If a vector 2hat i + 3hat j + 8hat k is perpe | vedclass

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(C) Let the two vectors be $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\overrightarrow{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.Since

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$-0.5$

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(C) Let the two vectors be $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\overrightarrow{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.Since the vectors are perpendicular,their dot product must be zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.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 = -0.5$.

If a vector $2\hat i + 3\hat j + 8\hat k$ is perpendicular to the vector $4\hat j - 4\hat i + \alpha \hat k$,then the value of $\alpha$ is:

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