{S}_1 = 20\text{ m}, 45^\circS1​=20 m,45∘ West of North; S⃗2=15 m,30∘\vec{S}_2 = 15\text{ m}, 30^\circS2​=15 m,30∘ North of East; S⃗3=20 m\vec{S}_3 = 20\text{ m}S3​=20 m due South.

Solution: Set the x-axis along West–East and the y-axis along South–North, with East and North as positive.

S⃗1=20cos⁡45∘(−i^)+20sin⁡45∘(j^)\vec{S}_1 = 20\cos45^\circ(-\hat{i}) + 20\sin45^\circ(\hat{j})S1​=20cos45∘(−i^)+20sin45∘(j^​)

S⃗2=15cos⁡30∘(i^)+15sin⁡30∘(j^)\vec{S}_2 = 15\cos30^\circ(\hat{i}) + 15\sin30^\circ(\hat{j})S2​=15cos30∘(i^)+15sin30∘(j^​)

S⃗3=0(i^)+20(−j^)\vec{S}_3 = 0(\hat{i}) + 20(-\hat{j})S3​=0(i^)+20(−j^​)

Adding the three:

S⃗=S⃗1+S⃗2+S⃗3=(−202+1532+0)i^+(202+152−20)j^=−1.15i^+1.64j^=Sxi^+Syj^\vec{S} = \vec{S}_1+\vec{S}_2+\vec{S}_3 = \left(-\frac{20}{\sqrt2}+\frac{15\sqrt3}{2}+0\right)\hat{i}+\left(\frac{20}{\sqrt2}+\frac{15}{2}-20\right)\hat{j} = -1.15\hat{i}+1.64\hat{j} = S_x\hat{i}+S_y\hat{j}S=S1​+S2​+S3​=(−2​20​+2153​​+0)i^+(2​20​+215​−20)j^​=−1.15i^+1.64j^​=Sx​i^+Sy​j^​

∣S⃗∣=Sx2+Sy2=(−1.15)2+(1.64)2=2 m|\vec{S}| = \sqrt{S_x^2+S_y^2}=\sqrt{(-1.15)^2+(1.64)^2}=2\text{ m}∣S∣=Sx2​+Sy2​​=(−1.15)2+(1.64)2​=2 m

The direction is θ=tan⁡−11.151.64=35∘\theta=\tan^{-1}\dfrac{1.15}{1.64}=35^\circθ=tan−11.641.15​=35∘ West of North.

Now Solve: Find the resultant of F⃗1=402 N\vec{F}_1=40\sqrt2\text{ N}F1​=402​ N North-East and F⃗2=50 N\vec{F}_2=50\text{ N}F2​=50 N at 37∘37^\circ37∘ north of west.
Answer: 70 N, due North.

Example 11: Finding an Unknown Vector from a Given Sum

Problem: The x and y components of vector A⃗\vec{A}A are 4 m and 6 m. If the x and y components of (A⃗+B⃗)(\vec{A}+\vec{B})(A+B) are 10 m and 9 m, calculate (a) the x, y components of B⃗\vec{B}B; (b) its length; (c) the angle it makes with the x-axis.

Solution: Since components add separately, Ax+Bx=10A_x+B_x=10Ax​+Bx​=10 and Ay+By=9A_y+B_y=9Ay​+By​=9. Given Ax=4,Ay=6A_x=4, A_y=6Ax​=4,Ay​=6:

(a) Bx=10−4=6 mB_x = 10-4 = 6\text{ m}Bx​=10−4=6 m, By=9−6=3 mB_y = 9-6 = 3\text{ m}By​=9−6=3 m

(b) B=Bx2+By2=62+32=45=35 mB=\sqrt{B_x^2+B_y^2}=\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt5\text{ m}B=Bx2​+By2​​=62+32​=45​=35​ m

(c) θ=tan⁡−1(ByBx)=tan⁡−1(12)\theta=\tan^{-1}\left(\frac{B_y}{B_x}\right)=\tan^{-1}\left(\frac12\right)θ=tan−1(Bx​By​​)=tan−1(21​)

Now Solve: A vector has x, y components 20 N and 10 N. Combined with another vector, the resultant has magnitude 50250\sqrt2502​ N, making equal angles with the x and y axes. Find the magnitude of the second vector.
Answer: 50 N.

11

Types of Vector Multiplication

Multiplying Vectors

We've mastered adding and subtracting vectors. Now, we venture into multiplication. With scalars, multiplication is simple: 3 times 4 is always 12. But with vectors, it's more complex. Why? Because we have to consider the directions.

In physics, we often need to combine two vectors to produce a third quantity. Sometimes, this new quantity is a scalar (like 'work'), and other times it's a new vector (like 'torque'). To handle these different physical situations, we need two distinct ways to multiply vectors.

The Two Products of Vectors

The two types of vector multiplication are defined by the kind of quantity they produce:

  1. The Scalar Product (or Dot Product):
    This method multiplies two vectors to produce a scalar quantity. It is a measure of how much one vector lies along the direction of another. A classic example is the calculation of Work, where W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}W=F⋅d. The result, work, is a scalar (energy).
    Notation: A⃗⋅B⃗\vec{A} \cdot \vec{B}A⋅B
  2. The Vector Product (or Cross Product):
    This method multiplies two vectors to produce a new vector. The resulting vector is perpendicular to the plane containing the original two vectors. A prime example is Torque, τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}τ=r×F, where the resulting torque vector's direction determines the axis of rotation.
    Notation: A⃗×B⃗\vec{A} \times \vec{B}A×B

The dot product collapses two vectors into a single number, AB cos(theta); the cross product builds a new vector, perpendicular to both, of magnitude AB sin(theta).
The dot product collapses two vectors into a single number, AB cos(theta); the cross product builds a new vector, perpendicular to both, of magnitude AB sin(theta).

