A vector vec{A} when added to the sum of the | vedclass

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(B) Let the given vectors be $\vec{B} = (\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{C} = (-2\hat{i} + \hat{j} - \hat{k})$.Their sum is $\vec{S} = \vec{

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$\sqrt{6}$

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(B) Let the given vectors be $\vec{B} = (\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{C} = (-2\hat{i} + \hat{j} - \hat{k})$.Their sum is $\vec{S} = \vec{B} + \vec{C} = (1-2)\hat{i} + (-2+1)\hat{j} + (2-1)\hat{k} = -\hat{i} - \hat{j} + \hat{k}$.The unit vector along the $Y$-axis is $\hat{j}$.According to the problem, $\vec{A} + \vec{S} = \hat{j}$.Therefore, $\vec{A} = \hat{j} - \vec{S} = \hat{j} - (-\hat{i} - \hat{j} + \hat{k}) = \hat{i} + 2\hat{j} - \hat{k}$.The magnitude of vector $\vec{A}$ is $|\vec{A}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.

$A$ vector $\vec{A}$ when added to the sum of the vectors $(\hat{i} - 2\hat{j} + 2\hat{k})$ and $(-2\hat{i} + \hat{j} - \hat{k})$ gives a unit vector along the $Y$-axis. The magnitude of vector $\vec{A}$ is

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