The angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\frac{5\pi}{6}$ and the projection of $\overrightarrow{\mathbf{a}}$ in the direction of $\overrightarrow{\mathbf{b}}$ is $\frac{- 6}{\sqrt{3}}$ then $|\overrightarrow{\mathbf{a}}|$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\because\ \ \ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \left| \overrightarrow{\mathbf{a}} ight||\overrightarrow{\mathbf{b}}|\ cos\ \frac{5\pi}{6}$
$= - \frac{\left| \overrightarrow{\mathbf{a}} ight||\overrightarrow{\mathbf{b}}|\sqrt{3}}{2}$
Since, the projection of $\overrightarrow{\mathbf{a}}$ in the direction of
$\overrightarrow{\mathbf{b}} = - \frac{6}{\sqrt{3}}$
$\Longrightarrow - \frac{\left| \overrightarrow{\mathbf{a}} ight|\left| \overrightarrow{\mathbf{b}} ight|\sqrt{3}}{2\left| \overrightarrow{\mathbf{b}} ight|} = - \frac{6}{\sqrt{3}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} ight| = \frac{6 	imes 2}{3} = 4$

Q. The angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\frac{5\pi}{6}$ and the projection of $\overrightarrow{\mathbf{a}}$ in the direction of $\overrightarrow{\mathbf{b}}$ is $\frac{- 6}{\sqrt{3}}$ then $|\overrightarrow{\mathbf{a}}|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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