The Defining Formulas

Here are the fundamental formulas for the two types of products. We will explore them in detail in the upcoming sections.

Scalar (Dot) Product:

The result is a scalar whose value is given by:

A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos\thetaA⋅B=∣A∣∣B∣cosθ

where θ\thetaθ is the angle between the two vectors.

Vector (Cross) Product:

The result is a vector. Its magnitude is given by:

∣A⃗×B⃗∣=∣A⃗∣∣B⃗∣sin⁡θ|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin\theta∣A×B∣=∣A∣∣B∣sinθ

The direction of the resultant vector is given by the Right-Hand Rule, which we will learn about soon.

If you need to calculate a physical quantity that is itself a vector (like torque) from two other vectors (like position and force), which type of multiplication would you use?

12

Scalar (Dot) Product

The Scalar Product

The dot product is a way of multiplying two vectors that tells you how much one vector 'goes along' the other. It measures the extent to which two vectors are aligned. The result is a single number, a scalar, hence the alternative name 'scalar product'.

Geometric Definition

The dot product of A⃗\vec{A}A and B⃗\vec{B}B is defined as the magnitude of A⃗\vec{A}A times the magnitude of B⃗\vec{B}B times the cosine of the angle θ\thetaθ between them.

A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos\thetaA⋅B=∣A∣∣B∣cosθ

You can think of the term ∣B⃗∣cos⁡θ|\vec{B}| \cos\theta∣B∣cosθ as the 'shadow' or projection of vector B⃗\vec{B}B onto the direction of vector A⃗\vec{A}A. So, the dot product is literally the magnitude of A⃗\vec{A}A multiplied by the component of B⃗\vec{B}B that is parallel to A⃗\vec{A}A.

The projection B cos(theta) is the part of B that lies along A; the dot product is the magnitude of A times this projection.
The projection B cos(theta) is the part of B that lies along A; the dot product is the magnitude of A times this projection.

Calculating with Components

Start with the simplest dot products of all: the unit vectors themselves. Using A⃗⋅B⃗=ABcos⁡θ\vec{A}\cdot\vec{B} = AB\cos\thetaA⋅B=ABcosθ with magnitudes of 1:

  • What is i^⋅i^\hat{i}\cdot\hat{i}i^⋅i^? The angle between a vector and itself is 0∘0^\circ0∘, so i^⋅i^=(1)(1)cos⁡0∘=1\hat{i}\cdot\hat{i} = (1)(1)\cos 0^\circ = 1i^⋅i^=(1)(1)cos0∘=1. Likewise j^⋅j^=1\hat{j}\cdot\hat{j} = 1j^​⋅j^​=1 and k^⋅k^=1\hat{k}\cdot\hat{k} = 1k^⋅k^=1.
  • What is i^⋅j^\hat{i}\cdot\hat{j}i^⋅j^​? The axes are perpendicular, so i^⋅j^=(1)(1)cos⁡90∘=0\hat{i}\cdot\hat{j} = (1)(1)\cos 90^\circ = 0i^⋅j^​=(1)(1)cos90∘=0. Likewise j^⋅k^=0\hat{j}\cdot\hat{k} = 0j^​⋅k^=0 and k^⋅i^=0\hat{k}\cdot\hat{i} = 0k^⋅i^=0.

So: a unit vector dotted with itself gives 1; two different unit vectors give 0.

Now compute a full dot product with the distributive law, multiplying term by term exactly as in algebra:

(3i^+4j^)⋅(2i^+4j^)=6 (i^⋅i^)+12 (i^⋅j^)+8 (j^⋅i^)+16 (j^⋅j^)(3\hat{i} + 4\hat{j}) \cdot (2\hat{i} + 4\hat{j}) = 6\,(\hat{i}\cdot\hat{i}) + 12\,(\hat{i}\cdot\hat{j}) + 8\,(\hat{j}\cdot\hat{i}) + 16\,(\hat{j}\cdot\hat{j})(3i^+4j^​)⋅(2i^+4j^​)=6(i^⋅i^)+12(i^⋅j^​)+8(j^​⋅i^)+16(j^​⋅j^​)

=6(1)+12(0)+8(0)+16(1)=22= 6(1) + 12(0) + 8(0) + 16(1) = 22=6(1)+12(0)+8(0)+16(1)=22

Every cross term dies, and only the i^\hat{i}i^-with-i^\hat{i}i^ and j^\hat{j}j^​-with-j^\hat{j}j^​ products survive. There is therefore no need to expand next time — multiply matching components and add:

A⃗⋅B⃗=AxBx+AyBy+AzBz\vec{A}\cdot\vec{B} = A_x B_x + A_y B_y + A_z B_zA⋅B=Ax​Bx​+Ay​By​+Az​Bz​

For the vectors above: (3)(2)+(4)(4)=22(3)(2) + (4)(4) = 22(3)(2)+(4)(4)=22, the same answer in one line.

