The line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$
The line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$
- intersect each other and point of intersection is $(4,3,-2)$.
- do not intersect.
- intersect each other and point of intersection is $(3,2,5)$.
- intersect each other and point of intersection is $(-2,-1,-1)$
Solution
$\begin{aligned}
& \text { S.D. }=\left|\frac{\left|\begin{array}{ccc}
1+2 & -1-1 & 1+1 \\
3 & 2 & 5 \\
4 & 3 & -2
\end{array}\right|}{\sqrt{(-4-15)^2+(20+6)^2+(9-8)^2}}\right| \\
& =\left|\frac{3(-4-15)-2(20+6)+2(9-8)}{\sqrt{19^2+26^2+1^2}}\right| \\
& =\frac{|-57-52+2|}{\sqrt{361+676+1}}=\frac{107}{\sqrt{1039}} \neq 0
\end{aligned}$
Hence, the two lines do not intersect each other.
Asked in: MHT CET 2022 (08 Aug Shift 2)
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