The angle made by a vector vec{B} = 3 hat{i} | vedclass

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(D) Given vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.The unit vector along the $y$-axis is $\hat{j} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$.The

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$\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$

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(D) Given vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.The unit vector along the $y$-axis is $\hat{j} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$.The angle $	heta$ between a vector $\vec{B}$ and the $y$-axis is given by the formula $\cos 	heta = \frac{\vec{B} \cdot \hat{j}}{|\vec{B}| |\hat{j}|}$.First,calculate the magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$.The magnitude of $\hat{j}$ is $1$.The dot product $\vec{B} \cdot \hat{j} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} + 0 \hat{k}) = 2$.Therefore,$\cos 	heta = \frac{2}{\sqrt{29} 	imes 1} = \frac{2}{\sqrt{29}}$.Thus,$	heta = \cos^{-1}\left(\frac{2}{\sqrt{29}}ight)$.

The angle made by a vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ with the $y$-axis is:

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