Key Properties and Applications

  • Commutative: The order doesn't matter. A⃗⋅B⃗=B⃗⋅A⃗\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}A⋅B=B⋅A.
  • Distributive over Addition: The dot product distributes across a sum of vectors just like ordinary multiplication distributes over addition: A⃗⋅(B⃗+C⃗)=A⃗⋅B⃗+A⃗⋅C⃗\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}A⋅(B+C)=A⋅B+A⋅C. This property is what lets us expand dot products of vector sums term by term.
  • Finding the Angle Between Vectors: By rearranging the definition, we can find the angle between two vectors: cos⁡θ=A⃗⋅B⃗∣A⃗∣∣B⃗∣\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}cosθ=∣A∣∣B∣A⋅B​.
  • Checking for Perpendicularity: This is a crucial application. If two non-zero vectors are perpendicular (orthogonal), the angle between them is 90°, and cos⁡(90°)=0\cos(90°) = 0cos(90°)=0. Therefore, the dot product of two perpendicular vectors is always zero.

Physical Application: Work
A perfect physical example of the dot product is the calculation of work done by a constant force. Only the component of the force in the direction of the displacement does work.

W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}W=F⋅d

The Vector Projection Formula

Earlier we described Acos⁡θA\cos\thetaAcosθ as the 'shadow' of A⃗\vec{A}A along B⃗\vec{B}B — but that shadow is just a number (a magnitude). We can go further and write the projection of A⃗\vec{A}A onto B⃗\vec{B}B as a full vector C⃗\vec{C}C, pointing along B^\hat{B}B^ with that shadow length:

C⃗=Acos⁡θ B^=A⃗⋅B⃗∣B⃗∣B^=A⃗⋅B⃗∣B⃗∣⋅B⃗∣B⃗∣\vec{C} = A\cos\theta \, \hat{B} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}\hat{B} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}\cdot\frac{\vec{B}}{|\vec{B}|}C=AcosθB^=∣B∣A⋅B​B^=∣B∣A⋅B​⋅∣B∣B​

Combining the two factors of ∣B⃗∣|\vec{B}|∣B∣ in the denominator gives the compact closed-form projection formula:

C⃗=A⃗⋅B⃗∣B⃗∣2B⃗\vec{C} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\vec{B}C=∣B∣2A⋅B​B

This formula is extremely useful whenever you need the actual vector component of A⃗\vec{A}A along a given direction B⃗\vec{B}B, rather than just its magnitude.

Vector A⃗=3i^+4j^\vec{A} = 3\hat{i} + 4\hat{j}A=3i^+4j^​. Vector B⃗\vec{B}B is perpendicular to A⃗\vec{A}A. Which of the following could be vector B⃗\vec{B}B?

Example 16: Angle Between Two 3D Vectors

Problem: If A⃗=3i^+4j^+53k^\vec{A} = 3\hat{i} + 4\hat{j} + 5\sqrt3\hat{k}A=3i^+4j^​+53​k^ and B⃗=3i^+4j^\vec{B} = 3\hat{i} + 4\hat{j}B=3i^+4j^​, find the angle between A⃗\vec{A}A and B⃗\vec{B}B.

Solution: We know A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\thetaA⋅B=∣A∣∣B∣cosθ, so

cos⁡θ=A⃗⋅B⃗∣A⃗∣∣B⃗∣\cos\theta = \frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}cosθ=∣A∣∣B∣A⋅B​

First compute the magnitudes: ∣A⃗∣=32+42+(53)2=9+16+75=100=10|\vec{A}|=\sqrt{3^2+4^2+(5\sqrt3)^2}=\sqrt{9+16+75}=\sqrt{100}=10∣A∣=32+42+(53​)2​=9+16+75​=100​=10, and ∣B⃗∣=32+42=25=5|\vec{B}|=\sqrt{3^2+4^2}=\sqrt{25}=5∣B∣=32+42​=25​=5.

The dot product is A⃗⋅B⃗=(3)(3)+(4)(4)+(53)(0)=9+16+0=25\vec{A}\cdot\vec{B} = (3)(3)+(4)(4)+(5\sqrt3)(0) = 9+16+0 = 25A⋅B=(3)(3)+(4)(4)+(53​)(0)=9+16+0=25.

cos⁡θ=25(10)(5)=12⇒θ=60∘\cos\theta = \frac{25}{(10)(5)} = \frac12 \quad \Rightarrow \quad \theta = 60^\circcosθ=(10)(5)25​=21​⇒θ=60∘

13

Vector (Cross) Product

The Vector Product

The cross product is the second way to multiply vectors. Unlike the dot product, which results in a scalar, the cross product of two vectors, A⃗\vec{A}A and B⃗\vec{B}B, produces a new vector, C⃗\vec{C}C. This new vector has a unique direction and a magnitude that represents the 'product of perpendicularity' between A⃗\vec{A}A and B⃗\vec{B}B.

The cross product is essential for describing rotational phenomena in physics, like torque and angular momentum.

Part 1: The Magnitude of the Cross Product

The magnitude of the cross product, ∣A⃗×B⃗∣|\vec{A} \times \vec{B}|∣A×B∣, is defined as the product of the magnitudes of the two vectors times the sine of the angle θ\thetaθ between them.

∣A⃗×B⃗∣=∣A⃗∣∣B⃗∣sin⁡θ|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin\theta∣A×B∣=∣A∣∣B∣sinθ

Geometric Interpretation: Area

The magnitude of the cross product has a direct geometric meaning: it is equal to the area of the parallelogram formed by the two vectors when they are placed tail-to-tail.

The magnitude of A x B equals the area of the parallelogram on A and B: base A times height B sin(theta).
The magnitude of A x B equals the area of the parallelogram on A and B: base A times height B sin(theta).

This makes sense because ∣B⃗∣sin⁡θ|\vec{B}|\sin\theta∣B∣sinθ is the height of the parallelogram with ∣A⃗∣|\vec{A}|∣A∣ as its base. The area is base times height, which is ∣A⃗∣(∣B⃗∣sin⁡θ)|\vec{A}|(|\vec{B}|\sin\theta)∣A∣(∣B∣sinθ).

