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?
Best Answer
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?