Question

The equation of the plane containing the lines $\mathbf{r} = \mathbf{a}_{1} + \lambda\mathbf{a}_{2}$ and $\mathbf{r} = \mathbf{a}_{2} + \lambda\mathbf{a}_{1}$ is

• A. $\lbrack\mathbf{r}\mspace{6mu}\mathbf{a}_{1}\mspace{6mu}\mathbf{a}_{2}\rbrack = 0$
• B. $\lbrack\mathbf{r}\mspace{6mu}\mathbf{a}_{1}\mspace{6mu}\mathbf{a}_{2}\rbrack = \mathbf{a}_{1}.\mspace{6mu}\mathbf{a}_{2}$
• C. $\lbrack\mathbf{r}\mspace{6mu}\mathbf{a}_{2}\mspace{6mu}\mathbf{a}_{1}\rbrack = \mathbf{a}_{1}.\mspace{6mu}\mathbf{a}_{2}$
• D. None of these

Answer 1

Answer: (A)

$\lbrack\mathbf{r}\mspace{6mu}\mathbf{a}_{1}\mspace{6mu}\mathbf{a}_{2}\rbrack = 0$

Answer 2

The required plane passes through a point having position vector $\mathbf{a}_{1}$ and is parallel to the vectors $\mathbf{a}_{1}$ and $\mathbf{a}_{2}$. If $\mathbf{r}$ is the position vector of any point on the plane, then $\mathbf{r} - \mathbf{a}_{1},\mathbf{a}_{1},\mathbf{a}_{2}$ are coplanar.

Therefore, $(\mathbf{r} - \mathbf{a}_{1}).(\mathbf{a}_{1} \times \mathbf{a}_{2}) = 0$

⇒ $\lbrack\mathbf{r}\mathbf{a}_{1}\mathbf{a}_{2}\rbrack$ =$\lbrack\mathbf{a}_{1}\mathbf{a}_{1}\mathbf{a}_{2}\rbrack \Rightarrow \lbrack\mathbf{r}\mathbf{a}_{1}\mathbf{a}_{2}\rbrack = 0$

Hence, the required plane is $\lbrack\mathbf{r}\mathbf{a}_{1}\mathbf{a}_{2}\rbrack = 0$.