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

Answer 1

With a plane defined as Ax + By +Cz +D = 0,

the distance between that plane and a given point (x₁, y₁, z₁) is

|Ax₁+By₁+Cz₁+D| / √(A²+B²+C²)

In this case, the plane is 0x + 5y -4z -56 = 0

So

A = 0, B = 5, C = -4, D = -56 and x₁ = -3, y₁ = -6, z₁ = -9

Putting these values in the distance formula above yields

|0(-3) + 5(-6) + -4(-9) + (-56)| / √(0²+5²+(-4)²)

|0 -30 +36 -56 | / √(25+16) = 50/√41

Answer 2

Normal equation of plane: (5y - 4z - 56)/√41 = 0

Dostance d = I5(-6) - 4(-9)I/√41 = 6/√41