Question

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}.$ If $\overrightarrow{c}$ is a vector such that $\overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}|\text{and}|\overrightarrow{c}-\overrightarrow{a}|=2\sqrt{2}$ and the angle between $(\overrightarrow{a}\times \overrightarrow{b})$ and $\overrightarrow{c}$ is ${30}^0$, then $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|$ is equal to

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

Answer 1

The correct option is B $\frac{3}{2}$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|=2\hat{i}-2\hat{j}+\hat{k}$
$\therefore |\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{4+4+1}=3$
We have
$|(\overrightarrow{c}-\overrightarrow{a}){|}^2=8$
$\Rightarrow |\overrightarrow{c}{|}^2-2\overrightarrow{a}.\overrightarrow{c}+|\overrightarrow{a}{|}^2=8$
$\Rightarrow |\overrightarrow{c}{|}^2-2|\overrightarrow{c}|+9=8$
$\Rightarrow |\overrightarrow{c}{|}^2-2|\overrightarrow{c}|+1\Rightarrow |\overrightarrow{c}|=1$
$\therefore |(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=|(\overrightarrow{a}\times \overrightarrow{b})||\overrightarrow{c}|\sin {30}^0=\frac{3}{2}$