Q. Suppose $\overrightarrow{\mathbf{a}} = \lambda\ \widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$, $\overrightarrow{\mathbf{b}} = \lambda\ \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\lambda\ \widehat{\mathbf{k}}.$ If the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is grater than $90{^\circ}$, then λ satisfies the inequality
A) $- 7 < \lambda < 1$
B) $\lambda > 1$
C) $1 < \lambda < 7$
D) $- 5 < \lambda < 1$
Explanation
(a)
Given, $\overrightarrow{\mathbf{a}} = \lambda\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}},\ \overrightarrow{\mathbf{b}} = \lambda\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\lambda\widehat{\mathbf{k}}$
$\therefore\cos{\theta = \frac{\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}}{\left| \overrightarrow{\mathbf{a}} \right|\left| \overrightarrow{\mathbf{b}} \right|}}$
$= \frac{\lambda^{2} - 7 + 6\lambda}{\sqrt{\lambda^{2} + 49 + 9}\sqrt{\lambda^{2} + 1 + 4\lambda^{2}}} < 0$
$\Longrightarrow (\lambda + 7)(\lambda - 1) < 0$
$\Longrightarrow \ - 7 < \lambda < 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.