Vector Algebra Question 267
Question: The vector equation of the plane passing through the origin and the line of intersection of the plane $ \mathbf{r}.\mathbf{a}=\lambda $ and $ \mathbf{r}.\mathbf{b}=\mu $ is
Options:
A) $ \mathbf{r}.(\lambda \mathbf{a}-\mu \mathbf{b})=0 $
B) $ \mathbf{r}.,(\lambda \mathbf{b}-\mu \mathbf{a})=0 $
C) $ \mathbf{r}.(\lambda \mathbf{a}+\mu \mathbf{b})=0 $
D) $ \mathbf{r}.(\lambda \mathbf{b}+\mu \mathbf{a})=0 $
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Answer:
Correct Answer: B
Solution:
- The equation of a plane through the line of intersection of the planes $ \mathbf{r}.\mathbf{a}=\lambda $ and $ \mathbf{r}.,\mathbf{b}=\mu $ can be written as $ (\mathbf{r}.\mathbf{a}-\lambda )+k(\mathbf{r}.\mathbf{b}-\mu )=0 $ or $ \mathbf{r}.(\mathbf{a}+k\mathbf{b})=\lambda +k\mu $ …..(i) This passes through the origin, therefore $ \mathbf{0}.(\mathbf{a}+k\mathbf{b})=\lambda +\mu k\Rightarrow k=\frac{-\lambda }{\mu } $ Putting the value of k in (i), we get the equation of the required plane as $ \mathbf{r}.(\mu \mathbf{a}-\lambda \mathbf{b})=0\ \Rightarrow \ \ \mathbf{r}\ .\ (\lambda \mathbf{b}-\mu \mathbf{a})=0 $ .
BETA
