A vector overrightarrow{A} when added to the | vedclass

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(B) Let the given vectors be $\vec{B} = (\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and $\vec{C} = (-2 \hat{\imath}+\hat{\jmath}-\hat{k})$.The sum of the

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$\sqrt{6}$

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(B) Let the given vectors be $\vec{B} = (\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and $\vec{C} = (-2 \hat{\imath}+\hat{\jmath}-\hat{k})$.The sum of these vectors is $\vec{B} + \vec{C} = (1-2)\hat{\imath} + (-2+1)\hat{\jmath} + (2-1)\hat{k} = -\hat{\imath} - \hat{\jmath} + \hat{k}$.According to the problem,$\vec{A} + (\vec{B} + \vec{C}) = \hat{\jmath}$ (unit vector along the $y$-axis).Substituting the sum: $\vec{A} + (-\hat{\imath} - \hat{\jmath} + \hat{k}) = \hat{\jmath}$.Solving for $\vec{A}$: $\vec{A} = \hat{\jmath} - (-\hat{\imath} - \hat{\jmath} + \hat{k}) = \hat{\jmath} + \hat{\imath} + \hat{\jmath} - \hat{k} = \hat{\imath} + 2\hat{\jmath} - \hat{k}$.The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.

$A$ vector $\overrightarrow{A}$ when added to the sum of the vectors $(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and $(-2 \hat{\imath}+\hat{\jmath}-\hat{k})$ gives a unit vector along the $y$-axis. The magnitude of the vector $\overrightarrow{A}$ is

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