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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 $61.9^\circ$.

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1 Answer 1

up vote 4 down vote accepted

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 $$p-2q+1=9\\p+q-1=-2$$

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subtracting the 2nd equation from the first gives -2q-q +1-(-1) = 9-2. Simplifying gives -3q+2=7 or 3q=-5. This implies q=-5/3 ? –  Ozzy Feb 21 '13 at 9:43
    
This also gives p = 2+1 -5/3 = 4/3. But the answer was given as p=2, q=-3 :S –  Ozzy Feb 21 '13 at 9:46
    
@Ozzy: There was a typo. 2 should have been -2. Now you can find them correctly. –  Babak S. Feb 21 '13 at 9:57
    
Oh! I should have looked back myself... Thanks :) –  Ozzy Feb 21 '13 at 9:58
    
Ozzy: :-)$~~~~~~$ –  Babak S. Feb 21 '13 at 9:59

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