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Here is the whole question.

Find an equation of the plane that passes through the points $P(1,0,-1)$ and $Q(2,1,0)$ and is parallel to the line of intersection of the planes $x+y+z=5$ and $x+y-z=1$.

And the answer by the book is $7x-5y-2z=9$.

As far as I know, the general strategy for finding an equation of a plane is to construct it via its normal vector.

And for the problem, I thought the intersection information could be used this way : Since the intersection line is parallel to the plane, the orthogonal vector of the intersection vector would be normal vector for the plane in question.

I'm not sure this strategy could work, and even if it works, how given information can be used in this process.

Please help me!

user35603
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user162475
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    You can find a normal vector for the plane by taking the cross-product of the vector from P to Q with a vector parallel to the line of intersection of the two planes. I believe the book has the incorrect answer, since the plane they give intersects the line of intersection of the planes. (I think it should be $x+y-2z=3$.) – user84413 Jul 09 '14 at 14:45

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You get one vector parallel to the plane from the line of intersection, as you point out. A second vector comes from the line through the two given points. So you have two vectors lying in the plane; I haven't done the computation, but they probably are not parallel. Given two nonparallel vectors, how can you find a vector perpendicular to both? This will be a normal vector to the plane.

rogerl
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  • Thanks rogerl. Maybe it is cross product, but it doesn't match with the answer from the book! The intersection is given by the equation '4y+3z=11' and to obtain a vector whose direction is parallel to this line, I simply subtract vector <0,2,1> from vector <0,5,-3>, which are solutions of the equation. But cross product between this vector and vector PQ doesn't really get closer to the answer. Can you point out what's gone wrong? – user162475 Jul 09 '14 at 14:58
  • $4y+3z=11$ is a plane, not a line. And the other direction vector is just $Q-P$. – rogerl Jul 09 '14 at 15:00
  • Then, how can I get the equation of the intersection line? or a vector parallel to the intersection line? – user162475 Jul 09 '14 at 15:02
  • @user162475: You don't need an equation of the intersection. It is sufficient to have a vector parallel to the intersection. Try to find the cross product of normal vectors of those planes. – user35603 Jul 09 '14 at 15:09
  • Thanks!!! It really helped! – user162475 Jul 09 '14 at 15:12
  • But still it didn't get close to the answer from the book.. Maybe I should throw away the trust to the writer's carefulness.. – user162475 Jul 09 '14 at 15:20
  • @user162475, see the comment of user84413 to your original post. – user35603 Jul 09 '14 at 15:50