The direction cosines of vector (A - B),if A | vedclass

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(A) Given vectors are $A = 2\hat{i} + 3\hat{j} + \hat{k}$ and $B = 2\hat{i} + 2\hat{j} + 3\hat{k}$.First,calculate the vector $C = A - B$:$C = (2 - 2)

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$0, \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}}$

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(A) Given vectors are $A = 2\hat{i} + 3\hat{j} + \hat{k}$ and $B = 2\hat{i} + 2\hat{j} + 3\hat{k}$.First,calculate the vector $C = A - B$:$C = (2 - 2)\hat{i} + (3 - 2)\hat{j} + (1 - 3)\hat{k} = 0\hat{i} + 1\hat{j} - 2\hat{k}$.Next,find the magnitude of vector $C$:$|C| = \sqrt{0^2 + 1^2 + (-2)^2} = \sqrt{0 + 1 + 4} = \sqrt{5}$.The direction cosines $(l, m, n)$ are given by the components of the vector divided by its magnitude:$l = \frac{C_x}{|C|} = \frac{0}{\sqrt{5}} = 0$.$m = \frac{C_y}{|C|} = \frac{1}{\sqrt{5}}$.$n = \frac{C_z}{|C|} = \frac{-2}{\sqrt{5}}$.Thus,the direction cosines are $0, \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}}$.

The direction cosines of vector $(A - B)$,if $A = 2\hat{i} + 3\hat{j} + \hat{k}$ and $B = 2\hat{i} + 2\hat{j} + 3\hat{k}$,are:

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