Let the line L pass through (1,1,1) and intersect the lines (x-1)/(2)=(y+1)/(3)=(z-1)/(4) …

Question

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 $L$ ?

Accepted Answer

$$\begin{aligned} & L: \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c} \ & L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \text { (say) } \end{aligned}$$ Any point on $L_1$ be $A(2 \lambda+1,3 \lambda+1,4 \lambda+1)$ $$L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu \text { (say) }$$ Any point on $L_2$ be $B(\mu+3,2 \mu-4, \mu)$ $D$ of $L$ be : $<2 \lambda, 3 \lambda-2,4 \lambda>$ or $<\mu+2,2 \mu+3$, $\mu-1>$ Now $\frac{2 \lambda}{\mu+2}=\frac{3 \lambda-7}{2 \mu+3}=\frac{4 \lambda}{\mu-1}$ $$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda=\frac{-6}{5} \quad \mu=-5 \ & \therefore< a, b, c >\equiv\langle-3,-7,-6 >\text { or }<3,7,6> \ & \therefore \quad L: \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6} \end{aligned}\ \&(7,15,13) \text { lies on the line. } \end{aligned}$$

Solutions · 1