Three Dimensional Geometry Question 23
Question 23 - 2024 (31 Jan Shift 2)
The shortest distance between lines $L _1$ and $L _2$, 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
$L _2=\frac{x+4}{3}=\frac{y-4}{2}=\frac{z-3}{0}$
$\therefore S . D=\frac{\left|\begin{array}{ccc}x _2-x _1 & y _2-y _1 & z _2-z _1 \\ 2 & -3 & 2 \\ 3 & 2 & 0\end{array}\right|}{\left|\overrightarrow{n _1} \times \overrightarrow{n _2}\right|}$
$=\frac{\left|\begin{array}{ccc}5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0\end{array}\right|}{\left|\overrightarrow{n _1 \times n _2}\right|}$
$=\frac{141}{-4 \hat{i}+6 \hat{j}+13 \hat{k}}$
$=\frac{\text { math } 141}{\sqrt{16+36+169}}$
$=\frac{141}{\sqrt{221}}$
