Motion in a Plane • Topic 1 of 3

Vectors & Their Operations

In Class 9 you described motion along a straight line, where direction could be handled with a simple plus or minus sign. Motion in a plane is richer: a cricket ball arcs across the field, a swimmer drifts across a river. To handle two dimensions we need a tool that carries both magnitude and direction together — and that tool is the vector.

Physical quantities fall into two families. A scalar has only magnitude: mass, time, temperature, speed, distance and energy are all scalars and obey ordinary arithmetic. A vector has both magnitude and direction and follows special rules of addition: displacement, velocity, acceleration, force and momentum are vectors. A vector is drawn as an arrow whose length is its magnitude and whose head points in its direction. We write it as $\vec{A}$ and its magnitude as $|\vec{A}|$ or simply $A$.

Two vectors are added head to tail. In the triangle law, the tail of $\vec{B}$ is placed at the head of $\vec{A}$, and the arrow drawn from the tail of $\vec{A}$ to the head of $\vec{B}$ is the resultant $\vec{R}=\vec{A}+\vec{B}$. The parallelogram law states that if two vectors acting at a point are the adjacent sides of a parallelogram, the diagonal through that point is their resultant. When $\vec{A}$ and $\vec{B}$ have an angle $\theta$ between them, the magnitude of the resultant is $R=\sqrt{A^2+B^2+2AB\cos\theta}$, and it makes an angle $\alpha$ with $\vec{A}$ where $\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}$.

Working with arrows geometrically is clumsy, so we resolve each vector into perpendicular components along the x and y axes. Using the unit vectors $\hat{i}$ (along x) and $\hat{j}$ (along y), a vector making angle $\theta$ with the x-axis becomes $\vec{A}=A\cos\theta\,\hat{i}+A\sin\theta\,\hat{j}$. A unit vector has magnitude exactly 1 and only fixes direction; $\hat{A}=\dfrac{\vec{A}}{|\vec{A}|}$. To add vectors by components, simply add the x-parts and the y-parts separately. Two ways of multiplying vectors also appear: the dot (scalar) product $\vec{A}\cdot\vec{B}=AB\cos\theta$ gives a scalar (used for work), while the cross (vector) product $\vec{A}\times\vec{B}=AB\sin\theta\,\hat{n}$ gives a vector perpendicular to both (used for torque).

  • Scalar: magnitude only (mass, speed, time, distance).
  • Vector: magnitude and direction (displacement, velocity, force).
  • Resultant: $R=\sqrt{A^2+B^2+2AB\cos\theta}$ with $\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}$.
  • Components: $A_x=A\cos\theta$, $A_y=A\sin\theta$; magnitude $A=\sqrt{A_x^2+A_y^2}$.
  • Products: dot $\vec{A}\cdot\vec{B}=AB\cos\theta$ (scalar); cross $|\vec{A}\times\vec{B}|=AB\sin\theta$ (vector).
Parallelogram law: resultant of two vectors A and BxyABR = A + BthetaR = sqrt(A^2 + B^2 + 2AB cos theta)
1
Worked Example
Two forces of $3\,\text{N}$ and $4\,\text{N}$ act at right angles ($\theta=90^\circ$) to each other at a point. Find the magnitude of their resultant.
Solution
  1. Use $R=\sqrt{A^2+B^2+2AB\cos\theta}$ with $\theta=90^\circ$, so $\cos 90^\circ=0$.
  2. $R=\sqrt{3^2+4^2+0}=\sqrt{9+16}$.
  3. $=\sqrt{25}=5\,\text{N}$.

Answer: Resultant $=5\,\text{N}$.

2
Worked Example
A vector $\vec{A}$ of magnitude $10\,\text{units}$ makes an angle of $30^\circ$ with the x-axis. Find its x and y components.
Solution
  1. $A_x=A\cos\theta=10\cos 30^\circ=10\times\dfrac{\sqrt{3}}{2}$.
  2. $=5\sqrt{3}\approx 8.66\,\text{units}$.
  3. $A_y=A\sin\theta=10\sin 30^\circ=10\times\dfrac{1}{2}=5\,\text{units}$.

