Q. If $\overrightarrow{\mathbf{a}}$ is a vector of magnitude 50, collinear with the vector $\overrightarrow{\mathbf{b}} = 6\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}} - \frac{15}{2}\ \widehat{\dot{\mathbf{k}}}$ and makes an acute angle with the positive direction of $z$-axis, then $\overrightarrow{\mathbf{a}}$ is equal to

A) $- 24\widehat{\dot{\mathbf{i}}} + 32\widehat{\dot{\mathbf{j}}} + \ 30\widehat{\dot{\mathbf{k}}}$
B) $24\widehat{\dot{\mathbf{i}}} - 32\widehat{\dot{\mathbf{j}}} - \ 30\widehat{\dot{\mathbf{k}}}$
C) $12\widehat{\dot{\mathbf{i}}} - 16\widehat{\dot{\mathbf{j}}} - 15\widehat{\dot{\mathbf{k}}}$
D) $- 12\widehat{\dot{\mathbf{i}}} + 16\widehat{\dot{\mathbf{j}}} - 15\widehat{\dot{\mathbf{k}}}$
Explanation

(a)
Since, $\overrightarrow{\mathbf{a}} = m\overrightarrow{\mathbf{b}}$ for some scalar $m\ ie,$
$\overrightarrow{\mathbf{a}} = m\left( 6\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}} - \frac{15}{2}\widehat{\mathbf{k}} \right)$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} \right| = |m|\sqrt{36 + 64 + \frac{225}{4}}$
$\Longrightarrow 50 = \frac{25}{2}|m| \Longrightarrow |m| = 4$
$\Longrightarrow m = \pm 4$
Since, $\overrightarrow{\mathbf{a}}$ makes an acute angle with the positive direction of $z$-axis, so its $z$ componant must be positive and hence, $m$ must be $- 4$
$\therefore\ \ \overrightarrow{\mathbf{a}} = - 4\left( 6\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}} - \frac{15}{2}\widehat{\mathbf{k}} \right)$
$= - 24\widehat{\dot{\mathbf{i}}} + 32\widehat{\dot{\mathbf{j}}} + 30\widehat{\mathbf{k}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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