If theta is the angle between the unit vector | vedclass

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(B) Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors,so $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$. The dot product $\mathbf{a} \cdot \mathbf{

Vedclass · Accepted Answer

$\frac{1}{2} |\mathbf{a} + \mathbf{b}|$

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(B) Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors,so $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$. The dot product $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 	heta = \cos 	heta$. Consider the magnitude squared of the sum of the vectors: $|\mathbf{a} + \mathbf{b}|^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b})$. Substituting the values: $|\mathbf{a} + \mathbf{b}|^2 = 1^2 + 1^2 + 2 \cos 	heta = 2 + 2 \cos 	heta$. Using the trigonometric identity $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$: $|\mathbf{a} + \mathbf{b}|^2 = 2(1 + \cos 	heta) = 2(2 \cos^2 \frac{	heta}{2}) = 4 \cos^2 \frac{	heta}{2}$. Taking the square root on both sides: $|\mathbf{a} + \mathbf{b}| = 2 \cos \frac{	heta}{2}$. Therefore,$\cos \frac{	heta}{2} = \frac{1}{2} |\mathbf{a} + \mathbf{b}|$.

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

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