# Finding the scalar equation of a plane

• Mar 26th 2009, 07:29 AM
Dudeface
Finding the scalar equation of a plane
Here's the question:
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.

Now I know that the normal to the given plane (3,2,6) should be perpendicular to the normal of the required plane but after that, I don't know how to proceed.

Any help would be much appreciated.

Thanks,
Dudeface
• Mar 26th 2009, 07:43 AM
HallsofIvy
Quote:

Originally Posted by Dudeface
Here's the question:
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.

Now I know that the normal to the given plane (3,2,6) should be perpendicular to the normal of the required plane but after that, I don't know how to proceed.

Any help would be much appreciated.

Thanks,
Dudeface

And the problem is that there are an infinite number of vectors perpedicular to that normal vector! However, it is true that the normal vector can be taken be lying in a plane perpedicular to the give plane.

The vector from (1,2,3) to (2,3,-1), (2-1,3-2,-1-3)= (1,1,-4), also lies in the plane. Find a vector perpendicular to both by taking the cross product. THAT will be normal to the plane you want.
• Mar 26th 2009, 12:19 PM
Dudeface
Surely you mean between (1,2,3) and (3,2,-1)?
You mean the cross product between MN and the normal vector of the perpendicular plane (ie (3,2,6))?
• Mar 26th 2009, 01:14 PM
Plato
Use $\displaystyle \left\langle {1,2, - 4} \right\rangle \times \left\langle {2,3,6} \right\rangle$ as the normal.