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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
A
$-1$
B
$-\dfrac{1}{2}$
C
$\dfrac{1}{2}$
D
1
Detailed Solution
Two vectors are perpendicular when their scalar (dot) product is zero.
$(2\hat{i} + 3\hat{j} + 8\hat{k})\cdot(-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$
$2 \times (-4) + 3 \times 4 + 8 \times \alpha = 0$
$-8 + 12 + 8\alpha = 0$
$8\alpha = -4$
$\alpha = -\dfrac{1}{2}$
$(2\hat{i} + 3\hat{j} + 8\hat{k})\cdot(-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$
$2 \times (-4) + 3 \times 4 + 8 \times \alpha = 0$
$-8 + 12 + 8\alpha = 0$
$8\alpha = -4$
$\alpha = -\dfrac{1}{2}$