Q39 / 66 · JEE Advanced 2008 (Paper 2) · Mathematics
Paragraph:
Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$Question:
The unit vector perpendicular to both $L_1$ and $L_2$ is
- A$\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}}{\sqrt{99}}$
- Correct answerB$\frac{-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$
- C$\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$
- D$\frac{7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{99}}$
Solution
The equations of given lines in vector form may be written as
$$
L_1: \mathbf{r}(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})
$$
and $L_2: \mathbf{r}=(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$
$\therefore$ The vector perpendicular to both $L_1$ and $L_2$ is
$$
\left|\begin{array}{rrr}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{array}\right|=-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}
$$
$$
\begin{aligned}
\therefore \text { Required unit vector } & =\frac{(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{\sqrt{(-1)^2+(-7)^2+(5)^2}} \\
& =\frac{1}{5 \sqrt{3}}(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})
\end{aligned}
$$
Hence, the correct answer is option: 'B'.