The angles which the vector vec{A} = 3hat{i} | vedclass

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(A) The magnitude of the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ is given by $|\vec{A}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt

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$\cos^{-1} \frac{3}{7}, \cos^{-1} \frac{6}{7}, 	ext{ and } \cos^{-1} \frac{2}{7}$

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(A) The magnitude of the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ is given by $|\vec{A}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.Let $\alpha, \beta, 	ext{ and } \gamma$ be the angles that the vector makes with the $x, y, 	ext{ and } z$ axes respectively.The direction cosines are given by $\cos \alpha = \frac{A_x}{|\vec{A}|}, \cos \beta = \frac{A_y}{|\vec{A}|}, 	ext{ and } \cos \gamma = \frac{A_z}{|\vec{A}|}$.Substituting the values: $\cos \alpha = \frac{3}{7}, \cos \beta = \frac{6}{7}, 	ext{ and } \cos \gamma = \frac{2}{7}$.Therefore,the angles are $\alpha = \cos^{-1}(\frac{3}{7}), \beta = \cos^{-1}(\frac{6}{7}), 	ext{ and } \gamma = \cos^{-1}(\frac{2}{7})$.

The angles which the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ makes with the coordinate axes are

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