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(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Consider the expression $|a - b|^2$: $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)

Vedclass · Accepted Answer

$\frac{1}{2}|a-b|$

Vedclass · Answer

(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Consider the expression $|a - b|^2$: $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$ Since $a \cdot b = |a||b| \cos 	heta = 1 	imes 1 	imes \cos 	heta = \cos 	heta$,we have: $|a - b|^2 = 1^2 + 1^2 - 2 \cos 	heta$ $|a - b|^2 = 2 - 2 \cos 	heta$ $|a - b|^2 = 2(1 - \cos 	heta)$ Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$: $|a - b|^2 = 2(2 \sin^2 \frac{	heta}{2})$ $|a - b|^2 = 4 \sin^2 \frac{	heta}{2}$ Taking the square root on both sides: $|a - b| = 2 \sin \frac{	heta}{2}$ Therefore,$\sin \frac{	heta}{2} = \frac{1}{2} |a - b|$.

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

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