The magnitude of the projection of the vector | vedclass

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(B) Let the unit vector along the line be $\hat{v}$. Since the line makes equal angles $\alpha$ with the coordinate axes,its direction cosines are $\c

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$\sqrt{3}$

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(B) Let the unit vector along the line be $\hat{v}$. Since the line makes equal angles $\alpha$ with the coordinate axes,its direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.We know that $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,which implies $3\cos^2 \alpha = 1$,so $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.Thus,the unit vector along the line is $\hat{v} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.The projection of vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line is given by the magnitude of the dot product $\vec{a} \cdot \hat{v}$.$|\vec{a} \cdot \hat{v}| = |(4\hat{i} - 3\hat{j} + 2\hat{k}) \cdot \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})|$$= |\pm \frac{1}{\sqrt{3}}(4 - 3 + 2)| = |\pm \frac{3}{\sqrt{3}}| = \sqrt{3}$.

The magnitude of the projection of the vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line which makes equal angles with the coordinate axes is

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