finding the line of intersection

Question

Find the line of intersection between two planes x+y+z=1 and x-2y+3z=1 ?

I found r1,r2,n1 and n2 but I don't know what are the other steps

Answer 1

If you subtract the first equation minus the second, you get $3y-2z=0$, so $3y=2z$ and $y=\frac{2}{3}z$.

Letting $z=3t$ gives $y=2t$, and then substituting in the first equation gives $x=1-5t$.

Therefore the line of intersection has parametric equations $x=1-5t, y=2t, z=3t$.

(You could also work this by taking the cross-product of the normal vectors for the two planes, to find a vector parallel to the line; and then finding a point on the line by letting $z=0$, say, and then solving for x and y.)

Answer 2

Well $(1,1,1)$ and $(1,-2,3)$ are the normals to the planes so the direction of the line will be their cross product,

$$\begin{vmatrix} \mathbf{i}&\mathbf{j}& \mathbf{k}\\ 1&1&1\\ 1&-2&3\\ \end{vmatrix}= 5\mathbf{i}-2\mathbf{j} -3\mathbf{k}$$ Now you just need a point on the line, $(1,0,0)$ will do, so we have

$$\frac{x-1}{5}=\frac{y}{-2}=\frac{z}{-3}$$