If the vector overrightarrow{A} = 2hat{i} + 4 | vedclass

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Question

(A) The magnitude of the vector $\overrightarrow{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ is given by:$|\overrightarrow{A}| = \sqrt{(2)^2 + (4)^2 + (-5)^2

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

Vedclass · Answer

(A) The magnitude of the vector $\overrightarrow{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ is given by:$|\overrightarrow{A}| = \sqrt{(2)^2 + (4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45}$.The direction cosines $(l, m, n)$ are defined as the ratios of the components of the vector to its 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 the vector $\overrightarrow{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}$,find its direction cosines.

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