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 $\left(\overset{ \rightarrow }{n}\right)=2\hat{i}-\hat{j}-\hat{k}$
A vector parallel to given line $\left(\overset{ \rightarrow }{b}\right)=2\hat{i}+\hat{j}+5\hat{k}$
A vector parallel to required line
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -1 \\ 2 & 1 & 5 \end{vmatrix}=\hat{i}\left(- 4\right)-\hat{j}\left(12\right)+\hat{k}\left(4\right)$
$=-4\left(\hat{i} + 3 \hat{j} - \hat{k}\right)$
Equation of required line is
$\frac{x + 5}{1}=\frac{y + \frac{1}{2}}{3}=\frac{z + \frac{27}{2}}{- 1}$