The angle between the vectors (hat{i} + hat{j | vedclass

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(C) Let $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.The dot product is given by $\vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.The dot product is given by $\vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = (1)(0) + (1)(1) + (0)(1) = 1$.The magnitudes are $|\vec{A}| = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $|\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.The angle $	heta$ between the vectors is given by $\cos 	heta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.Substituting the values,$\cos 	heta = \frac{1}{\sqrt{2} 	imes \sqrt{2}} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

The angle between the vectors $(\hat{i} + \hat{j})$ and $(\hat{j} + \hat{k})$ is ....... $^\circ$.

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