Vector Algebra Question 282
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 point?s P and Q. then $ \lambda $ equals
Options:
A) $ -,1/2 $
B) 1/2
C) 1
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] A vector perpendicular to the plane of O, P and Q is $ \overrightarrow{OP}\times \overrightarrow{OQ}. $ Now, $ \overrightarrow{OP}\times \overrightarrow{OQ}= \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 4 & 1 & \lambda \\ 2 & -1 & \lambda \\ \end{vmatrix} =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} $
$ \therefore \lambda =-\frac{1}{2} $
BETA
