When vector overrightarrow{A} = 2hat{i} + 3ha | vedclass

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(B) Given that $\vec{B} - \overrightarrow{A} = 2\hat{j}$.Substituting the value of $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ into the equa

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$\sqrt{33}$

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(B) Given that $\vec{B} - \overrightarrow{A} = 2\hat{j}$.Substituting the value of $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ into the equation:$\vec{B} - (2\hat{i} + 3\hat{j} + 2\hat{k}) = 2\hat{j}$$\vec{B} = 2\hat{j} + (2\hat{i} + 3\hat{j} + 2\hat{k})$$\vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}$The magnitude of vector $\vec{B}$ is given by $|\vec{B}| = \sqrt{x^2 + y^2 + z^2}$.$|\vec{B}| = \sqrt{2^2 + 5^2 + 2^2}$$|\vec{B}| = \sqrt{4 + 25 + 4}$$|\vec{B}| = \sqrt{33}$

When vector $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ 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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