The angles which a vector hat{i} + hat{j} + s | vedclass

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(C) Let the vector be $\vec{R} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.Comparing this with $\vec{R} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k}$,we get

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$60^\circ, 60^\circ, 45^\circ$

Vedclass · Answer

(C) Let the vector be $\vec{R} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.Comparing this with $\vec{R} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k}$,we get $R_x = 1, R_y = 1, R_z = \sqrt{2}$.The magnitude of the vector is $|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2} = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.The direction cosines are given by $\cos \alpha = \frac{R_x}{|\vec{R}|}, \cos \beta = \frac{R_y}{|\vec{R}|}, \cos \gamma = \frac{R_z}{|\vec{R}|}$.For the $X$-axis: $\cos \alpha = \frac{1}{2} \Rightarrow \alpha = 60^\circ$.For the $Y$-axis: $\cos \beta = \frac{1}{2} \Rightarrow \beta = 60^\circ$.For the $Z$-axis: $\cos \gamma = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow \gamma = 45^\circ$.Thus,the angles are $60^\circ, 60^\circ, 45^\circ$.

The angles which a vector $\hat{i} + \hat{j} + \sqrt{2} \hat{k}$ makes with $X, Y$ and $Z$ axes respectively are

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