Answer: $A_x\approx 8.66$, $A_y=5$ (so $\vec{A}=8.66\,\hat{i}+5\,\hat{j}$).

3
Worked Example
Two forces of equal magnitude $F$ act at an angle of $60^\circ$. The resultant has magnitude $\sqrt{3}\,F$. Verify this using the parallelogram law.
Solution
  1. Here $A=B=F$ and $\theta=60^\circ$, so $\cos 60^\circ=\dfrac{1}{2}$.
  2. $R=\sqrt{F^2+F^2+2F\cdot F\cdot\tfrac{1}{2}}=\sqrt{2F^2+F^2}$.
  3. $=\sqrt{3F^2}=\sqrt{3}\,F$.

Answer: $R=\sqrt{3}\,F$, as required.

4
Worked Example
Find the magnitude of the vector $\vec{A}=6\,\hat{i}+8\,\hat{j}$ and the unit vector along it.
Solution
  1. Magnitude $A=\sqrt{A_x^2+A_y^2}=\sqrt{6^2+8^2}=\sqrt{100}=10$.
  2. Unit vector $\hat{A}=\dfrac{\vec{A}}{A}=\dfrac{6\,\hat{i}+8\,\hat{j}}{10}$.
  3. $=0.6\,\hat{i}+0.8\,\hat{j}$.

Answer: $A=10$; $\hat{A}=0.6\,\hat{i}+0.8\,\hat{j}$.

5
Worked Example
If $\vec{A}=2\,\hat{i}+3\,\hat{j}$ and $\vec{B}=4\,\hat{i}-\hat{j}$, find $\vec{A}\cdot\vec{B}$.
Solution
  1. For component form, $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$.
  2. $=(2)(4)+(3)(-1)$.
  3. $=8-3=5$.

Answer: $\vec{A}\cdot\vec{B}=5$.

6
Worked Example
Two vectors of magnitudes $5\,\text{units}$ and $5\,\text{units}$ act with an angle of $90^\circ$ between them. Find the magnitude of their cross product.
Solution
  1. $|\vec{A}\times\vec{B}|=AB\sin\theta$ with $\theta=90^\circ$, so $\sin 90^\circ=1$.
  2. $=5\times 5\times 1$.
  3. $=25\,\text{units}$.

Answer: $|\vec{A}\times\vec{B}|=25\,\text{units}$.

Key Points

  • A scalar has magnitude only; a vector has both magnitude and direction.
  • Vectors add head to tail using the triangle or parallelogram law: $R=\sqrt{A^2+B^2+2AB\cos\theta}$.
  • Any vector resolves into components $A_x=A\cos\theta$, $A_y=A\sin\theta$, written with unit vectors $\hat{i},\hat{j}$.
  • A unit vector $\hat{A}=\dfrac{\vec{A}}{|\vec{A}|}$ has magnitude 1 and fixes only direction.
  • Dot product $\vec{A}\cdot\vec{B}=AB\cos\theta$ gives a scalar; cross product $|\vec{A}\times\vec{B}|=AB\sin\theta$ gives a vector.
Tap an option to check your answer0 / 4
Q1.Which of the following is a vector quantity?
Explanation: Force has both magnitude and direction, so it is a vector. Speed, mass and temperature are scalars.
Q2.Two forces of 6 N and 8 N act at right angles. Their resultant is:
Explanation: At 90 degrees, R = sqrt(6^2 + 8^2) = sqrt(100) = 10 N.
Q3.The x-component of a vector A making angle theta with the x-axis is:
Explanation: The component along the x-axis is the adjacent side, A cos theta.
Q4.The magnitude of a unit vector is always:
Explanation: A unit vector has magnitude exactly 1 and only specifies direction.