If $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ are the unit vectors such that $\overrightarrow{\mathbf{a}}$ is perpendicular to the plane $\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ and the angle between $\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ is $\frac{\pi}{3}$ then $|\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{c}} = 0$
Now, $|\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^{2} = |{\overrightarrow{\mathbf{a}}|}^{2} + \left| \overrightarrow{\mathbf{b}} ight|^{2} + \left| \overrightarrow{\mathbf{c}} ight|^{2} + 2(\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}})$
$= {(1)}^{2} + {(1)}^{2} + {(1)}^{2} + 2\left( 0 + \left| \overrightarrow{\mathbf{b}} ight|\left| \overrightarrow{\mathbf{c}} ight|cos\ \frac{\pi}{3} + 0 ight)$
$= 3 + 2 	imes 1 	imes 1 	imes \frac{1}{2} = 4$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} ight| = \pm 2$

Explanation

Reviewed by Dr. Abubacker

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