The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.
What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
On
The normal to the plane is perpendicular to $ (3,4,1) - (3,1,-6) = (0, 3, 7) $ and also perpendicular to $(7,9,4) $, so it is along their cross product:
$N = (0, 3, 7) \times (7, 9, 4) = (-51, 49, -21) $
The plane passes through $(3,4,1)$ hence its equation is
$ (-51, 49, -21) \cdot ( (x, y, z) - (3, 4, 1) ) = 0 $
and this simplifies to
$ (-51, 49, -21) \cdot (x, y, z) = 22$
That is,
$ 51 x - 49 y + 21 z + 22 = 0 $
make the ansatz the plane has the equation $$ax+by+cz+d=0$$ then we have $$3a+4b+c+d=0$$(1) $$3a+b-6c+d=0$$ (2) and $$[a,b,c]\cdot[7,9,4]=0$$ (3) can you proceed?