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If $\vec{A}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ the direction of cosines of the vector $\vec{A}$ are
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$\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$
$\vec{A}=2 \hat{i}+4 \hat{j}-5 \hat{k} \therefore|\vec{A}|=\sqrt{(2)^2+(4)^2+(-5)^2}=\sqrt{45}$
$\therefore \cos \alpha=\frac{2}{\sqrt{45}}, \quad \cos \beta=\frac{4}{\sqrt{45}}, \quad \cos \gamma=\frac{-5}{\sqrt{45}}$
$\therefore \cos \alpha=\frac{2}{\sqrt{45}}, \quad \cos \beta=\frac{4}{\sqrt{45}}, \quad \cos \gamma=\frac{-5}{\sqrt{45}}$
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