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Three Dimensional Geometry — MATHS STD 12 Science — Question

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

Given lines
$\vec{r} =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})$
$\vec{r} =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$
Comparing equations $(1)$ and $(2)$ with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and
$\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}$
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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