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

Question

Dr. Abubacker · Accepted Answer

(d)
$|\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^{2} = |\overrightarrow{\mathbf{a}}|^{2} + |\overrightarrow{\mathbf{b}}|^{2} - 2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|\cos	heta$
$= 1^{2} + 1^{2}2 \bullet 1 \bullet 1 \bullet \cos{60{^\circ}}\ \ \ \ \ \ \ \lbrack\because\left| \overrightarrow{\mathbf{a}} ight| = |\overrightarrow{\mathbf{b}}| = 1brack$
$= 2 - 2 \bullet \frac{1}{2} = 1$

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

Explanation

Reviewed by Dr. Abubacker

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