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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:
$
\begin{array}{l}
\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})
\end{array}
$
✓

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
$
\begin{array}{l}
\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} \text { and } \overrightarrow{b_2}=4 \hat{i}+6 \hat{j}+8 \hat{k} \\
\quad \therefore \overrightarrow{b_2}=\vec{b}
\end{array}
$
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 $\vec{b} \times\left(a_2-a_1\right)$,
$
\begin{aligned}
\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}
\end{aligned}
$
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}
$
Ans.

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