If two vectors vec{A} = 4hat{i} + nhat{j} + 2 | vedclass

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(C) Two vectors $\vec{A}$ and $\vec{B}$ are mutually perpendicular if their dot product is zero, i.e.,$\vec{A} \cdot \vec{B} = 0$.Given $\vec{A} = 4\h

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$-3$

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(C) Two vectors $\vec{A}$ and $\vec{B}$ are mutually perpendicular if their dot product is zero, i.e.,$\vec{A} \cdot \vec{B} = 0$.Given $\vec{A} = 4\hat{i} + n\hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} + 2\hat{j} - \hat{k}$.Calculating the dot product:$(4\hat{i} + n\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 0$$(4)(2) + (n)(2) + (2)(-1) = 0$$8 + 2n - 2 = 0$$6 + 2n = 0$$2n = -6$$n = -3$Therefore, the value of '$n$' is $-3$.

If two vectors $\vec{A} = 4\hat{i} + n\hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} + 2\hat{j} - \hat{k}$ are mutually perpendicular to each other, then the value of '$n$' is:

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