Vector Algebra Question 373
Question: The vector equation of the plane through the point (2, 1, ?1) and passing through the line of intersection of the plane $ \mathbf{r}.(\mathbf{i}+3\mathbf{j}-\mathbf{k})=0 $ and $ \mathbf{r}.(\mathbf{j}+2\mathbf{k})=0 $ is
Options:
A) $ \mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=0 $
B) $ \mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=6 $
C) $ \mathbf{r}.(\mathbf{i}-3\mathbf{j}-13\mathbf{k})=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The vector equation of a plane through the line of intersection of the planes $ \mathbf{r}.(\mathbf{i}+3\mathbf{j}-\mathbf{k})=0 $ and $ \mathbf{r}.(\mathbf{j}+2\mathbf{k}) $ =0 can be written as $ (\mathbf{r}.(\mathbf{i}+3\mathbf{j}-\mathbf{k}))+\lambda (\mathbf{r}.(\mathbf{j}+2\mathbf{k}))=0 $ …..(i) This passes through $ 2\mathbf{i}+\mathbf{j}-\mathbf{k} $ \ $ (2\mathbf{i}+\mathbf{j}-\mathbf{k}).(\mathbf{i}+3\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}-\mathbf{k}).(\mathbf{j}+2\mathbf{k})=0 $ or $ (2+3+1)+\lambda (0+1-2)=0\Rightarrow \lambda =6 $ Put the value of $ \lambda $ in (i) we get $ \mathbf{r}.(\mathbf{i}+9\mathbf{j}+11\mathbf{k})=0 $ , which is the required plane.
BETA