Part 2: The Direction of the Cross Product

The resultant vector C⃗=A⃗×B⃗\vec{C} = \vec{A} \times \vec{B}C=A×B has a direction that is perpendicular to the plane containing both A⃗\vec{A}A and B⃗\vec{B}B. But there are two perpendicular directions (e.g., 'up' or 'down'). Which one is it?

We use the Right-Hand Rule to decide.

How to Use the Right-Hand Rule:

  1. Point the fingers of your right hand in the direction of the first vector (A⃗\vec{A}A).
  2. Curl your fingers in the direction of the second vector (B⃗\vec{B}B). You should curl through the smaller of the two angles between the vectors.
  3. Your thumb will then point in the direction of the resultant cross product vector C⃗\vec{C}C.

The right-hand rule on the axes: point your fingers along A, curl them toward B (the gray sweep), and your thumb gives C = A x B, perpendicular to both, along the z-axis. Drag the scene to turn it, and zoom with the buttons.

Key Properties

  • Anti-Commutative: The order of multiplication matters and inverts the direction. A⃗×B⃗=−(B⃗×A⃗)\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A})A×B=−(B×A). This is a direct consequence of the Right-Hand Rule.
  • Distributive over Addition: Just like the dot product, the cross product distributes over vector addition: A⃗×(B⃗+C⃗)=A⃗×B⃗+A⃗×C⃗\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}A×(B+C)=A×B+A×C. This lets us expand cross products of vector sums term by term, which is exactly how the component formula for the cross product is derived.
  • Parallel Vectors: If two vectors are parallel (θ=0°\theta = 0°θ=0°) or antiparallel (θ=180°\theta = 180°θ=180°), the sine of the angle is 0. Therefore, the cross product of two parallel vectors is the zero vector (0⃗\vec{0}0).

In the given diagram, vector A points East and vector B points North. ___img[m2_cross_product_quiz.svg, 1.0, Vectors A and B drawn in the plane of the page.] What is the direction of A⃗×B⃗\vec{A} \times \vec{B}A×B?

Example 12: Angle Between a Cross Product and a Sum

Problem: Find the angle between (A⃗×B⃗)(\vec{A}\times\vec{B})(A×B) and (A⃗+B⃗)(\vec{A}+\vec{B})(A+B).

Solution: The vector A⃗×B⃗\vec{A}\times\vec{B}A×B is, by definition, perpendicular to the entire plane containing both A⃗\vec{A}A and B⃗\vec{B}B. Since A⃗+B⃗\vec{A}+\vec{B}A+B is itself a vector that lies within that same plane (it's a combination of A⃗\vec{A}A and B⃗\vec{B}B), it must be perpendicular to A⃗×B⃗\vec{A}\times\vec{B}A×B. The angle is 90∘90^\circ90∘.

Example 13: A Chain of Perpendicularity and Parallelism

Problem: If P⃗⋅Q⃗=0\vec{P}\cdot\vec{Q}=0P⋅Q​=0 and P⃗×R⃗=0\vec{P}\times\vec{R}=0P×R=0, what is the angle between Q⃗\vec{Q}Q​ and R⃗\vec{R}R?

Solution: A zero dot product tells us P⃗⋅Q⃗=0⇒Q⃗\vec{P}\cdot\vec{Q}=0 \Rightarrow \vec{Q}P⋅Q​=0⇒Q​ is perpendicular to P⃗\vec{P}P. A zero cross product tells us P⃗×R⃗=0⇒R⃗\vec{P}\times\vec{R}=0 \Rightarrow \vec{R}P×R=0⇒R is parallel (or antiparallel) to P⃗\vec{P}P. Since Q⃗\vec{Q}Q​ is perpendicular to P⃗\vec{P}P, and R⃗\vec{R}R points along the very same line as P⃗\vec{P}P, Q⃗\vec{Q}Q​ must also be perpendicular to R⃗\vec{R}R. The angle between Q⃗\vec{Q}Q​ and R⃗\vec{R}R is 90∘90^\circ90∘.

Example 15: Proving Three Vectors Form a Triangle and Finding Its Area

Problem: Prove that the vectors A⃗=3i^+j^−2k^\vec{A}=3\hat{i}+\hat{j}-2\hat{k}A=3i^+j^​−2k^, B⃗=−i^+3j^+4k^\vec{B}=-\hat{i}+3\hat{j}+4\hat{k}B=−i^+3j^​+4k^, and C⃗=4i^−2j^−6k^\vec{C}=4\hat{i}-2\hat{j}-6\hat{k}C=4i^−2j^​−6k^ can form the three sides of a triangle. Also find the area of that triangle.

Solution: Three vectors can be arranged head-to-tail to form a closed triangle exactly when one of them equals the sum of the other two (recall the closed-polygon corollary of the Triangle Law). Adding C⃗\vec{C}C and B⃗\vec{B}B:

C⃗+B⃗=(4i^−2j^−6k^)+(−i^+3j^+4k^)=3i^+j^−2k^=A⃗\vec{C}+\vec{B} = (4\hat{i}-2\hat{j}-6\hat{k})+(-\hat{i}+3\hat{j}+4\hat{k}) = 3\hat{i}+\hat{j}-2\hat{k} = \vec{A}C+B=(4i^−2j^​−6k^)+(−i^+3j^​+4k^)=3i^+j^​−2k^=A

Since C⃗+B⃗=A⃗\vec{C}+\vec{B}=\vec{A}C+B=A, the three vectors do indeed close up into a triangle.

