Find the angle in degrees between the vector | vedclass

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(B) Given vector $\vec{A} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.The unit vector along the $Z$-axis is $\hat{k}$,so $\vec{B} = 0\hat{i} + 0\hat{j} +

Vedclass · Accepted Answer

$45$

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(B) Given vector $\vec{A} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.The unit vector along the $Z$-axis is $\hat{k}$,so $\vec{B} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.The angle $	heta$ between two vectors is given by the dot product formula: $\cos 	heta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.First,calculate the dot product: $\vec{A} \cdot \vec{B} = (1)(0) + (1)(0) + (\sqrt{2})(1) = \sqrt{2}$.Next,calculate the magnitude of $\vec{A}$: $|\vec{A}| = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.The magnitude of $\vec{B}$ is $|\vec{B}| = \sqrt{0^2 + 0^2 + 1^2} = 1$.Substitute these values into the formula: $\cos 	heta = \frac{\sqrt{2}}{2 	imes 1} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \cos^{-1}(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

Find the angle in degrees between the vector $\vec{A} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$ and the $Z$-axis.

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