Question

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 $-\frac{1}{x}$. Find $x$.

Answer 1

Given,
$\overrightarrow{A}=2\hat{i}+3\hat{j}+8\hat{k}$
$\overrightarrow{B}=4\hat{j}-4\hat{i}+\alpha \hat{k}$
Dot product of these vectors should be equal to zero because they are perpendicular.
$\therefore \overrightarrow{A}.\overrightarrow{B}=-8+12+8\alpha =0$
$\Rightarrow 8\alpha =-4$
$\Rightarrow \alpha =-\frac{1}{2}$
Final Answer: $\left(2\right)$