There are two planes intersecting at a line.
Plane 1: $x - 2y + z - 9 = 0 $ Plane 2: $x + y - z + 2 = 0$
There is a point $A = (p, q, 1)$ on the line of intersection.
How can I find $p ~\text{and}~ q$?
I tried to used orthogonal conditions ($a.n1=0, a.n2=0)$
but I got the values $p=1/3, q=2/3$ and the answer given is $p=2, q=-3$
Please show me the correct way to solve this.
I'm not sure if the answer is related to the first part of the question, for which I had to solve the angle between the two planes as $61.9^\circ$.
If that point wants to lie on the line of intersection of two planes, so the coordinates of the point $$(p,q,1)$$ should satisfy the equations of the planes simultaneously. This means that you should just to solve the system of $$p-2q+1=9\\p+q-1=-2$$