Question

Let $a=2\hat{i}+\hat{j}-2\hat{k}$ and $b=\begin{array}{rcl}& & \hat{i}\end{array}+\begin{array}{rcl}& & \hat{j}\end{array}$. If $c$ is a vector such that $a.c=\left|c\right|$,$\left|c-a\right|=2\sqrt{2}$ and the angle between $a\times b$ and $c$ is the $30°$, then $\left|\left(a\times b\right)\times c\right|$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$\frac{3}{2}$

Explanation for the correct option:

Finding $\left|\left(a\times b\right)\times c\right|$:

Let $a=2\begin{array}{rcl}& & \hat{i}\end{array}+\begin{array}{rcl}& & \hat{j}\end{array}-2\begin{array}{rcl}& & \hat{k}\end{array}$ and $b=\begin{array}{rcl}& & \hat{i}\end{array}+\begin{array}{l}\hat{j}\end{array}$.

Given that the angle between $a\times b$ and $c$ is $30°$,

$\left|\left(a\times b\right)\times c\right|=\left|\left(a\times b\right)\right|\left|c\right|\sin \left(30\right)°\dots \dots \left(i\right)$

$\begin{array}{rcl}\overrightarrow{a}\times \overrightarrow{b}& =& \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|\\ & =& \hat{i}\left(0+2\right)-\hat{j}\left(0+2\right)+\hat{k}\left(2-1\right)\\ & =& 2\hat{i}-2\hat{j}+\hat{k}\end{array}$

$\begin{array}{rcl}\left|\overrightarrow{a}\times \overrightarrow{b}\right|& =& \sqrt{2^2+{\left(-2\right)}^2+1}\\ & =& \sqrt{9}\\ & =& 3\end{array}$

Given $a=2\hat{i}+\hat{j}-2\hat{k}$

$\begin{array}{rcl}\left|a\right|& =& \sqrt{{\left(2\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}\\ & =& \sqrt{4+1+4}\\ & =& \sqrt{9}\\ \left|a\right|& =& 3\dots \dots \left(ii\right)\end{array}$

Given,$a.c=\left|c\right|$,$\left|c-a\right|=2\sqrt{2}$

Squaring both sides of the equation $\left|c-a\right|=2\sqrt{2}$, we get

$$\begin{gathered} \Rightarrow {\left(\left|c-a\right|\right)}^2={\left(2\sqrt{2}\right)}^2 \\ \Rightarrow {\left|c\right|}^2+{\left|a\right|}^2-2a.c=4\times 2\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right] \\ \Rightarrow {\left|c\right|}^2+{\left|a\right|}^2-2\left|c\right|=8\left[\because a.c=\left|c\right|\right] \\ \Rightarrow {\left|c\right|}^2-2\left|c\right|+9=8\mathbf{}\left[\because {\left|a\right|}^2=9\text{from equation}\left(ii\right)\right] \\ \Rightarrow {\left|c\right|}^2-2\left|c\right|+1=0 \\ \Rightarrow {\left(\left|c\right|-1\right)}^2=0 \\ \Rightarrow \left|c\right|=1 \end{gathered}$$

Substituting all these values in the equation $\left(i\right)$, we get

$$\begin{gathered} \left|\left(a\times b\right)\times c\right|=\left|\left(a\times b\right)\right|\left|c\right|\sin \left(30\right)° \\ =3\times 1\times \frac{1}{2}\mathbf{}\left[\because \left|\left(a\times b\right)\right|=3;\left|c\right|=1;sin\left(30\right)°=\frac{1}{2}\right] \\ =\frac{3}{2} \end{gathered}$$

Hence, option (B) is the correct answer.