If the projection of the vector $\overrightarrow{\mathbf{a}}\ on\ \overrightarrow{\mathbf{b}}$ is $|\overrightarrow{\mathbf{a}} imes \overrightarrow{\mathbf{b}}|$ and if $3\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$, then the angle between$\ \overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ is

Question

If the projection of the vector $\overrightarrow{\mathbf{a}}\ on\ \overrightarrow{\mathbf{b}}$ is $|\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}|$ and if $3\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$, then the angle between$\ \overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ is

Dr. Abubacker · Accepted Answer

(a)
Given projection of $\overrightarrow{\mathbf{a}}\ on\ \overrightarrow{\mathbf{b}} = |\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}|$
$\Longrightarrow \ \frac{\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|} = |\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}|$
$\Longrightarrow \ \frac{\left| \overrightarrow{\mathbf{a}} ight||\overrightarrow{\mathbf{b}}|\ cos\ 	heta}{|\overrightarrow{\mathbf{b}}|} = \left| \overrightarrow{\mathbf{a}} ight||\overrightarrow{\mathbf{b}}|\ sin\ 	heta$
$\Longrightarrow 	an	heta = \frac{1}{|\overrightarrow{\mathbf{b}}|}$
$\Longrightarrow 	an	heta = \frac{1}{\frac{1}{3}\sqrt{1^{2} + 1^{2} + 1^{2}}}$
$\Longrightarrow 	an{	heta = \sqrt{3}}$
$\Longrightarrow \ 	heta = \frac{\pi}{3}$

Explanation

Reviewed by Dr. Abubacker

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