Question

Let $a=2\hat{i}+\hat{j}-2\hat{k},b=\hat{i}+\hat{j}$ and $c$ be a vector such that $|c-a|=3,|(a\times b)\times c|=3$ and the angle between $c$ and $a\times b$ is ${30}^{\circ }$. Then, $a\cdot c$ is equal to

• A. $\frac{25}{8}$
• B. 2
• C. 5
• D. $\frac{1}{8}$

Answer 1

The correct option is B

2

We have,
$a=2\hat{i}+\hat{j}-2\hat{k}\Rightarrow |a|=\sqrt{4+1+4}=3\text{and}b=\hat{i}+\hat{j}\Rightarrow |b|=\sqrt{1+1}=\sqrt{2}\text{Now},|c-a|=3\Rightarrow |c-a{|}^2=9\Rightarrow (c-a).(c-a)=9\Rightarrow |c{|}^2+|a{|}^2-2c.a=9\text{Again},|(a\times b)\times c|=3\Rightarrow |a\times b||c|sin{30}^o=3\Rightarrow |c|=\frac{6}{|a\times b|}\text{But}a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|=2\hat{i}-2\hat{j}+\hat{k}\therefore |c|=\frac{6}{\sqrt{4+4+1}}=2$
FromEqs. (i) and (ii), we get
$\Rightarrow (2{)}^2+(3{)}^2-2c.a=9\Rightarrow c.a=2$