Q. The plane through the point $( - 1, - 1, - 1)$ and containing the line of intersection of the planes $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} \right) = 0$ and $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) = 0$ is
A) $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}} \right) = 0$
B) $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 0$
C) $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}} \right) = 0$
D) $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} \right) = 0$
Explanation
(a)
The cartesian form of an equation of planes are
$x + 3y - z = 0\ and\ y + 2z = 0$
The line of intersection of two planes is
$(x + 3y - z) + \lambda(y + 2z) = 0$ ….(i)
Since, it is passing through $( - 1,\ - 1,\ - 1)$
$\therefore( - 1 - 3 + 1) + \lambda( - 1 - 2) = 0$
$\Longrightarrow \ \lambda = - 1$
On putting the value of λ in Eq. (i), we get
$x + 2y - 3z = 0$
Hence, vector equation of plane is
$\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}} \right) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.