Question

Let $a=2\hat{i}+\hat{j}-2\hat{k}$ and $b=\hat{i}+\hat{j}$. If c is a vector such that $a.c=|c|,|c-a|=2\sqrt{2}$ and the angle between $a\times b$ and c is ${30}^o$. Then, $\left[\right(a\times b)\times c]$ is equal to

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

Answer 1

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

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

$⟹|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{2^2+(-2{)}^2+1^2}=3$

Also, $|\overrightarrow{c}-\overrightarrow{a}|=2\sqrt{2}⟹|\overrightarrow{c}-\overrightarrow{a}{|}^2=8⟹(\overrightarrow{c}-\overrightarrow{a})\cdot (\overrightarrow{c}-\overrightarrow{a})=8$

$⟹\overrightarrow{c}\cdot \overrightarrow{c}-\overrightarrow{c}\cdot \overrightarrow{a}-\overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{a}\cdot \overrightarrow{a}=|\overrightarrow{c}{|}^2+|\overrightarrow{a}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{c}=8$

$⟹|\overrightarrow{c}{|}^2+9-2\overrightarrow{c}=8[\because \overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}|]$

$⟹|\overrightarrow{c}{|}^2-2\overrightarrow{c}+1=0$

$⟹(|\overrightarrow{c}{|}^2-1{)}^2=0⟹|\overrightarrow{c}|=1$

Now, $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=|(\overrightarrow{a}\times \overrightarrow{b})||\overrightarrow{c}|\sin {30}^o$

$\therefore |(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=3\times 1\times \sin {30}^o=\frac{3}{2}$