Q54 / 160 · TS EAMCET 2015 · Mathematics
The shortest distance between the skew-lines $\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}, \frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}$ is
- A6
- B7
- C$3 \sqrt{3}$
- Correct answerD$\sqrt{35}$
Solution
Equations of given lines in vector form can be written as
$$
\begin{aligned}
L_1: \mathbf{r}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}+\lambda(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
\text { and } L_2: \mathbf{r}=\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}+\mu(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
\quad \text { Here, } \\
\mathbf{a}_1=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{a}_2=\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\
\mathbf{b}_1=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}_2=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}
$$
Here,
Now, $\mathbf{a}_2-\mathbf{a}_1=-2 \hat{\mathbf{i}}-11 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}$
Also, $\quad \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 2 & 1 \\ 1 & 3 & 2\end{array}\right|=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$
Now, $\quad\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{1+9+25}=\sqrt{35}$
$\therefore$ Distance between the above two skew-lines
$$
\begin{array}{r}
=\left|\frac{\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|}\right| \\
=\frac{(-2 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})}{\sqrt{35}}
\end{array}
$$
$=\left|\frac{-2-33+0}{\sqrt{35}}\right|=\sqrt{35}$
Hence, the correct answer is option: 'D'.