Q. If the planes $\overrightarrow{\mathbf{r}} \bullet \left( 2\widehat{\dot{\mathbf{i}}} - \lambda\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} \right) = 0$ and $\overrightarrow{\mathbf{r}} \bullet \left( \lambda\ \widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} \right) = 5$ are perpendicular to each other, then the value of $\lambda^{2} + \lambda$ is
A) 0
B) 2
C) 1
D) 3
Explanation
(a)
Since, given planes are perpendicular, it means its normal are perpendicular.
$\therefore 2(\lambda) - \lambda(5) + 3( - 1) = 0$
$\Longrightarrow - 3\lambda - 3 = 0$
$\Longrightarrow \ \lambda = - 1$
$\therefore\ \lambda^{2} + \lambda = {( - 1)}^{2} - 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.