Question

A plane passing through origin $P(4,1,\lambda )$ and $Q(2,-1$ , $\lambda$ $)$ is perpendicular to a line with direction ratios $(1,-1,6)$, then $\lambda$ is equal to

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $-\frac{1}{2}$

Given : Plane passes through the Points $P(4,1,\lambda ),Q(2,-1,\lambda )$ and $O(0,0,0)$
So, $D.r^{\prime }s$ of normal is $:\overrightarrow{OP}\times \overrightarrow{OQ}$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 1& \lambda \\ 2& -1& \lambda \end{array}\right|=2\lambda \hat{i}-2\lambda \hat{j}-6\hat{k}$
$\Rightarrow D.r^{\prime }s$ of normal to the plane $=(2\lambda ,-2\lambda ,-6)$
Also, the given line is perpendicular to the plane, $i.e.$ the line is normal to the plane($D.r^{\prime }s$ will be proportional).
$D.r^{\prime }s$ of lines is $(1,-1,6)$
$\Rightarrow \frac{2\lambda }{1}=\frac{-2\lambda }{-1}=\frac{6}{-6}\Rightarrow \lambda =-\frac{1}{2}$