The triangle formed by vectors A, B and C: the head-to-tail path along C then B lands on the head of A
The triangle formed by vectors A, B and C: the head-to-tail path along C then B lands on the head of A

The area of a triangle formed by two of its side vectors, say A⃗\vec{A}A and B⃗\vec{B}B, is half the magnitude of their cross product:

Area=12∣A⃗×B⃗∣=12∣(3i^+j^−2k^)×(−i^+3j^+4k^)∣\text{Area}=\frac12|\vec{A}\times\vec{B}| = \frac12\left|(3\hat{i}+\hat{j}-2\hat{k})\times(-\hat{i}+3\hat{j}+4\hat{k})\right|Area=21​∣A×B∣=21​​(3i^+j^​−2k^)×(−i^+3j^​+4k^)​

Expanding the cross product term by term using the distributive property:

Area=12∣(3i^)×(−i^+3j^+4k^)+(j^)×(−i^+3j^+4k^)+(−2k^)×(−i^+3j^+4k^)∣\text{Area}=\frac12\left|(3\hat{i})\times(-\hat{i}+3\hat{j}+4\hat{k})+(\hat{j})\times(-\hat{i}+3\hat{j}+4\hat{k})+(-2\hat{k})\times(-\hat{i}+3\hat{j}+4\hat{k})\right|Area=21​​(3i^)×(−i^+3j^​+4k^)+(j^​)×(−i^+3j^​+4k^)+(−2k^)×(−i^+3j^​+4k^)​

⇒Area=12∣(9k^−12j^)+(k^+4i^)+(2j^+6i^)∣\Rightarrow \text{Area}=\frac12\left|(9\hat{k}-12\hat{j})+(\hat{k}+4\hat{i})+(2\hat{j}+6\hat{i})\right|⇒Area=21​​(9k^−12j^​)+(k^+4i^)+(2j^​+6i^)​

⇒Area=12∣10i^−10j^+10k^∣\Rightarrow \text{Area}=\frac12\left|10\hat{i}-10\hat{j}+10\hat{k}\right|⇒Area=21​​10i^−10j^​+10k^​

⇒Area=53 units\Rightarrow \text{Area} = 5\sqrt3 \text{ units}⇒Area=53​ units

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Scalar and Vector Triple Products

Combining Three Vectors

We've seen how to multiply two vectors. But what happens when we combine three? Two important constructions arise: the scalar triple product, which results in a scalar, and the vector triple product, which results in a vector.

Scalar Triple Product: Measuring Volume

The scalar triple product is defined as the dot product of one vector with the cross product of two other vectors.

Scalar Triple Product=A⃗⋅(B⃗×C⃗)\text{Scalar Triple Product} = \vec{A} \cdot (\vec{B} \times \vec{C})Scalar Triple Product=A⋅(B×C)

Geometric Interpretation: Volume

This product has a profound geometric meaning: its absolute value, ∣A⃗⋅(B⃗×C⃗)∣|\vec{A} \cdot (\vec{B} \times \vec{C})|∣A⋅(B×C)∣, represents the volume of the parallelepiped formed by the three vectors A⃗\vec{A}A, B⃗\vec{B}B, and C⃗\vec{C}C when they are placed tail-to-tail.

This is because ∣B⃗×C⃗∣|\vec{B} \times \vec{C}|∣B×C∣ gives the area of the base parallelogram, and A⃗⋅(...)\vec{A} \cdot (\text{...})A⋅(...) then projects the height of the parallelepiped (given by A⃗\vec{A}A) onto the normal of the base, effectively calculating Area × Height.

The parallelepiped on A, B and C: the shaded base has area |B x C|, and the height h is the projection of A onto the base normal, so the volume is |A . (B x C)|. Drag the scene to turn it, and zoom with the buttons.

Test for Coplanarity

A crucial application of this property is to check if three vectors are coplanar (lie on the same plane). If the three vectors lie on the same plane, they cannot form a 3D shape, so the volume of the parallelepiped is zero. Therefore:
If A⃗⋅(B⃗×C⃗)=0\vec{A} \cdot (\vec{B} \times \vec{C}) = 0A⋅(B×C)=0, the vectors are coplanar.

If three vectors A, B, and C are coplanar, what is the value of their scalar triple product?

Vector Triple Product

The vector triple product involves the cross product of one vector with the result of another cross product.

A⃗×(B⃗×C⃗)\vec{A} \times (\vec{B} \times \vec{C})A×(B×C)

The result is a new vector. Unlike the scalar triple product, its geometric meaning is less direct, but it can be simplified using a very useful identity known as the 'BAC-CAB' rule.

A⃗×(B⃗×C⃗)=B⃗(A⃗⋅C⃗)−C⃗(A⃗⋅B⃗)\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})A×(B×C)=B(A⋅C)−C(A⋅B)

Notice that the result is a linear combination of vectors B⃗\vec{B}B and C⃗\vec{C}C. This means the resulting vector lies in the same plane as B⃗\vec{B}B and C⃗\vec{C}C.

Because A x (B x C) must be perpendicular to B x C, it falls back into the plane containing B and C. Drag the scene to turn it, and zoom with the buttons.

