Distance Between Point and Plane

The distance between point and plane is the length of the perpendicular to the plane passing through the given point. In other words, we can say that the distance between point and plane is the length of the normal vector dropped from the given point onto the given plane. If we want to determine the distance between the point P with coordinates (x_o, y_o, z_o) and the given plane with equation Ax + By + Cz = D, then the distance between point P and the given plane is given by |Ax_o + By_o+ Cz_o + D|/√(A² + B² + C²).

The distance between point and plane is the shortest perpendicular distance from the point to the given plane. In simple words, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the given point to the given plane. Let us now see the formula for the distance between point and plane.

The shortest distance between a point and plane is equal to the length of the normal vector which starts from the given point and touches the plane. Consider a point P with coordinates (x_o, y_o, z_o) and the given plane π with equation Ax + By + Cz = D. Then, the distance between the point P and the plane π is given by, |Ax_o + By_o+ Cz_o + D|/√(A² + B² + C²).

Now that we know the formula for the distance between point and plane, let us derive its formula using various formulas of three-dimensional geometry. Consider a point P with coordinates (x_o, y_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, y₁, 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 - By₁ - Cz₁) = 0 ⇒ Ax + By + Cz + D = 0, where D = - (Ax₁ + By₁ + Cz₁). Hence, we have:

• Equation of plane: Ax + By + Cz + D = 0
• Point P: (x_o, y_o, z_o)
• Normal Vector: Ai + Bj + Ck

Let w be the vector joining points P(x_o, y_o, z_o) and Q(x₁, y₁, z₁). Then, w = (x_o - x₁, y_o - y₁, z_o - z₁). Now, let calculate the unit normal vector, i.e., normal vector with magnitude equal to 1 which is given by the division of the normal vector v divided by its magnitude. The unit normal vector is given by,

n = v/||v||

= (A, B, C)/√(A² + B² + C²)

Now, the distance between point P and the given plane is nothing but the length of the projection of vector w onto the unit normal vector n. As we know, the length of the vector n is equal to one, the distance from point P to the plane is the absolute value of the dot product of the vectors w and n, i.e.,

Distance, d = |w.n|

= | (x_o - x₁, y_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)

= | Ax_o + By_o + Cz_o - (Ax₁ + By₁ + Cz₁) |/√(A² + B² + C²)

= | Ax_o + By_o + Cz_o + D |/√(A² + B² + C²) [Because D = - (Ax₁ + By₁ + Cz₁)]

Since the point Q with coordinates (x₁, y₁, z₁) is an arbitrary point on the given plane and D = - (Ax₁ + By₁ + Cz₁), therefore the formula remains the same for any point Q on the plane and hence, does not depend on the point Q, i.e., wherever the point Q lies on the plane, the formula for the distance between point and plane remains the same. Hence, the distance between point P(x_o, y_o, z_o) and plane π: Ax + By + Cz + D = 0 is given by, d = |Ax_o + By_o + Cz_o + D |/√(A² + B² + C²)

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 point and plane.

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

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

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

Substituting the values in the formula, we have

d = |Ax_o + By_o + Cz_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 Distance Between Point and Plane

• Distance Between Point and Plane Formula: |Ax_o + By_o + Cz_o + D |/√(A² + B² + C²)
• Distance Between Point and Plane is zero if the given point lies on the given plane.

Related Topics on Distance Between Point and Plane

• Distance Formula
• Distance Between Two Points
• Euclidean Distance Formula

What is Distance Between Point and Plane in Geometry?

The distance between point and plane is the length of the perpendicular to the plane passing through the given point. In other words, the distance between point and plane is the shortest perpendicular distance from the point to the given plane.

What is the Formula for the Distance Between Point and Plane?

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

How to Find the Shortest Distance Between Point and Plane?

To find the shortest distance between point and plane, we use the formula d = |Ax_o + By_o + Cz_o + D |/√(A² + B² + C²), where (x_o, y_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, y_o, z_o) and x-z plane is given by the y-coordinate, i.e., y_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 that is parallel to the normal vector to the plane, that drops from the given point onto the given plane.

What is the Distance Between Point and Plane When the Point Lies on the Given Plane?

The distance between point and plane is equal to zero if the given point lies on the given plane.