Mathematics › Vectors › Product of 2 vectors
Paragraph: Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2:…
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$\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}}{\sqrt{99}}$ $\frac{-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$ $\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$ $\frac{7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{99}}$ 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}
$
Asked in: JEE Advanced 2008 (Paper 2)
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