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 $\left|\begin{array}{ccc}x-0& y+5& z+1\\ 1-0& -2+5& 5+1\\ -3-0& 5+5& 0+1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& y+5& z+1\\ 1& 3& 6\\ -3& 10& 1\end{array}\right|=0$
$\Rightarrow 3x+y-z+4=0$
$\Rightarrow r(3\hat{i}+\hat{j}-\hat{k})+4=0$
$\Rightarrow r(-3\hat{i}-\hat{j}+\hat{k})=4$
$\Rightarrow r\left(\frac{-3}{\sqrt{11}}-\frac{\hat{j}}{\sqrt{11}}+\frac{\hat{k}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}}$
$\therefore$ 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
$=\left|\frac{\frac{-3}{\sqrt{11}}\frac{-5}{\sqrt{11}}\frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11}\sqrt{62}}\right|=\frac{14}{\sqrt{682}}$