Question

If the line, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then find the value of $k$.

Answer 1

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\text{and}\frac{x-3}{1}=\frac{q-k}{2}=\frac{z}{1}$

$\text{Line 1}=t\left(say\right)\text{Line 2}=p\left(say\right)$

$x=2t+1x=p+3$

$y=3t-1y=2p+k$

$z=4t+1z=p$

equating $x\&z$ coordinates

$2t+1=p+3\Rightarrow 2t-p=2\left(1\right)$

$4t+1=p+3\Rightarrow 4t-p=-1\left(2\right)$

Adding $\left(1\right)\&\left(2\right)$ by Multiple (2) by `-1’

$-2t=3\Rightarrow t=-3/2$

$\Rightarrow p=4\times \left(\frac{-3}{2}\right)+1$

$p=-5$

$\therefore y=2p+k$

$\Rightarrow y=-10+k$

$\&y=3t-1\Rightarrow y=\frac{-9}{2}-1=\frac{-11}{2}$

$\Rightarrow -10+k=\frac{-11}{2}$

$\Rightarrow k=\frac{-11}{2}+10$

$k=\frac{+9}{2}$