Find the distance from the point (-3,-6,-9) to the plane 5y-4z=56

Answer 1

There is a standard formula for this.

If you have an equation of the plane in the form ax + by + cz = d, then the (perpendicular) distance from the point (x0, y0, z0) to this plane is given by:

| a(x0) + b(y0) + c(z0) - d| / sqrt(a^2 + b^2 +c^2)

Note the minus sign on the constant term! If my memory is correct, there are quite a few places on the internet that get this wrong.

In your case, the numerator is:

| 0(-3) +5(-6) -4(-9) - 56| = 50

and the denominator is:

sqrt((0)^2 + (4)^2 + (5)^2) = sqrt(41)