Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two unit vectors such that angle between them is $60{^\circ}$. Then, $|\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}|$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\left| \overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} ight|^{2} = \left| \overrightarrow{\mathbf{a}}\left. \  ight|^{2} + ight|\overrightarrow{\mathbf{b}}\left. \  ight|^{2} - 2\ \left| \overrightarrow{\mathbf{a}} ight|\left| \overrightarrow{\mathbf{b}} ight|\cos	heta$
$\Rightarrow \ \left| \overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} ight|^{2} = 1 + 1 - 2\cos{60{^\circ} = 2 - 1}$
$\Rightarrow |\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}| = 1$

Q. Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two unit vectors such that angle between them is $60{^\circ}$. Then, $|\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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