Question

A line $L$ lies in the plane $2x-y-z=4$ such that it is perpendicular to the line $\frac{x-2}{2}=\frac{y-3}{1}=\frac{z-4}{5}.$ The line $L$ passes through the point of intersection of the given line and given plane. Which of the following points does not satisfy line $L?$

• A. $\left(-5,-\frac{1}{2},-\frac{27}{2}\right)$
• B. $\left(3,1,6\right)$
• C. $\left(0,\frac{29}{2},\frac{-37}{2}\right)$
• D. $\left(-4,\frac{5}{2},\frac{-29}{2}\right)$

Answer 1

Answer: (B)

Any point on given line is $\left(2\lambda +2,\lambda +3,5\lambda +4\right)$
This must satisfy the equation of the plane
$\Rightarrow 4\lambda +4-\lambda -3-5\lambda -4=4$
$\Rightarrow -2\lambda =7\Rightarrow \lambda =-\frac{7}{2}$
So, the coordinates of point are $\left(-5,-\frac{1}{2},\frac{-27}{2}\right)$
A normal vector to given plane $\left(\overrightarrow{n}\right)=2\hat{i}-\hat{j}-\hat{k}$
A vector parallel to given line $\left(\overrightarrow{b}\right)=2\hat{i}+\hat{j}+5\hat{k}$
A vector parallel to required line
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& -1\\ 2& 1& 5\end{array}\right|=\hat{i}\left(-4\right)-\hat{j}\left(12\right)+\hat{k}\left(4\right)$
$=-4\left(\hat{i}+3\hat{j}-\hat{k}\right)$
Equation of required line is
$\frac{x+5}{1}=\frac{y+\frac{1}{2}}{3}=\frac{z+\frac{27}{2}}{-1}$