Question

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

• A. $\frac{1}{2}$
• B. $- \frac{1}{2}$
• C. 1
• D. -1

Answer 1

Answer: (A)

$\frac{1}{2}$

Answer 2

Given,
$A =2 \hat{ i }+3 \hat{ j }+8 \hat{ k }$
$B =4 \hat{ j }-4 \hat{ i }+ \alpha \hat{ k }$
$=-4 \hat{ i }+4 \hat{ j }+\alpha \hat{ k }$
$A \perp B$ Hence, $A \cdot B =0$
$\Rightarrow (2 \hat{ i }+3 \hat{ j }+8 \hat{ k }) \cdot(-4 \hat{ i }+4 \hat{ j }+\alpha \hat{ k })=0$
$\Rightarrow -8+12+8 \alpha=0$
$\Rightarrow 8 \alpha+4=0$
$\Rightarrow 8 \alpha=-4$
$\alpha =-\frac{4}{8}=-\frac{1}{2}$