Question

The perpendicular distance of the point $\hat{i}+\hat{j}+\hat{k}$ from the line joining the points $3\hat{i}+4\hat{j}-\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ is

• A. $\frac{3}{\sqrt{10}}$
• B. $\frac{\sqrt{10}}{2}$
• C. $\frac{7}{\sqrt{10}}$
• D. $2\sqrt{10}$

Answer 1

Answer: (C)

$\Theta$ Area of $△PAB=\frac{1}{2}|\overrightarrow{PA}\times \overrightarrow{PB}|=\frac{1}{2}(AB)(PM)$
$\Theta \overrightarrow{PA}=2\hat{i}+3\hat{j}-2\hat{k}$ and $\overrightarrow{PB}=\hat{i}-2\hat{k}$

$\overrightarrow{PA}\times \overrightarrow{PB}=\left|\begin{array}{ccc}<br/>\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -2\\ 1& 0& -2\end{array}\right|$
$=-6\hat{i}+2\hat{j}-3\hat{k}\text{ and }\overrightarrow{AB}=-\hat{i}-3\hat{j}$
$\therefore 7=\sqrt{10}\cdot PM\Rightarrow PM=\frac{7}{\sqrt{10}}$