Question

Solve :
$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-1}{1};\frac{x}{2}=\frac{z+3}{3},y=-1$

Answer 1

$L_1$

$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-1}{1}$ and

$L_2$

$\frac{x}{2}=\frac{z+3}{3},y=-1$

intersection of $L_1$ and $L_2$

$\frac{x-3}{1}=\frac{y+z}{-1}=\frac{z-1}{1}=t$

$x=t+3,y=-t+z,z=t+1$

point on $L_1$ can be represented as $(t+3,-t-2,t+1)$

$\frac{x}{2}=\frac{z+3}{3}=p$ $y=-1$

$x=2p$ $z=3p-3$

point on $L_2$ can be represented as $(2p,-1,3p-3)$

for there to be an intersection point.

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

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

$\Rightarrow p=1$

intersection point $(x,y,z)$

$=(t+3,-t-2,t+1)$

$=(-1+3,1-2,-1+1)$

$intersectionpoint=(2,-1,0)$