Vector Algebra Question 395
Question: The equation of the plane containing the line $ \mathbf{r}=\mathbf{i}+\mathbf{j}+\lambda (2\mathbf{i}+\mathbf{j}+4\mathbf{k}) $ is
Options:
A) $ \mathbf{r}.(\mathbf{i}+2\mathbf{j}-\mathbf{k})=3 $
B) $ \mathbf{r}.(\mathbf{i}+2\mathbf{j}-\mathbf{k})=6 $
C) $ \mathbf{r}.(-\mathbf{i}-2\mathbf{j}+\mathbf{k})=3 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The position vector of any point on the given line is $ \mathbf{i}+\mathbf{j}+\lambda (2\mathbf{i}+\mathbf{j}+4\mathbf{k}) $ or $ (2\lambda +1)\mathbf{i}+(\lambda +1)\mathbf{j}+4\lambda \mathbf{k} $ which lies on $ \mathbf{r}.(\mathbf{i}+2\mathbf{j}-\mathbf{k})=3 $ . Hence, the plane $ \mathbf{r}.(\mathbf{i}+2\mathbf{j}-\mathbf{k})=3 $ contains the given line.
BETA
