Chapter 10: Vectors

In our day to day life, we come across many queries such as “What is your height?”, “How should a football player hit the ball to give a pass to another player?”, “What should be the speed and direction of an aeroplane?”. The answers to these questions require quantities having either magnitude alone or magnitude as well as direction.

Vectors and Scalars

A quantity that has magnitude as well as direction is called a vector. Directed line segments are vectors. A quantity that has only magnitude and no direction is called a scalar.

• Position Vector: Consider a point $P(x, y, z)$ in space. The vector $\vec{OP}$ having initial point $O$ (origin) and terminal point $P$ is called the position vector of point $P$. Its magnitude is given by $\sqrt{x^2 + y^2 + z^2}$.
• Direction Cosines: If $\vec{r}$ makes angles $\alpha, \beta, \gamma$ with the positive direction of $x, y$ and $z$-axes, then $\cos \alpha, \cos \beta, \cos \gamma$ are called direction cosines (typically denoted as $l, m, n$). $l^2 + m^2 + n^2 = 1$.

Operations on Vectors

1. Addition of Vectors: Add corresponding components. Triangle law and Parallelogram law.
2. Scalar Multiplication: Multiply every component by scalar $\lambda$.

Product of Two Vectors

1. Scalar (Dot) Product: For two non-zero vectors $\vec{a}$ and $\vec{b}$, the scalar product is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between them. If they are perpendicular, $\vec{a} \cdot \vec{b} = 0$.
2. Vector (Cross) Product: The vector product is $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta , \hat{n}$, where $\hat{n}$ is a unit vector perpendicular to both. If they are parallel, $\vec{a} \times \vec{b} = 0$. Area of a parallelogram with adjacent sides $\vec{a}, \vec{b}$ is $|\vec{a} \times \vec{b}|$.

Competency-Based Questions

1. [Sample Paper 2023] If $\vec{a}, \vec{b}$ and $\vec{c}$ are three mutually perpendicular vectors of equal magnitude, find the angle between $\vec{a}$ and $\vec{a} + \vec{b} + \vec{c}$.

Solution:

Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = x$.

Since they are mutually perpendicular, $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.

Now, $|\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$

$= |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})$

$= x^2 + x^2 + x^2 + 0 = 3x^2$.

So, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}x$.

Let $\theta$ be the angle between $\vec{a}$ and $(\vec{a} + \vec{b} + \vec{c})$.

Then $\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}| |\vec{a} + \vec{b} + \vec{c}| \cos \theta$

$\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = x \cdot (\sqrt{3}x) \cdot \cos\theta$

$|\vec{a}|^2 + 0 + 0 = \sqrt{3}x^2 \cos\theta$

$x^2 = \sqrt{3}x^2 \cos\theta \implies \cos\theta = \frac{1}{\sqrt{3}}$.

Therefore, $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

2. [CBSE 2021] Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$.

Solution:

Let the resultant vector be $\vec{R} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} + \hat{k})$

$\vec{R} = 3\hat{i} + \hat{j} + 0\hat{k}$

Magnitude of $\vec{R} = |\vec{R}| = \sqrt{3^2 + 1^2 + 0^2} = \sqrt{10}$.

Unit vector parallel to $\vec{R} = \hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{3\hat{i} + \hat{j}}{\sqrt{10}}$.

Vector of magnitude 5 parallel to $\vec{R} = \pm 5 \hat{R} = \pm 5 \left(\frac{3\hat{i} + \hat{j}}{\sqrt{10}}\right) = \pm \sqrt{\frac{5}{2}} (3\hat{i} + \hat{j})$.

3. [CBSE 2019] Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.

Solution:

The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$

$= \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2))$

$= \hat{i}(20) - \hat{j}(-5) + \hat{k}(-5) = 20\hat{i} + 5\hat{j} - 5\hat{k}$.

Magnitude = $\sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450}$

$= 15\sqrt{2} \text{ sq. units}$.