The angle subtended by the vector vec{A} = 4 | vedclass

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(C) The vector is given by $\vec{A} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k}$.The magnitude of the vector $\vec{A}$ is $|\vec{A}| = \sqrt{4^2 + 3^2 + 12^2

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$\cos^{-1}(\frac{4}{13})$

Vedclass · Answer

(C) The vector is given by $\vec{A} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k}$.The magnitude of the vector $\vec{A}$ is $|\vec{A}| = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.The unit vector along the $x$-axis is $\hat{i} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k}$.The angle $	heta$ between the vector $\vec{A}$ and the $x$-axis is given by the dot product formula: $\cos 	heta = \frac{\vec{A} \cdot \hat{i}}{|\vec{A}| |\hat{i}|}$.Calculating the dot product: $\vec{A} \cdot \hat{i} = (4 	imes 1) + (3 	imes 0) + (12 	imes 0) = 4$.Substituting the values: $\cos 	heta = \frac{4}{13 	imes 1} = \frac{4}{13}$.Therefore,$	heta = \cos^{-1}(\frac{4}{13})$.

The angle subtended by the vector $\vec{A} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k}$ with the $x$-axis is

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