If vec{A} = 2hat{i} + 3hat{j} + 8hat{k} is pe | vedclass

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(B) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.Given $\vec{A} = 2\hat{i} + 3\

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$-1/2$

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(B) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.Given $\vec{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\vec{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.Calculating the dot product: $(2\hat{i} + 3\hat{j} + 8\hat{k}) \cdot (-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$.$(2)(-4) + (3)(4) + (8)(\alpha) = 0$.$-8 + 12 + 8\alpha = 0$.$4 + 8\alpha = 0$.$8\alpha = -4$.$\alpha = -4/8 = -1/2$.

If $\vec{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ is perpendicular to $\vec{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$,then the value of $\alpha$ is:

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