Question

Let the position vectors of the points $P$ and $Q$ be $4\hat{i}+\hat{j}+\lambda \hat{k}$ and $2\hat{i}-\hat{j}+\lambda \hat{k}$, respectively. Vector $\hat{i}-\hat{j}+6\hat{k}$ is perpendicular to the plane containing the origin and the points $P$ and $Q$, then $\lambda$ equals

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$-\frac{1}{2}$

A vector perpendicular to the plane of $O,P$ and $Q$ is $\overrightarrow{OP}\times \overrightarrow{OQ}$.

Now, $\overrightarrow{OP}\times \overrightarrow{OQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 1& \lambda \\ 2& -1& \lambda \end{array}\right|=2\lambda \hat{i}-2\lambda \hat{j}-6\hat{k}$

Therefore, $\hat{i}-\hat{j}+6\hat{k}$ is parallel to
$2\lambda \hat{i}-2\lambda \hat{j}-6\hat{k}$
Hence, $\frac{1}{2\lambda }=\frac{-1}{-2\lambda }=\frac{6}{-6}$
$\Rightarrow \lambda =-\frac{1}{2}$