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Key Concepts Summary

Summary: Scalars, Vectors and Their Representation

  • Physical Quantities: Classified into scalars (completely described by magnitude and unit, e.g., mass, time, speed) and vectors (require both magnitude and direction, e.g., force, displacement, velocity).
  • Insufficiency of Numbers: Vectors cannot be added using ordinary arithmetic. Their sum depends on their relative orientations.
  • Vector Representation: Represented graphically as an arrow (length = magnitude, arrowhead = direction) and algebraically as a letter with a top arrow (e.g., A⃗\vec{A}A). Magnitude is written as AAA or ∣A⃗∣|\vec{A}|∣A∣.
  • Vector Properties:
    • Equality: Two vectors are equal if they have the same magnitude and direction , regardless of their starting point.
    • Unit Vector: A vector of magnitude 1 indicating direction, computed as A^=A⃗/∣A⃗∣\hat{A} = \vec{A} / |\vec{A}|A^=A/∣A∣.
    • Negative Vector: Same magnitude, opposite direction (180∘180^\circ180∘ flip).
    • Zero Vector: Magnitude of zero, indeterminate direction.
    • Parallel & Antiparallel: Parallel vectors point in the same direction (0∘0^\circ0∘ angle); antiparallel vectors point in opposite directions (180∘180^\circ180∘ flip).

Summary: Components and Vector Addition

  • Vector Resolution (Components): Any vector A⃗\vec{A}A can be split into orthogonal components: Ax=Acos⁡θA_x = A \cos\thetaAx​=Acosθ Ay=Asin⁡θA_y = A \sin\thetaAy​=Asinθ A⃗=Axi^+Ayj^\vec{A} = A_x\hat{i} + A_y\hat{j}A=Ax​i^+Ay​j^​
  • Reconstructing a Vector: From perpendicular components, magnitude and direction are calculated as: A=Ax2+Ay2A = \sqrt{A_x^2 + A_y^2}A=Ax2​+Ay2​​ θref=tan⁡−1∣AyAx∣\theta_{\text{ref}} = \tan^{-1}\left| \frac{A_y}{A_x} \right|θref​=tan−1​Ax​Ay​​​ (Apply quadrant adjustments based on signs of AxA_xAx​ and AyA_yAy​ to find θ\thetaθ).
  • Vector Addition Laws:
    • Triangle Law: Add head-to-tail; the resultant R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B points from the tail of A⃗\vec{A}A to the head of B⃗\vec{B}B.
    • Parallelogram Law: Place vectors tail-to-tail. The resultant is the diagonal of the parallelogram: R=A2+B2+2ABcos⁡θR = \sqrt{A^2 + B^2 + 2AB\cos\theta}R=A2+B2+2ABcosθ​ α=tan⁡−1(Bsin⁡θA+Bcos⁡θ)(angle with A⃗)\alpha = \tan^{-1}\left( \frac{B\sin\theta}{A + B\cos\theta} \right) \quad (\text{angle with } \vec{A})α=tan−1(A+BcosθBsinθ​)(angle with A)
  • Vector Subtraction: Defined as adding the negative of a vector: D⃗=A⃗−B⃗=A⃗+(−B⃗)\vec{D} = \vec{A} - \vec{B} = \vec{A} + (-\vec{B})D=A−B=A+(−B) D=A2+B2−2ABcos⁡θD = \sqrt{A^2 + B^2 - 2AB\cos\theta}D=A2+B2−2ABcosθ​
  • Component Method: To add multiple vectors, sum their components independently: Rx=∑Ax,Ry=∑AyR_x = \sum A_x, \quad R_y = \sum A_yRx​=∑Ax​,Ry​=∑Ay​ R=Rx2+Ry2,θ=tan⁡−1(RyRx)R = \sqrt{R_x^2 + R_y^2}, \quad \theta = \tan^{-1}\left( \frac{R_y}{R_x} \right)R=Rx2​+Ry2​​,θ=tan−1(Rx​Ry​​)

Summary: Products of Vectors

  • Scalar (Dot) Product: Yields a scalar quantity representing alignment: A⃗⋅B⃗=ABcos⁡θ=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = AB\cos\theta = A_x B_x + A_y B_y + A_z B_zA⋅B=ABcosθ=Ax​Bx​+Ay​By​+Az​Bz​ Properties: Commutative, distributive. Perpendicular vectors have a dot product of zero.
  • Vector (Cross) Product: Yields a vector perpendicular to both input vectors: C⃗=A⃗×B⃗\vec{C} = \vec{A} \times \vec{B}C=A×B ∣A⃗×B⃗∣=ABsin⁡θ|\vec{A} \times \vec{B}| = AB\sin\theta∣A×B∣=ABsinθ Direction is found via the right-hand rule. In component form: A⃗×B⃗=(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} + (A_z B_x - A_x B_z)\hat{j} + (A_x B_y - A_y B_x)\hat{k}A×B=(Ay​Bz​−Az​By​)i^+(Az​Bx​−Ax​Bz​)j^​+(Ax​By​−Ay​Bx​)k^ Properties: Anti-commutative. Parallel vectors have a cross product of zero.
  • Triple Products:
    • Scalar Triple Product: A⃗⋅(B⃗×C⃗)\vec{A} \cdot (\vec{B} \times \vec{C})A⋅(B×C). Its absolute value represents the volume of the parallelepiped. If it equals zero, the vectors are coplanar.
    • Vector Triple Product: A⃗×(B⃗×C⃗)\vec{A} \times (\vec{B} \times \vec{C})A×(B×C). Simplified using the BAC-CAB identity: A⃗×(B⃗×C⃗)=B⃗(A⃗⋅C⃗)−C⃗(A⃗⋅B⃗)\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})A×(B×C)=B(A⋅C)−C(A⋅B)
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Worked Examples

Example 1: Drone Flight Components

Problem: A search-and-rescue drone flies with a constant speed of 20 m/s20 \text{ m/s}20 m/s at an angle of 30∘30^\circ30∘ North of East. Calculate the Eastward (horizontal) and Northward (vertical) components of its velocity vector.

