Q. If $\theta$ be the angle between the vectors $\overrightarrow{\mathbf{a}} = 2\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} - \widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}} = 6\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + 2\widehat{\mathbf{k}}$, then

A) $\cos{\theta = \frac{4}{21}}$
B) $\cos{\theta = \frac{3}{19}}$
C) $\cos{\theta = \frac{2}{19}}$
D) $\cos{\theta = \frac{5}{21}}$
Explanation

(a)
We know, if $\theta$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, then
$\cos{\theta = \frac{\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}}{\left| \overrightarrow{\mathbf{a}} \right|\mathbf{|}\overrightarrow{\mathbf{b}}\mathbf{|}\ }}$
$= \frac{\left( 2\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} - \widehat{\mathbf{k}} \right) \bullet (6\widehat{\mathbf{i}} - 3\widehat{\mathbf{j}} + 2\widehat{\mathbf{k}})}{\sqrt{2^{2} + 2^{2} + ( - 1)^{2}}\sqrt{6^{3} + ( - 3)^{2} + 2^{2}}}$
$= \frac{12 - 6 - 2}{\sqrt{4 + 4 + 1}\sqrt{36 + 9 + 4}}$
$= \frac{4}{\sqrt{9}\sqrt{49}} = \frac{4}{21}$

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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