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

• 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. $\mathbf{r}.(\lambda\mathbf{b} + \mu\mathbf{a}) = 0$

Answer 1

Answer: (B)

$\mathbf{r}.(\lambda\mathbf{b} - \mu\mathbf{a}) = 0$

Answer 2

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 $\mathbf{r}.(\mu\mathbf{a} - \lambda\mathbf{b}) = 0\mspace{6mu} \Rightarrow \mspace{6mu}\mspace{6mu}\mathbf{r}\mspace{6mu}.\mspace{6mu}(\lambda\mathbf{b} - \mu\mathbf{a}) = 0$.