![]() ![]() What is the Distance Between Point and Plane When the Point Lies on the Given 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. 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? What is the Distance Between a Point and the x-z Plane? To find the shortest distance between point and plane, we use the formula d = |Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2), where (x o, y o, z o) is the given point and Ax + By + Cz + D = 0 is the equation of the given 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 |/√(A 2 + B 2 + C 2) How to Find the Shortest Distance Between Point and Plane? What is the Formula for the Distance Between Point and Plane? In other words, the distance between point and plane is the shortest perpendicular distance from the point to the given plane. The distance between point and plane is the length of the perpendicular to the plane passing through the given point. ![]() Related Topics on Distance Between Point and PlaneįAQs on Distance Between Point and Plane What is Distance Between Point and Plane in Geometry? Distance Between Point and Plane is zero if the given point lies on the given plane.Distance Between Point and Plane Formula: |Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2).Important Notes on Distance Between Point and Plane Substituting the values in the formula, we haveĭ = |Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2) Solution: We know that the formula for distance between point and plane is: d = |Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2) 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.Įxample: Determine the distance between the point P = (1, 2, 5) and the plane π: 3x + 4y + z + 7 = 0 How To Apply Distance From Point to Plane Formula? 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 |/√(A 2 + B 2 + C 2) Since the point Q with coordinates (x 1, y 1, z 1) is an arbitrary point on the given plane and D = - (Ax 1 + By 1 + Cz 1), 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. = | Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2) 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., 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. 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. Then, w = (x o - x 1, y o - y 1, z o - z 1). Let w be the vector joining points P(x o, y o, z o) and Q(x 1, y 1, z 1). Equation of plane: Ax + By + Cz + D = 0.This equation can be rewritten as Ax + By + Cz + (- Ax 1 - By 1 - Cz 1) = 0 ⇒ Ax + By + Cz + D = 0, where D = - (Ax 1 + By 1 + Cz 1). Then the equation of the plane is given by A(x - x 1) + B(y - y 1) + C(z - z 1) = 0. 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 (x 1, y 1, z 1) on the plane. 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. ![]()
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