Question

Let the position vectors of the points $P$ and $Q$ be $4i+j+nk$ and $2i-j+nk$ respectively vector $i-j+6k$ is perpendicular to the plane containing the origin and the points $P$ and $Q$ then $n$ equals

• A. $-1/2$
• B. $1/2$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $-1/2$

A plane contains point $\overline{O},\overline{P},\overline{Q}$and is perpendicular to line

$\overline{L}=\hat{i}-\hat{j}+6\hat{k}$

We know that this means that

$\left(\overline{L}\right).\left(\overline{OP}\right)=0$

$\left(\overline{L}\right).\left(\overline{OQ}\right)=0$ Dot products of two perpendicular lines is zero

$\left(\overline{L}\right).\left(\overline{QP}\right)=0$

$\therefore$ taking eqn $\left(1\right)\Rightarrow$

$(\hat{i}-\hat{j}+6\hat{k}).\left(\right(4-0)\hat{i}+(1-0)\hat{j}+(x-0\left)\hat{k}\right)=0$

$4-1+6x=0$

$\Rightarrow m=\frac{-1}{2}$