Solution:

  1. Identify the given variables:
    The magnitude of the velocity is v=20 m/sv = 20 \text{ m/s}v=20 m/s. The angle with the East direction (positive x-axis) is θ=30∘\theta = 30^\circθ=30∘.
  2. Calculate the horizontal (Eastward) component vxv_xvx​:
    vx=vcos⁡θ=20cos⁡(30∘)v_x = v \cos\theta = 20 \cos(30^\circ)vx​=vcosθ=20cos(30∘)vx=20×0.866=17.32 m/sv_x = 20 \times 0.866 = 17.32 \text{ m/s}vx​=20×0.866=17.32 m/s
  3. Calculate the vertical (Northward) component vyv_yvy​:
    vy=vsin⁡θ=20sin⁡(30∘)v_y = v \sin\theta = 20 \sin(30^\circ)vy​=vsinθ=20sin(30∘)vy=20×0.500=10.00 m/sv_y = 20 \times 0.500 = 10.00 \text{ m/s}vy​=20×0.500=10.00 m/s
  4. State the final vector in component form:
    v⃗=17.32i^+10.00j^ m/s\vec{v} = 17.32\hat{i} + 10.00\hat{j} \text{ m/s}v=17.32i^+10.00j^​ m/s

Example 2: Net Force on a Sailboat

Problem: A sailboat is subjected to two perpendicular forces: the wind pushes it Westward with a force of 40 N40 \text{ N}40 N, while the ocean current pushes it Southward with a force of 30 N30 \text{ N}30 N. Find the magnitude and direction of the net force vector acting on the sailboat.

Solution:

  1. Establish the coordinate system:
    Let East be the positive x-direction and North be the positive y-direction. Therefore, West is the negative x-direction, and South is the negative y-direction.
  2. State the force components:
    Fx=−40 N(Westward)F_x = -40 \text{ N} \quad (\text{Westward})Fx​=−40 N(Westward)Fy=−30 N(Southward)F_y = -30 \text{ N} \quad (\text{Southward})Fy​=−30 N(Southward)
  3. Calculate the magnitude of the net force FFF:
    F=Fx2+Fy2=(−40)2+(−30)2F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-40)^2 + (-30)^2}F=Fx2​+Fy2​​=(−40)2+(−30)2​F=1600+900=2500=50 NF = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ N}F=1600+900​=2500​=50 N
  4. Calculate the reference direction angle θref\theta_{\text{ref}}θref​:
    θref=tan⁡−1∣FyFx∣=tan⁡−1∣−30−40∣=tan⁡−1(0.75)≈36.87∘\theta_{\text{ref}} = \tan^{-1}\left| \frac{F_y}{F_x} \right| = \tan^{-1}\left| \frac{-30}{-40} \right| = \tan^{-1}(0.75) \approx 36.87^\circθref​=tan−1​Fx​Fy​​​=tan−1​−40−30​​=tan−1(0.75)≈36.87∘
  5. Apply quadrant correction:
    Since both Fx<0F_x < 0Fx​<0 and Fy<0F_y < 0Fy​<0, the net force points in the third quadrant (South-West). The physical angle measured counter-clockwise from the positive x-axis is:
    θ=180∘+θref=180∘+36.87∘=216.87∘\theta = 180^\circ + \theta_{\text{ref}} = 180^\circ + 36.87^\circ = 216.87^\circθ=180∘+θref​=180∘+36.87∘=216.87∘

Answer: The net force is 50 N50 \text{ N}50 N at an angle of 216.87∘216.87^\circ216.87∘ counter-clockwise from the positive x-axis (or 36.87∘36.87^\circ36.87∘ South of West).

Two perpendicular forces of 40extN40 ext{ N}40extN and 30extN30 ext{ N}30extN act on the same object. What is the magnitude of the net force?

Example 3: Robot Limb Direction

Problem: A robotic manipulator arm exerts a force described by the component vector F⃗=−8i^+6j^ N\vec{F} = -8\hat{i} + 6\hat{j} \text{ N}F=−8i^+6j^​ N. Find the unit vector F^\hat{F}F^ that represents the direction of this force.

Solution:

  1. Calculate the magnitude of the force vector ∣F⃗∣|\vec{F}|∣F∣:
    ∣F⃗∣=Fx2+Fy2=(−8)2+62|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{(-8)^2 + 6^2}∣F∣=Fx2​+Fy2​​=(−8)2+62​∣F⃗∣=64+36=100=10 N|\vec{F}| = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}∣F∣=64+36​=100​=10 N
  2. Divide the vector by its magnitude to obtain the unit vector F^\hat{F}F^:
    F^=F⃗∣F⃗∣=−8i^+6j^10\hat{F} = \frac{\vec{F}}{|\vec{F}|} = \frac{-8\hat{i} + 6\hat{j}}{10}F^=∣F∣F​=10−8i^+6j^​​F^=−0.8i^+0.6j^\hat{F} = -0.8\hat{i} + 0.6\hat{j}F^=−0.8i^+0.6j^​
  3. Verify that the unit vector's magnitude is exactly 1:
    ∣F^∣=(−0.8)2+(0.6)2=0.64+0.36=1.0=1|\hat{F}| = \sqrt{(-0.8)^2 + (0.6)^2} = \sqrt{0.64 + 0.36} = \sqrt{1.0} = 1∣F^∣=(−0.8)2+(0.6)2​=0.64+0.36​=1.0​=1

Answer: The unit vector indicating the direction of the force is F^=−0.8i^+0.6j^\hat{F} = -0.8\hat{i} + 0.6\hat{j}F^=−0.8i^+0.6j^​.

