Vector Algebra Question 262
Question: The vector equation of the plane through the point $ 2\mathbf{i}-\mathbf{j}-4\mathbf{k} $ and parallel to the plane $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})-7=0 $ is
Options:
A) $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=0 $
B) $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=32 $
C) $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=12 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- The equation of a plane parallel to the plane $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})-7=0 $ is $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})+\lambda =0 $ . This passes through $ 2\mathbf{i}-\mathbf{j}-4\mathbf{k} $ . Therefore, $ (2\mathbf{i}-\mathbf{j}-4\mathbf{k}).(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})+\lambda =0 $
Þ $ 8+12+12+\lambda =0\Rightarrow \lambda =-32 $ So, the required plane is $ \mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})-32=0 $ .
BETA
