Question

A unit vector perpendicular to the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ is

• A. $\frac{-\hat{i}+7\hat{j}+7\hat{k}}{\sqrt{99}}$
• B. $\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$
• C. $\frac{-\hat{i}+7\hat{j}+5\hat{k}}{5\sqrt{3}}$
• D. $\frac{7\hat{i}-7\hat{j}-\hat{k}}{\sqrt{99}}$

Answer 1

Answer: (B)

$\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$

Vector perpendicular to both lines $=$ cross product of their direction ratios

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& 2& 3\end{array}\right|$

$=\hat{i}(3-4)-\hat{j}(9-2)+\hat{b}(6-1)$

$=-\hat{i}-7\hat{j}+5\hat{k}$

unit vector $=\frac{-\hat{i}-7\hat{j}+5\hat{k}}{\sqrt{(1{)}^2-(7{)}^2+(5{)}^5}}=\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$