Q. If the vectors $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}$ are collinear and $\left| \overrightarrow{\mathbf{b}} \right| = 21,\ then\ \overrightarrow{\mathbf{b}}$ is equal to
A) $\pm (2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}})$
B) $\pm 3(2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}})$
C) $(\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}})$
D) $\pm 21(2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}})$
Explanation
(b)
Since, $\overrightarrow{\mathbf{a}}\ and\overrightarrow{\mathbf{b}}$ are collinear.
$\therefore\ \overrightarrow{\mathbf{b}} = m\overrightarrow{\mathbf{a}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{b}} \right| = m\left| \overrightarrow{\mathbf{a}} \right|$
$\Longrightarrow \left| \overrightarrow{\mathbf{b}} \right| = m\sqrt{4 + 9 + 36} = \pm 7m$
$\Longrightarrow 21 = \pm 7m \Longrightarrow m = \pm 3$
$\therefore\ \overrightarrow{\mathbf{b}} = \pm 3\overrightarrow{\mathbf{a}} = \pm (2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.