Dot Product Of Two Vectors - Practice Questions & MCQ

Multiplication (or product) of two vectors is defined in two ways, namely, dot (or scalar) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, we have various applications in geometry, mechanics and engineering.

Dot (scalar) Product

If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are two non-zero vectors, then their scalar product (or dot product) is denoted by $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}$ and is defined as

Observations:

1. $\overrightarrow{a}\cdot \overrightarrow{b}$ is a real number.
2. $\overrightarrow{a}\cdot \overrightarrow{b}$ is positive if $\theta$ is acute.
3. $\overrightarrow{a}\cdot \overrightarrow{b}$ is negative if $\theta$ is obtuse.
4. $\overrightarrow{a}\cdot \overrightarrow{b}$ is zero if $\theta$ is ${90}^{\circ }$.
5. $\overrightarrow{a}\cdot \overrightarrow{b}\le |\overrightarrow{a}||\overrightarrow{b}|$

For any two non-zero vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, then $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}=0$ if and only if $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ perpendicular to each other. i.e.
$\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}=0\iff \overrightarrow{\mathbf{a}}\perp \overrightarrow{\mathbf{b}}$

As $\hat{\mathbf{i}},\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are mutually perpendicular unit vectors along the coordinate axes, therefore,
$\hat{\mathbf{i}}\cdot \hat{\mathbf{j}}=\hat{\mathbf{j}}\cdot \hat{\mathbf{i}}=0,\hat{\mathbf{j}}\cdot \hat{\mathbf{k}}=\hat{\mathbf{k}}\cdot \hat{\mathbf{j}}=0;\hat{\mathbf{k}}\cdot \hat{\mathbf{i}}=\hat{\mathbf{i}}\cdot \hat{\mathbf{k}}=0$

If $\theta =0$, then $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{a}||\overrightarrow{b}|$
In particular, $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{a}}=|\overrightarrow{\mathbf{a}}{|}^2$
As $\hat{\mathbf{i}},\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are unit vectors along the coordinate axes, therefore
$\hat{\mathbf{i}}\cdot \hat{\mathbf{i}}=|\hat{\mathbf{i}}{|}^2=1,\hat{\mathbf{j}}\cdot \hat{\mathbf{j}}=|\hat{\mathbf{j}}{|}^2=1\text{ and }\hat{\mathbf{k}}\cdot \hat{\mathbf{k}}=|\hat{\mathbf{k}}{|}^2=1$

Properties of Dot (Scalar) Product

1. $\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{a}}$ ( commutative )
2. $\overrightarrow{\mathbf{a}}\cdot (\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})=\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{c}}$ (distributive)
3. $(m\overrightarrow{\mathbf{a}})\cdot \overrightarrow{\mathbf{b}}=m(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{a}}\cdot (m\overrightarrow{\mathbf{b}});$ where $m$ is a scalar and $\overrightarrow{a},\overrightarrow{b}$ are any two vectors
4. $(l\overrightarrow{\mathbf{a}})\cdot (m\overrightarrow{\mathbf{b}})=\mathrm{lm}(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}})$; where $l$ and $m$ are scalars

For any two vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, we have
(i)
$\begin{array}{rl}|\overrightarrow{a}\pm \overrightarrow{b}{|}^2& =|\overrightarrow{a}\pm \overrightarrow{b}|\cdot |\overrightarrow{a}\pm \overrightarrow{b}|\\ & =|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2\pm 2\overrightarrow{a}\cdot \overrightarrow{b}\\ & =|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2\pm 2|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \end{array}$
(ii) $|\overrightarrow{a}+\overrightarrow{b}|\cdot |\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2$
(iii) $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|+|\overrightarrow{b}|\Rightarrow \overrightarrow{a}$ and $\overrightarrow{b}$ are like vectors
(iv) $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|\Rightarrow \overrightarrow{a}\perp \overrightarrow{b}$

If $\overrightarrow{\mathbf{a}}=a_1\hat{\mathbf{i}}+a_2\hat{\mathbf{j}}+a_3\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=b_1\hat{\mathbf{i}}+b_2\hat{\mathbf{j}}+b_3\hat{\mathbf{k}}$, then
$\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

Proof:
$\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}=& (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\\ =& a_1\hat{i}\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})+a_2\hat{j}\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})+a_3\hat{k}\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\\ =& a_1{b}_1(\hat{i}\cdot \hat{i})+a_1{b}_2(\hat{i}\cdot \hat{j})+a_1{b}_3(\hat{i}\cdot \hat{k})+a_2{b}_1(\hat{j}\cdot \hat{i})+a_2{b}_2(\hat{j}\cdot \hat{j})+a_2{b}_3(\hat{j}\cdot \hat{k})\\ & +a_3{b}_1(\hat{k}\cdot \hat{i})+a_3{b}_2(\hat{k}\cdot \hat{j})+a_3{b}_3(\hat{k}\cdot \hat{k})\\ =& a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$
The angle between two vectors

$\begin{array}{rl}\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}& =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\cos \theta \\ \Rightarrow & \cos \theta & =\frac{\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\\ \Rightarrow & & ={\cos }^{-1}\left(\frac{\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\right)\end{array}$

If $\overrightarrow{\mathbf{a}}=a_1\hat{\mathbf{i}}+a_2\hat{\mathbf{j}}+a_3\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=b_1\hat{\mathbf{i}}+b_2\hat{\mathbf{j}}+b_3\hat{\mathbf{k}}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{a_1{b}_1+a_2{b}_2+a_3{b}_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}\right)$

Geometrical Interpretation of Scalar Product

Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two vectors represented by OA and OB, respectively.
Draw $\mathrm{BL}\perp \mathrm{OA}$ and $\mathrm{AM}\perp \mathrm{OB}$.
From triangles OBL and OAM we have $\mathrm{OL}=\mathrm{OB}\cos \theta$ and $\mathrm{OM}=\mathrm{OA}\cos \theta$.
Here OL and OM are known as projections of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}$ respectively.

Now,
$\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}& =|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \\ & =|\overrightarrow{a}|(OB\cos \theta )\\ & =|\overrightarrow{a}|(OL)\\ & =(\text{ magnitude of }\overrightarrow{a})(\text{ projection of }\overrightarrow{b}\text{ on }\overrightarrow{a})\end{array}$

Again,
$\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}& =|\overrightarrow{a}||\overrightarrow{b}|\cos \theta \\ & =|\overrightarrow{b}|(|\overrightarrow{a}|\cos \theta )\\ & =|\overrightarrow{b}|(OA\cos \theta )\\ & =|\overrightarrow{b}|(OM)\\ & =(\text{ magnitude of }\overrightarrow{b})(\text{ projection of }\overrightarrow{a}\text{ on }\overrightarrow{b})\end{array}$

Thus. geometrically interpreted, the scalar product of two vectors is the product of the modulus of either vector and the projection of the other in its direction. 

Thus, 

Projection of $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}=\frac{\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}=\overrightarrow{\mathbf{a}}\cdot \frac{\overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}=\overrightarrow{\mathbf{a}}\cdot \hat{\mathbf{b}}$ Projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}=\frac{\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}|}=\overrightarrow{\mathbf{b}}\cdot \frac{\overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\overrightarrow{\mathbf{b}}\cdot \hat{\mathbf{a}}$