Vector Algebra Question 252
Question: The position vectors of two points P and Q are $ 3\mathbf{i}+\mathbf{j}+2\mathbf{k} $ and $ \mathbf{i}-2\mathbf{j}-4\mathbf{k} $ respectively. The equation of the plane through Q and perpendicular to PQ is
Options:
A) $ \mathbf{r}.(2\mathbf{i}+3\mathbf{j}+6\mathbf{k})=28 $
B) $ \mathbf{r}.(2\mathbf{i}+3\mathbf{j}+6\mathbf{k})=32 $
C) $ \mathbf{r}.(2\mathbf{i}+3\mathbf{j}+6\mathbf{k})+28=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- The required plane is $ {\mathbf{r}-(\mathbf{i}-2\mathbf{j}-4\mathbf{k})}.\overrightarrow{PQ}=0 $ .
BETA
