If bar{a}=2 hat{i}-hat{j}+hat{k},bar{b}=hat{i | vedclass

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(A) First,calculate the vector $\vec{v} = 3 \bar{a} + \bar{b} - 2 \bar{c}$.$\vec{v} = 3(2 \hat{i} - \hat{j} + \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k

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$\frac{1}{\sqrt{6}}(-\hat{i}+2 \hat{j}-\hat{k})$

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(A) First,calculate the vector $\vec{v} = 3 \bar{a} + \bar{b} - 2 \bar{c}$.$\vec{v} = 3(2 \hat{i} - \hat{j} + \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k}) - 2(4 \hat{i} - 2 \hat{j} + \hat{k})$$\vec{v} = (6 \hat{i} - 3 \hat{j} + 3 \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k}) - (8 \hat{i} - 4 \hat{j} + 2 \hat{k})$$\vec{v} = (6 + 1 - 8) \hat{i} + (-3 + 1 + 4) \hat{j} + (3 - 2 - 2) \hat{k}$$\vec{v} = -\hat{i} + 2 \hat{j} - \hat{k}$Now,find the magnitude of $\vec{v}$:$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$The unit vector in the direction of $\vec{v}$ is given by $\frac{\vec{v}}{|\vec{v}|} = \frac{-\hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}(-\hat{i} + 2 \hat{j} - \hat{k})$.

If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\bar{b}=\hat{i}+\hat{j}-2 \hat{k}$ and $\bar{c}=4 \hat{i}-2 \hat{j}+\hat{k}$,then the unit vector in the direction of $3 \bar{a}+\bar{b}-2 \bar{c}$ is

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