Q99 / 160 · JIPMER 2007 · Physics
If a vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$ is perpendicular to the vector $4 \hat{\mathbf{j}}-4 \hat{\mathbf{i}}+\alpha \hat{\mathbf{k}}$, then the value of $\alpha$ is
- A$-1$
- B$\frac{1}{2}$
- Correct answerC$-\frac{1}{2}$
- D$1$
Solution
Two vectors must be perpendicular if their dot product is zero.
Let,
$\begin{aligned} \overrightarrow{\mathbf{a}} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} \\ \overrightarrow{\mathbf{b}} & =4 \hat{\mathbf{j}}-4 \hat{\mathbf{i}}+\alpha \hat{\mathbf{k}} \\ & =-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}\end{aligned}$
According to the above hypothesis $\overrightarrow{\mathbf{a}} \perp \overrightarrow{\mathbf{b}}$
$\Rightarrow\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0$
$\begin{array}{lc}\Rightarrow & (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \cdot(-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=0 \\ \Rightarrow & -8+12+8 \alpha=0 \\ \Rightarrow & 8 \alpha=-4 \\ \therefore & \alpha=-\frac{4}{8}=-\frac{1}{2}\end{array}$
NOTE: $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=a b \cos \theta$. Here, $a$ and $b$ are always positive as they are the magnitudes $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$.
Hence, the correct answer is option: 'C'.