Vector Algebra Question 127
Question: The vector equation of the plane through the point $ \mathbf{i}+2\mathbf{j}-\mathbf{k} $ and perpendicular to the line of intersection of the planes $ \mathbf{r}.(3\mathbf{i}-\mathbf{j}+\mathbf{k})=1 $ and $ \mathbf{i}+4\mathbf{j}-2\mathbf{k}=2 $ is
Options:
A) $ \mathbf{r}.(2\mathbf{i}+7\mathbf{j}-13\mathbf{k})=1 $
B) $ \mathbf{r}.(2\mathbf{i}-7\mathbf{j}-13\mathbf{k})=1 $
C) $ \mathbf{r}.(2\mathbf{i}+7\mathbf{j}+13\mathbf{k})=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- The line of intersection of the planes $ \mathbf{r}.(3\mathbf{i}-\mathbf{j}+\mathbf{k})=1 $ and $ \mathbf{r}.(\mathbf{i}+4\mathbf{j}-2\mathbf{k})=2 $ is common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e., $ {{\mathbf{n}}_1}=3\mathbf{i}-\mathbf{j}+\mathbf{k} $ and $ {{\mathbf{n}}_2}=\mathbf{i}+4\mathbf{j}-2\mathbf{k} $ . Hence it is parallel to the vector $ {{\mathbf{n}}_1}\times {{\mathbf{n}}_2}=-2\mathbf{i}+7\mathbf{j}+13\mathbf{k}. $ Thus, we have to find the equation of the plane passing through $ \mathbf{a}=\mathbf{i}+2\mathbf{j}-\mathbf{k} $ and normal to the vector $ \mathbf{n}={{\mathbf{n}}_1}\times {{\mathbf{n}}_2} $ . The equation of the required plane is $ (\mathbf{r}-\mathbf{a}).\ \mathbf{n}=0 $ or $ \mathbf{r}.,\mathbf{n}=\mathbf{a}.\mathbf{n} $ or $ \mathbf{r}.(-2\mathbf{i}+7\mathbf{j}+13\mathbf{k}) $ = $ (\mathbf{i}+2\mathbf{j}-\mathbf{k}).(-2\mathbf{i}+7\mathbf{j}+13\mathbf{k}) $ or $ \mathbf{r}.(2\mathbf{i}-7\mathbf{j}-13\mathbf{k})=1 $ .
BETA
