When vector $\vec{A} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ is subtracted from vector $\vec{B}$ , it gives a vector equal to $2 \hat{j}$ . Then the magnitude of vector $\vec{B}$ will be:

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$\sqrt{33}$

$\sqrt{5}$

$\sqrt{6}$

answer is B.

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Detailed Solution

Let $\vec{B} = x \hat{i} + y \hat{j} + z \hat{k}$
Then $\vec{B} - \vec{A} = (x - 2) \hat{i} + (y - 3) \hat{j} + (z - 2) \hat{k}$
On solving, $\vec{B} = 2 \hat{i} + 5 \hat{j} + 2 \hat{k}$

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