Planes in 3D Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
Find the distance from the point (3,−1,2)(3, -1, 2) to the plane 2x+y−2z=42x + y - 2z = 4.

Solution

  1. 1
    Distance formula: d=∣Ax0+By0+Cz0−D∣A2+B2+C2d = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2+B^2+C^2}} where the plane is Ax+By+Cz=DAx + By + Cz = D.
  2. 2
    d=∣2(3)+1(−1)+(−2)(2)−4∣4+1+4=∣6−1−4−4∣3=∣−3∣3=1d = \frac{|2(3) + 1(-1) + (-2)(2) - 4|}{\sqrt{4+1+4}} = \frac{|6-1-4-4|}{3} = \frac{|-3|}{3} = 1.

Answer

d=1d = 1
The point-to-plane distance formula is analogous to the point-to-line distance formula in 2D. It projects the vector from any point on the plane to the given point onto the unit normal vector, yielding the perpendicular distance.

About Planes in 3D

A flat, infinite surface in 3D space described by ax+by+cz=dax + by + cz = d, where ⟨a,b,c⟩\langle a, b, c \rangle is the normal vector.

Learn more about Planes in 3D →

More Planes in 3D Examples

Example 1 easy

Write the equation of the plane with normal vector [formula] passing through the point [formula].

Example 2 medium

Find the equation of the plane through the points [formula], [formula], and [formula].

Example 4 hard

Find the angle between the planes [formula] and [formula].

← All Planes in 3D Examples Planes in 3D Concept

Related Concepts

Lines in 3D