Question

Let $\Pi$ be a plane containing the points $(0,-5,-1)$, $(1,-2,5),(-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\bar{i}+5 \bar{j}-6 \bar{k}$.
Then the length of the projection of the unit normal vector to the plan $\Pi$ on the line $L$ is examsnet.com

• A. $\frac{133 \sqrt{2}}{\sqrt{31}}$
• B. $\frac{14}{\sqrt{682}}$
• C. $\frac{133}{\sqrt{31}}$
• D. $\frac{268}{2 \sqrt{32}}$

Answer 1

Answer: (B)

Equation of plane $(\pi)$ containing the points
$(0,-5,-1),(1,-2,5),(-3,5,0)$
is $\begin{vmatrix}x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1\end{vmatrix}=0$
$\Rightarrow \begin{vmatrix}x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1\end{vmatrix}=0$
$\Rightarrow 3 x+y-z+4=0$
$\Rightarrow r(3 \hat{i}+\hat{j}-\hat{k})+4=0 $
$\Rightarrow r(-3 \hat{i}-\hat{j}+\hat{k})=4$
$\Rightarrow r\left(\frac{-3}{\sqrt{11}}-\frac{\hat{j}}{\sqrt{11}}+\frac{\hat{k}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}}$
$\therefore $ Unit normal vector to plane $\pi$ is
$\hat{n}=\frac{-3}{\sqrt{11}} \hat{i}-\frac{\hat{j}}{\sqrt{11}}+\frac{\hat{k}}{\sqrt{11}}$
$\therefore $ length of theprojection of the unit normal vector hat $n$ to plane $\pi$ on the line $L$ is
$=\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}}$