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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry3 Marks
Question
Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
✓

Answer

Given lines
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
Comparing equations (1) and (2) with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and
$
\begin{aligned}
\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} & \\
& \overrightarrow{a_1}=\hat{i}+\hat{j}, \overrightarrow{b_1}=2 \hat{i}-\hat{j}+\hat{k} \\
& \overrightarrow{a_2}=2 \hat{i}+\hat{j}-\hat{k} \text { and } \overrightarrow{b_2}=3 \hat{i}-5 \hat{j}+2 \hat{k}
\end{aligned}
$
So, $\quad \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and
$
\begin{aligned}
\overrightarrow{b_1} \times \overrightarrow{b_2} & =(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k}) \\
& =\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|
\end{aligned}
$
$
\begin{array}{l}
=(-2+5) \hat{i}-(4-3) \hat{j}+(-10+3) \hat{k} \\
=3 \hat{i}-\hat{j}-\hat{k}
\end{array}
$
Thus $\quad\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{9+1+49}=\sqrt{59}$
Therefore the shortest distance $
d=\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2} \times \overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|
$
Putting the values
$
\begin{aligned}
d & =\left|\frac{(3 \hat{i}-\hat{j}-7 \hat{k}) \cdot(\hat{i}-\hat{k})}{\sqrt{59}}\right| \\
& =\left|\frac{3-0+7}{\sqrt{59}}\right|=\frac{10}{\sqrt{59}}
\end{aligned}
$

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