Question

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

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

Answer 1

Answer: (A)

$\frac{-\overline{i}+7\overline{j}+7\overline{k}}{\sqrt{99}}$

Vector perpendicular to the lines $=(3i+j+2k)\times (i+2j+3k)$

$=-i+7j+7k$

$\Rightarrow$unit vector $=\frac{-i+7j+7k}{\sqrt{1^2+7^2+7^2}}$

$\Rightarrow \frac{-i+7j+7k}{\sqrt{99}}$