Question

The vector equation of the plane passing through the origin and the line of intersection of the planes $\overrightarrow{r}\cdot \overrightarrow{a}=\lambda$ and $\overrightarrow{r}\cdot \overrightarrow{b}=\mu$ is-

• A. $\overrightarrow{r}\cdot (\lambda \overrightarrow{a}-\mu \overrightarrow{b})=0$
• B. $\overrightarrow{r}\cdot (\lambda \overrightarrow{b}-\mu \overrightarrow{a})=0$
• C. $\overrightarrow{r}\cdot (\lambda \overrightarrow{a}+\mu \overrightarrow{b})=0$
• D. $\overrightarrow{r}\cdot (\lambda \overrightarrow{b}+\mu \overrightarrow{a})=0$

Answer 1

Answer: (B)

$\overrightarrow{r}\cdot (\lambda \overrightarrow{b}-\mu \overrightarrow{a})=0$

Equation of plane passing through line of intersection of plane $\overrightarrow{r}\cdot \overrightarrow{a}=\lambda$ and $\overrightarrow{r}\cdot \overrightarrow{b}=\mu$
is given by,
$(\overrightarrow{r}\cdot \overrightarrow{a}-\lambda )+k(\overrightarrow{r}\cdot \overrightarrow{b}-\mu )=0$, where $k$ is any constant
Also given this plane is passing through origin,
$\Rightarrow (-\lambda )+k(-\mu )=0\Rightarrow k=\frac{-\lambda }{\mu }$
So our plane becomes,
$(\overrightarrow{r}\cdot \overrightarrow{a}-\lambda )-\frac{\lambda }{\mu }(\overrightarrow{r}\cdot \overrightarrow{b}-\mu )=0$
$\Rightarrow (\mu \overrightarrow{r}\cdot \overrightarrow{a})-\lambda (\overrightarrow{r}\cdot \overrightarrow{b})=0$
$\Rightarrow \overrightarrow{r}\cdot (\mu \overrightarrow{a}-\lambda \overrightarrow{b})=0$

$\Rightarrow \overrightarrow{r}\cdot (\lambda \overrightarrow{b}-\mu \overrightarrow{a})=0$