Question

If the projection of the vector $a$ on $b$ is $|a\times b|$ and if $3b=i+j+k$, then the angle between $a$ and $b$ is:

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{6}$
• E. $0$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{3}$

Step 1: Use the concept of projection of a vector on another vector

Given: The projection of the vector $a$ on $b$ is $|a\times b|$.

We know that the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by $\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

Thus, $|a\times b|=\frac{a.b}{\left|b\right|}$………………..$\left(1\right)$

And, $3b=i+j+k$

$\Rightarrow$ $\left|3b\right|=\sqrt{1^2+1^2+1^2}$

$\Rightarrow$ $\left|b\right|=\frac{\sqrt{3}}{3}$

$\Rightarrow$ $\left|b\right|=\frac{1}{\sqrt{3}}$

Step 2: Use the concept of dot and cross product of two vectors

We know that, $|a\times b|=\left|a\right|\left|b\right|\sin \left(\theta \right)$ and $a.b=\left|a\right|\left|b\right|\cos \left(\theta \right)$

From equation $\left(1\right)$,

$\frac{a.b}{\left|b\right|}=\left|a\right|\left|b\right|\sin \left(\theta \right)$

$\Rightarrow$$\frac{\left|a\right|\left|b\right|\cos \left(\theta \right)}{\left|b\right|}=\left|a\right|\left|b\right|\sin \left(\theta \right)$

$\Rightarrow$ $\frac{\cos \left(\theta \right)}{\left|b\right|}=\sin \left(\theta \right)$

$\Rightarrow$ $\tan \left(\theta \right)=\frac{1}{\left|b\right|}$ $\left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \right]$

$\Rightarrow$ $\tan \left(\theta \right)=\sqrt{3}$

$\Rightarrow$ $\theta =\frac{\mathrm{\pi }}{3}$

Therefore the angle between $a$ and $b$ is $\frac{\mathrm{\pi }}{3}$.

Hence option $\left(A\right)$ is the correct option.