Example 4: Resultant Force Using Parallelogram Law

Problem: Two forces of magnitude F1=10 NF_1 = 10 \text{ N}F1​=10 N and F2=6 NF_2 = 6 \text{ N}F2​=6 N act on a particle. The angle between their directions is 60∘60^\circ60∘. Find the magnitude of the resultant force and its angle relative to the F1F_1F1​ force.

Solution:

  1. Identify the given variables:
    F1=10 NF_1 = 10 \text{ N}F1​=10 N, F2=6 NF_2 = 6 \text{ N}F2​=6 N, and θ=60∘\theta = 60^\circθ=60∘.
  2. Calculate the magnitude of the resultant force RRR:
    R=F12+F22+2F1F2cos⁡θR = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos\theta}R=F12​+F22​+2F1​F2​cosθ​R=102+62+2(10)(6)cos⁡(60∘)R = \sqrt{10^2 + 6^2 + 2(10)(6)\cos(60^\circ)}R=102+62+2(10)(6)cos(60∘)​R=100+36+120(0.5)=196=14 NR = \sqrt{100 + 36 + 120(0.5)} = \sqrt{196} = 14 \text{ N}R=100+36+120(0.5)​=196​=14 N
  3. Calculate the direction angle α\alphaα relative to F1F_1F1​:
    α=tan⁡−1(F2sin⁡θF1+F2cos⁡θ)\alpha = \tan^{-1}\left( \frac{F_2 \sin\theta}{F_1 + F_2 \cos\theta} \right)α=tan−1(F1​+F2​cosθF2​sinθ​)α=tan⁡−1(6sin⁡(60∘)10+6cos⁡(60∘))=tan⁡−1(5.19613)≈21.79∘\alpha = \tan^{-1}\left( \frac{6 \sin(60^\circ)}{10 + 6 \cos(60^\circ)} \right) = \tan^{-1}\left( \frac{5.196}{13} \right) \approx 21.79^\circα=tan−1(10+6cos(60∘)6sin(60∘)​)=tan−1(135.196​)≈21.79∘

Answer: The resultant force has a magnitude of 14 N14 \text{ N}14 N and points at an angle of 21.79∘21.79^\circ21.79∘ relative to the F1F_1F1​ force.

Example 5: Work Done by a Force (Scalar Product)

Problem: A crate is dragged along a horizontal floor by a force represented by F⃗=4i^+3j^−2k^ N\vec{F} = 4\hat{i} + 3\hat{j} - 2\hat{k} \text{ N}F=4i^+3j^​−2k^ N. The displacement of the crate is d⃗=5i^+2j^ m\vec{d} = 5\hat{i} + 2\hat{j} \text{ m}d=5i^+2j^​ m. Calculate the work done by this force.

Solution:

  1. Recall the definition of Work:
    Work is the scalar (dot) product of force and displacement: W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}W=F⋅d.
  2. Perform the dot product:
    W=Fxdx+Fydy+FzdzW = F_x d_x + F_y d_y + F_z d_zW=Fx​dx​+Fy​dy​+Fz​dz​W=(4)(5)+(3)(2)+(−2)(0)W = (4)(5) + (3)(2) + (-2)(0)W=(4)(5)+(3)(2)+(−2)(0)W=20+6+0=26 JoulesW = 20 + 6 + 0 = 26 \text{ Joules}W=20+6+0=26 Joules

Answer: The work done by the force is 26 J26 \text{ J}26 J.

Example 6: Torque on a Bolt (Vector Product)

Problem: A mechanic applies a force F⃗=20j^ N\vec{F} = 20\hat{j} \text{ N}F=20j^​ N at the end of a wrench. The position vector from the bolt to the point of application of the force is r⃗=0.15i^+0.05j^ m\vec{r} = 0.15\hat{i} + 0.05\hat{j} \text{ m}r=0.15i^+0.05j^​ m. Find the torque vector τ⃗\vec{\tau}τ acting on the bolt.

Solution:

  1. Recall the definition of Torque:
    Torque is the vector (cross) product of the position vector and the force vector: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}τ=r×F.
  2. Perform the cross product calculation:
    τ⃗=(0.15i^+0.05j^)×(20j^)\vec{\tau} = (0.15\hat{i} + 0.05\hat{j}) \times (20\hat{j})τ=(0.15i^+0.05j^​)×(20j^​)
    Using properties of unit vectors (i^×j^=k^\hat{i} \times \hat{j} = \hat{k}i^×j^​=k^ and j^×j^=0\hat{j} \times \hat{j} = 0j^​×j^​=0):
    τ⃗=0.15×20(i^×j^)+0.05×20(j^×j^)\vec{\tau} = 0.15 \times 20 (\hat{i} \times \hat{j}) + 0.05 \times 20 (\hat{j} \times \hat{j})τ=0.15×20(i^×j^​)+0.05×20(j^​×j^​)τ⃗=3k^ N⋅m\vec{\tau} = 3\hat{k} \text{ N}\cdot\text{m}τ=3k^ N⋅m

Answer: The torque vector is 3k^ N⋅m3\hat{k} \text{ N}\cdot\text{m}3k^ N⋅m, which points along the positive z-axis (perpendicularly out of the plane of rotation).

In Example 6 the torque is r⃗imesF⃗=3k^extN⋅extm\vec{r} imes \vec{F} = 3\hat{k} ext{ N}\cdot ext{m}rimesF=3k^extN⋅extm. What is F⃗imesr⃗\vec{F} imes \vec{r}Fimesr, with the factors reversed?

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