Three Dimensional Geometry Question 23
Question 23 - 2024 (31 Jan Shift 2)
The shortest distance between lines $\mathrm{L}{1}$ and $\mathrm{L}{2}$, where $L_{1}: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_{2}$ is the line passing through the points $A(-4,4,3) \cdot B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is
(1) $\frac{121}{\sqrt{221}}$
(2) $\frac{24}{\sqrt{117}}$
(3) $\frac{141}{\sqrt{221}}$
(4) $\frac{42}{\sqrt{117}}$
Show Answer
Answer (3)
Solution
$L_{2}=\frac{x+4}{3}=\frac{y-4}{2}=\frac{z-3}{0}$
$\therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}\mathrm{x}{2}-\mathrm{x}{1} & \mathrm{y}{2}-\mathrm{y}{1} & \mathrm{z}{2}-\mathrm{z}{1} \ 2 & -3 & 2 \ 3 & 2 & 0\end{array}\right|}{\left|\overrightarrow{\mathrm{n}{1}} \times \overrightarrow{\mathrm{n}{2}}\right|}$
$=\frac{\left|\begin{array}{ccc}5 & -5 & -7 \ 2 & -3 & 2 \ 3 & 2 & 0\end{array}\right|}{\left|\overrightarrow{\mathrm{n}{1} \times \mathrm{n}{2}}\right|}$
$=\frac{141}{-4 \hat{i}+6 \hat{j}+13 \hat{k}}$
$=\frac{\text { math } 141}{\sqrt{16+36+169}}$
$=\frac{141}{\sqrt{221}}$
