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Physics · MOTION IN A PLANE
हिन्दीA force vector is specified as \( \mathbf{F} = (6\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}})\text{ N} \). What is the cosine of the angle that this force vector makes with the positive y-axis?
Correct answer · Option 2
2/7
Explanation
The magnitude is \( |\mathbf{F}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{49} = 7\text{ N} \). The direction cosine with the y-axis is \( \cos\beta = \frac{F_y}{|\mathbf{F}|} = \frac{2}{7} \).
