# Find the line of intersection between the two planes `z-x-y=0` and `z-2x+y=0` .

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First, know that the vector that is in the direction of the line of intersection `(r)` is the cross product between the normal vectors of the planes since it is perpendicular to each of them. A plane in the form `ax+by+cz=0` has a normal vector `n=lta,b,cgt` .

Let `z-x-y=0` be plane `1` and `z-2x+y=0` be plane `2` . Then the two normal vectors are `n_1=lt-1,-1,1gt` and `n_2=lt-2,1,1gt` .

`r=n_1 xx n_2=(-1*-1)i-(-1+2)j+(-1-2)k=-2i-j-3k=lt-2,-1,-3gt`

A line has the parametric equation

`L(t)=(a,b,c)-t*r`

Where point `(a,b,c)` is any point on the line in the direction of `r`. So all we need now is a point on the line. The vector `r` has a `z` component of `-3` which means at some value of t it must go through the plane `z=0` . Therefore, we can set `z=0` in the plane `1` and `2` equations then solve for the `x` and `y` coordinates.

`(1):-> 0-x-y=0`

`(2):-> 0-2x+y=0`

Solving these equations gives `x=0` and `y=0` . So a point that the plane goes through is `(0,0,0)` . Then an equation has for the line must be

`L(t)=(0,0,0)-t*r`

`L(t)=-t*lt-2,-1,-3gt`

`L(t)=t*lt2,1,3gt`

The explicit parametric equations are:

`X(t)=2t`

`Y(t)=t`

`Z(t)=3t`