If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k=$

Question

Aicharya · Accepted Answer

The coordinates of any point on the first line are given by
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$
i.e $x=2 \lambda+1 y=3 \lambda-1 z=4 \lambda+1$
Thus, the coordinates of any point on this line are $(2 \lambda+1,3 \lambda-1,4 \lambda+1)$.
The coordinates of any point on the second line are given by
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$
i.e.
$\begin{aligned}
& x=\mu+3 \\
& y=2 \mu+k \\
& z=\mu
\end{aligned}$
Thus, the coordinates of any point on this line are $(\mu+3,2 \mu+k, \mu)$.
If these two lines intersect each other, then
$\begin{aligned}
& 2 \lambda+1=\mu+3,3 \lambda-1=2 \mu+k, 4 \lambda+1=\mu \\
& \Rightarrow 2 \lambda-\mu=2,3 \lambda-2 \mu=k+1,4 \lambda-\mu=-1
\end{aligned}$
Solving $2 \lambda-\mu=2$ and $4 \lambda-\mu=-1$, we get
$\lambda=-3 / 2 \text { and } \mu=-5$
By substituting the values $\lambda=-3 / 2$ and $\mu=-5$ in $3 \lambda-2 \mu=k+1$, we get
$k=9/ 2 `$

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$…

Solution