Q113 / 160 · TS EAMCET 2014 · Mathematics
The shortest distance between the skew lines $\hat{\mathbf{r}}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+t(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ and $\hat{\mathbf{r}}=(4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ is
- A$sqrt{6}$
- B$3$
- C$2 \sqrt{3}$
- Correct answerD$sqrt{3}$
Solution
Given skew lines are
$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+t(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
and $\mathrm{r}=(4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
Here,
$\begin{aligned}<br />& \mathbf{a}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{j}} \\<br />& \mathrm{b}_1=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\<br />& \mathrm{a}_2=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\<br />& \mathrm{b}_2=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}<br />\end{aligned}$
and
$\mathbf{b}_2=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
Now, 0
$\begin{aligned}<br />\mathbf{a}_2-\mathbf{a}_1 & =4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}-(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\<br />& =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}<br />\end{aligned}$
and
$\begin{aligned}<br />\mathrm{b}_1 \times \mathrm{b}_2 & =\left|\begin{array}{ccc}<br />\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\<br />1 & 3 & 2 \\<br />2 & 3 & 1<br />\end{array}\right| \\<br />& =\hat{\mathbf{i}}(3-6)-\hat{\mathbf{j}}(1-4)+\hat{\mathbf{k}}(3-6) \\<br />& =-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}<br />\end{aligned}$
$\therefore$ The shortest distance between skew lines
$\begin{aligned}<br />& =\frac{\left|\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right|}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} \\<br />& =\frac{|(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})|}{\sqrt{(-3)^2+(3)^2+(-3)^2}} \\<br />& =\frac{|-9+9-9|}{\sqrt{9+9+9}}=\frac{9}{3 \sqrt{3}}=\sqrt{3}<br />\end{aligned}$
$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+t(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
and $\mathrm{r}=(4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
Here,
$\begin{aligned}<br />& \mathbf{a}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{j}} \\<br />& \mathrm{b}_1=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\<br />& \mathrm{a}_2=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\<br />& \mathrm{b}_2=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}<br />\end{aligned}$
and
$\mathbf{b}_2=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
Now, 0
$\begin{aligned}<br />\mathbf{a}_2-\mathbf{a}_1 & =4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}-(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\<br />& =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}<br />\end{aligned}$
and
$\begin{aligned}<br />\mathrm{b}_1 \times \mathrm{b}_2 & =\left|\begin{array}{ccc}<br />\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\<br />1 & 3 & 2 \\<br />2 & 3 & 1<br />\end{array}\right| \\<br />& =\hat{\mathbf{i}}(3-6)-\hat{\mathbf{j}}(1-4)+\hat{\mathbf{k}}(3-6) \\<br />& =-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}<br />\end{aligned}$
$\therefore$ The shortest distance between skew lines
$\begin{aligned}<br />& =\frac{\left|\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right|}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} \\<br />& =\frac{|(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})|}{\sqrt{(-3)^2+(3)^2+(-3)^2}} \\<br />& =\frac{|-9+9-9|}{\sqrt{9+9+9}}=\frac{9}{3 \sqrt{3}}=\sqrt{3}<br />\end{aligned}$
Hence, the correct answer is option: 'D'.