The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \quad$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$
The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \quad$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$
- intersect each other and point of intersection is $(2,1,3)$
- intersect each other and point of intersection is $(3,2,4)$
- intersect each other and point of intersection is $(-2,3,3)$
- do not intersect.
Solution
The given lines are $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$
$\begin{aligned}
& \text { Here, } \\
& \qquad\left(x_1, y_1, \mathrm{z}_1\right) \equiv(1,-1,1) \\
& \qquad\left(x_2, y_2, \mathrm{z}_2\right) \equiv(-2,1,-1) \\
& \qquad\left(\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1\right) \equiv(3,2,5) \\
& \text { Consider }\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right| \\
& =\left|\begin{array}{ccc}
-3 & 2 & -2 \\
3 & 2 & 5 \\
4 & 3 & 2
\end{array}\right| \\
& =-3(-11)-2(-14)-2(1) \\
& =33+28-2 \\
& =59 \neq 0
\end{aligned}$
$\therefore \quad$ The lines are not intersecting.
Asked in: MHT CET 2023 (11 May Shift 1)
Practice more Line and Plane questions on Aicharya