Distance between point and plane - formula, proof, examples (2023)

HeDistance between point and planeis the length of the perpendicular to the plane that passes through the given point. In other words, we can say that the distance between the point and the plane is the length of the normal vector that falls from the given point to the given plane. If we want to calculate the distance between the coordinate point P (x_o, j_o, z_o) and the given plane of equation Ax + By + Cz = D, then the distance between the point P and the given plane is given by |Ax_o+ bis_o+ ch_o+ D|/√(A²+B²+C²).

What is the distance between the point and the plane?

The distance between the point and the plane is the shortest perpendicular distance from the point to the given plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector traced by the given point in the given plane. Now let's see the formula for the distance between the point and the plane.

Distance between the point and the plane Formula

The shortest distance between a point and a plane is equal to the length of the normal vector emanating from the given point and tangent to the plane. Consider aPointP with coordinates (x_o, j_o, z_o) and the π plane given by the equation Ax + By + Cz = D. Then the distance between the point P and the π plane is given by |Ax_o+ bis_o+ ch_o+ D|/√(A²+B²+C²).

Distance test between point and plane

Now that we know the formula for the distance between the point and the plane, let's derive its formula using various three-dimensional geometry formulas. Consider a point P with coordinates (x_o, j_o, z_o) in a three-dimensional space and a plane with the normal vector, say v = (A, B, C) and the point Q with coordinates (x₁, j₁, z₁) on the plane. Then the equation of the plane is given by A(x - x₁) + B(y - y₁) + C(z - z₁) = 0. This equation can be rewritten as Ax + By + Cz + (- Ax₁- Von₁- Ch₁) = 0 ⇒ Ax + By + Cz + D = 0, wobei D = - (Ax₁+ bis₁+ ch₁). Therefore we have:

• Plane equation: Ax + By + Cz + D = 0
• Point P: (x_o, j_o, z_o)
• Vector normal: Ai + Bj + Ck

be thatVectorConnection points P(x_o, j_o, z_o) y Q(x₁, j₁, z₁). Then w = (x_o- X₁, j_o- y₁, z_o-z₁). Now let us calculate the unit normal vector, i.e. h the normal vector with magnitude equal to 1 given by dividing the normal vector v by its magnitude. The unit normal vector is given by

n = v/||v||

= (A, B, C)/√(A²+B²+C²)

Now, the distance between the point P and the given plane is no more than the length of the projection of the vector w onto the normal unit vector n. As we know, the length of the vector n is equal to one, the distance from the point P to the plane is the absolute value of the scalar product of the vectors w and n, that is

Abstact, d = |w.n|

= | (X_o- X₁, j_o- y₁, z_o-z₁). [(A, B, C)/√(A²+B²+C²)] |

= |A(x_o- X₁) + B(y_o- y₁) + C(z_o-z₁)|/√(A²+B²+C²)

= | Axe_o+ bis_o+ ch_o- (Axe₁+ bis₁+ ch₁) |/√(A²+B²+C²)

= | Axe_o+ bis_o+ ch_o+ D |/√(A²+B²+C²) [Fine D = - (Ax₁+ bis₁+ ch₁)]

Since the coordinate point Q (x₁, j₁, z₁) is an arbitrary point in the given plane and D = - (Ax₁+ bis₁+ ch₁), thus the formula remains the same for any point Q in the plane, and therefore does not depend on point Q, i.e. wherever point Q lies in the plane, the formula for distance between the point and the plane remains the same. Therefore, the distance between the point P(x_o, j_o, z_o) and π plane: Ax + By + Cz + D = 0 is given by, d = |Ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²)

How to apply the formula for the distance from the point to the plane?

We have derived the formula for the distance from a point to a plane, we will solve an example using the formula to understand its application and determine the distance between the point and the plane.

Example:Find the distance between the point P = (1, 2, 5) and the π plane: 3x + 4y + z + 7 = 0

Solution:We know that the formula for the distance between a point and a plane is: d = |Ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²)

Here A=3, B=4, C=1, D=7, x_o= 1, j_o= 2, z_o= 5

If we substitute the values ​​into the formula, we get

d = |ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²)

= |3 × 1 + 4 × 2 + 1 × 5 + 7|/√(3²+ 4²+ 1²)

= |3 + 8 + 5|/√(9 + 16 + 1)

= |16|/√26

= 8√26/13 units

Important notes on the distance between point and plane

• Distance between the point and the plane Formula: |Ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²)
• The distance between the point and the plane is zero if the given point lies in the given plane.

Related topic about distance between point and plane

• distance formula
• Distance between two points
• Euclidean distance formula

Frequently asked questions about the distance between the point and the plane

What is the distance between the point and the plane in geometry?

HeDistance between point and planeis the length of the perpendicular to the plane that passes through the given point. In other words, the distance between the point and the plane is the shortest perpendicular distance between the point and the given plane.

What is the formula for the distance between a point and a plane?

The distance between the point P(x_o, j_o, z_o) and π plane: Ax + By + Cz + D = 0 is given by, d = |Ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²)

How to find the shortest distance between the point and the plane?

To find the shortest distance between the point and the plane, we use the formula d = |Ax_o+ bis_o+ ch_o+ D |/√(A²+B²+C²), where (x_o, j_o, z_o) is the given point and Ax + By + Cz + D = 0 is the equation of the given plane.

What is the distance between a point and the x-z plane?

The distance between a point (x_o, j_o, z_o) and the x-z plane is given by the y coordinate, that is, h and_o

How do you find the shortest distance from a point to a plane?

To calculate the shortest distance from a point to a plane, we consider the length of the vector parallel to the normal vector to the plane that falls from the given point to the given plane.

What is the distance between the point and the plane if the point lies in the given plane?

The distance between the point and the plane is zero if the given point lies in the given plane.