Question

Unit vector perpendicular to $\hat{i}-2\hat{j}+2\hat{k}$ and lying in the plane containing $\hat{i}+\hat{j}-2\hat{k}$ and $-\hat{i}+2\hat{j}+\hat{k}$ is

• A. $8\hat{i}-7\hat{j}+11\hat{k}$
• B. $8\hat{i}+7\hat{j}-11\hat{k}$
• C. $8\hat{i}-7\hat{j}-11\hat{k}$
• D. $\frac{1}{\sqrt{234}}(8\hat{i}-7\hat{j}-11\hat{k})$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{234}}(8\hat{i}-7\hat{j}-11\hat{k})$

Let the vector be $(\hat{i}+\hat{j}-2\hat{k})+a(-\hat{i}+2\hat{j}+\hat{k})$

$=(1-a)\hat{i}+(1+2a)\hat{j}+(a-2)\hat{k}$

This is perpendicular to $\hat{i}-2\hat{j}+2\hat{k}$

$\Rightarrow (1-a)-2(1+2a)+2(a-2)=0$

$\Rightarrow 1-a-2-4a+2a-4=0$

$\Rightarrow -3a-5=0$ or $a=\frac{-5}{3}$

The vector becomes $(1+\frac{5}{3})\hat{i}+(1-\frac{10}{3})\hat{j}+(\frac{-5}{3}-2)\hat{k}$

$=\frac{8}{3}\hat{i}-\frac{7}{3}\hat{j}-\frac{11}{3}\hat{k}$

The unit vector can be written as $\frac{1}{\sqrt{64+49+121}}(8\hat{i}-7\hat{j}-11\hat{k})$

$=\frac{1}{\sqrt{234}}(8\hat{i}-7\hat{j}-11\hat{k})$