Calculus 3 lines and planes

Question

Find the distance from the point (-8,-4,-7) to the plane 2x−5y+z=31.

Answer 1

In general, distance between a plane a.x + b.y + c.z + d = 0 and point (x1,y1,z1) is given by:

D = | a.x1 + b.x2 + c.z1 + d | / √a²+b²+c². Therefore we have:

D = | 2*-8 + -5*-4 + 1*-7 + -31 | / √2²+(-5)²+1² = 34/√30 = 6.2.