Posted by Lulu on Monday, July 7, 2008 at 12:39pm.
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. I know that normal of the latter equation is (3,2,6) but now what do I do? Thanks.

Calculus  Reiny, Monday, July 7, 2008 at 5:23pm
You are on the right track, so (3,2,6) must be a direction vector on your new plane.
you also have two points M(1,2,3) and N(3,2,1), so the vector MN or (2,0,1) is another direction vector.
So by taking the crossproduct of vectors (3,2,6) and (2,0,1) I got (2,6,1)
which must be the normal to the new plane.
So the new plane has equation
2x  6y + z = k
but M(1,2,3) lies on it, so 2  12  1 = k
k = 7
The new plane has equation 2x  6y + x = 7
Check:
1. Both M and N satisfy the new equation
2. Is the dotproduct of their normals zero? (3,2,6)∙(2,6,1) = 612+6 = 0
There you go! 
typo correction  Reiny, Monday, July 7, 2008 at 5:28pm
my line
<but M(1,2,3) lies on it, so 2  12  1 = k >
should say :
but M(1,2,3) lies on it, so 2  12 + 3 = k 
Calculus  Lulu, Monday, July 7, 2008 at 9:16pm
I appreciate the help, but you made another mistake. You got a wrong answer, it should be (2,0,4) not (2,0,1) so thus, you used the wrong information. But thanks, I know what to do now :)