How do I find the point-normal form of the equation of the plane containing the point (-3,-4,3) and perpendicular to (4,1,-2)?
I know to use #(r-p)*n)=0# but I keep getting #4x+y-2z=-10# and being told that's incorrect. Where am I going wrong? Have I just not simplified enough?
Thanks :)
I know to use
Thanks :)
1 Answer
Oct 2, 2017
The point-normal form of the equation of a plane is:
where
Explanation:
Given the normal vector
The above equation is in the requested point-normal form and it seems that you are trying to write the equation in the scalar form:
Ax+By+Cz = D
This is not the point-normal form.