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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 following lines : $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+\mu(4 \hat{i}+6 \hat{j}+8 \hat{k})$
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Answer

Comparing the equations of the given line with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$, we get
$\overrightarrow{a_1}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+4 \hat{k}$
$\overrightarrow{a_2}=2 \hat{i}+4 \hat{j}+5 \hat{k} $ and $ \overrightarrow{b_2}=4 \hat{i}+6 \hat{j}+8 \hat{k}$
$ \therefore \overrightarrow{b_2}=\vec{b}$
So $\overrightarrow{a_2}-\vec{a}_1=2 \hat{i}+4 \hat{j}+5 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k}=\hat{i}+2 \hat{j}+2 \hat{k}$
Here both the lines are parallel to each other.
Hence the distance between the parallel lines
$d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|$
Now finding the value of $\vec{b} \times\left(a_2-a_1\right)$,
$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 2 & 2 \end{array}\right| $
$ =(6-8) \hat{i}-(4-4) \hat{j}+(4-3) \hat{k}$
$ =-2 \hat{i}+\hat{k}$
$\therefore\left|\vec{b} \times\left(\vec{a}_2-\overrightarrow{a_1}\right)\right| $
$=\sqrt{(-2)^2+(1)^2}=\sqrt{5}$
$|\vec{b}| =\sqrt{(2)^3+(3)^2+(4)^2}$
​​​​​​​$=\sqrt{4+9+16}$
$ =\sqrt{29}$
Hence the distance between the parallel lines.
$\begin{array}{l} d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\ d=\frac{\sqrt{5}}{\sqrt{29}} \\ d=\frac{\sqrt{5}}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}}=\frac{\sqrt{145}}{29} \end{array}$

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