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

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

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

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{aligned}
\vec{a} \cdot \vec{b}= & \left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \\
= & a_1 \hat{i} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_2 \hat{j} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_3 \hat{k} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \\
= & 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}) \\
& \quad+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{aligned}
$
The angle between two vectors

$
\begin{array}{rlrl}
\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta \\
\Rightarrow \quad & \cos \theta & =\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|} \\
\Rightarrow \quad & & =\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 \quad \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{aligned}
\vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos \theta \\
& =|\vec{a}|(O B \cos \theta) \\
& =|\vec{a}|(O L) \\
& =(\text { magnitude of } \vec{a})(\text { projection of } \vec{b} \text { on } \vec{a})
\end{aligned}
$

Again,
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos \theta \\
& =|\vec{b}|(|\vec{a}| \cos \theta) \\
& =|\vec{b}|(O A \cos \theta) \\
& =|\vec{b}|(O M) \\
& =(\text { magnitude of } \vec{b})(\text { projection of } \vec{a} \text { on } \vec{b})
\end{aligned}
$

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