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

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

Answer 1

Answer: (C)

Two vectors must be perpendicular if their dot product is zero.
Let $\vec{a}=2\hat{i}+3\hat{j}+8k$
$\vec{b}=4\hat{j}-4\hat{i}+\alpha \hat{k}$
$=-4\hat{i}+4\hat{j}+\alpha \hat{k}$
According to the above hypothesis
$\vec{a}\perp \vec{b}$
$\Rightarrow \vec{a}\cdot \vec{b}=0$
$\Rightarrow (2\hat{i}+3\hat{j}+8\hat{k}).(-4\hat{i}+4\hat{j}+\alpha \hat{k})=0$
$\Rightarrow -8+12+8\alpha =0$
$\Rightarrow 8\alpha =-4$
$\therefore \alpha =-\frac{4}{8}=-\frac{1}{2}$
NOTE: $\vec{a}\cdot \vec{b}=ab\cos \theta$. Here, $a$ and $b$ are always positive as they are the magnitudes $a\vec{a}$ and $\vec{b}$.