Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$ . Then, which of the following points lies on the line L ?

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Dr's of } \mathrm{AC} \Rightarrow 2 \lambda, 3 \lambda-2,4 \lambda \ & 	ext { Dr's of } \mathrm{BC} \Rightarrow \mu+2,2 \mu+3, \mu-1 \ & \Rightarrow \frac{\mu+2}{2 \lambda}=\frac{2 \mu+3}{3 \lambda-2}=\frac{\mu-1}{4 \lambda} \ & \Rightarrow 2(\mu+2)=\mu-1 \Rightarrow \mu=-5 \ & \Rightarrow 	ext { Dr's of } \mathrm{BC} \Rightarrow 3,7,6 \ & \Rightarrow 	ext { equation of } \mathrm{L} \Rightarrow \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-1}{7}=\frac{\mathrm{z}-1}{6} \end{aligned}$ $(7,15,13) 	ext { satisfies. }$ *

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$…

Solution