If overrightarrow A = 2hat i + 4hat j - 5hat | vedclass

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Question

(A) Given vector $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$.First,calculate the magnitude of the vector $\overrightarrow A$:$|\overrightarrow A

Vedclass · Accepted Answer

$\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}, 	ext{and } \frac{-5}{\sqrt{45}}$

Vedclass · Answer

(A) Given vector $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$.First,calculate the magnitude of the vector $\overrightarrow A$:$|\overrightarrow A| = \sqrt{(2)^2 + (4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45}$.The direction cosines $(l, m, n)$ are given by the ratios of the components to the magnitude:$l = \cos \alpha = \frac{A_x}{|\overrightarrow A|} = \frac{2}{\sqrt{45}}$$m = \cos \beta = \frac{A_y}{|\overrightarrow A|} = \frac{4}{\sqrt{45}}$$n = \cos \gamma = \frac{A_z}{|\overrightarrow A|} = \frac{-5}{\sqrt{45}}$Thus,the direction cosines are $\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}, 	ext{and } \frac{-5}{\sqrt{45}}$.

If $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$,the direction cosines of the vector $\overrightarrow A$ are:

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