Question

Let $a=\hat{i}+2\hat{j}-\hat{k}$ and $b=\hat{i}+\hat{j}+\hat{k}.$ If $p$ is a
unit vector such that $[abp]$ is maximum. then $p=$

• A. $\frac{1}{\sqrt{6}}(\hat{i}-2\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})$
• C. $\frac{1}{\sqrt{14}}(3\hat{i}-2\hat{j}-\hat{k})$
• D. $\frac{1}{\sqrt{14}}(\hat{i}+2\hat{j}+3\hat{k})$

Answer 1

Answer: (C)

Given, $a=\hat{i}+2\hat{j}-\hat{k},b=\hat{i}+\hat{j}+\hat{k}$
$a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 1& 1& 1\end{array}\right|=3\hat{i}-2\hat{j}-\hat{k}$
$[abp]=p\cdot (a\times b)=p\cdot (3\hat{i}-2\hat{j}-\hat{k})$
$[abp]=|P||a\times b|\cos \theta$
$[abp]$ is maximum
$\therefore p=\frac{a\times b}{|a\times b|}$
$p=\frac{1}{\sqrt{14}}(3\hat{i}-2\hat{j}-\hat{k})$