Profile
Settings
Refer your friends
Sign out
The length of the perpendicular to the plane that passes through the given point can be used to calculate the distance between the point and the plane. To put it another way, we could say that the distance between the point and the plane is equal to the length of the normal vector when it is dropped from the given point onto the given plane. If we want to determine the distance between the point P with the coordinates (x0, y0, z0) and the given plane with the equation Ax + By + Cz = D, then the distance between point P and the given plane is given by the expression |Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2).
Distance Between Point and Plane
The shortest possible perpendicular distance from a point to a given plane is used to calculate the distance between the point and the plane. To put it another way, the length of the perpendicular parallel to the normal vector that is dropped from the given point to the given plane is the length of the shortest distance that can be travelled from a point to a plane. Now, let’s take a look at the equation that describes the distance between a point and a plane.
The Formula for the Distance Between a Point and a Plane
The length of the normal vector that originates from the given point and ends up touching the plane is equal to the shortest distance that can be travelled between the point and the plane. Take into account a point P with the coordinates (x0, y0, z0) in conjunction with the plane that has the equation Ax + By + Cz = D. The formula for calculating the distance from the point P to the plane is as follows:
|Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2).
Proof of the Difference in Distance Between a Point and a Plane
Now that we have the formula for the distance between a point and a plane, we can use the other formulas of three-dimensional geometry to derive the formula for the distance between a point and a plane. Consider a point P in a three-dimensional space with the coordinates (x0, y0, z0), a plane with the normal vector, say v = (A, B, C), and a point Q on the plane with the coordinates (x1, y1, z1) on the plane. Then, consider the relationship between these three points. If this is the case, the equation of the plane can be written as follows: A(x – x1) + B(y – y1) + C(z – z1) = 0. It is possible to rewrite this equation as follows: Ax + By + Cz + (- Ax1 – By1 – Cz1) = 0 Ax + By + Cz + D = 0, where D = – (Ax1 + By1 + Cz1). Because of this, we have:
•The plane has the following equation:
Ax + By + Cz + D = 0
•Point P: (x0, y0, z0)
•Ai + Bj + Ck is the normal vector.
Let’s say that the vector joining points P(x0, y0, z0) and Q are denoted by the letter w. (x1, y1, z1). Then, w = (x0 – x1, y0 – y1, z0– z1). Now let’s figure out how to calculate the unit normal vector, also known as the normal vector with a magnitude of 1, which can be found by dividing the normal vector v by its magnitude. This will give us the unit normal vector. The formula for the unit normal vector is as follows:
n = v/||v||
= (A, B, C)/√(A2 + B2 + C2)
Now, the distance between point P and the plane in question is simply the length of the projection of vector w onto the unit normal vector n. This is how we determine the distance between the two points. The length of the vector n is known to be equal to one, and it is common knowledge that the distance from point P to the plane is equal to the absolute value of the dot product of the vectors w and n, which can be written as the following:
Distance, d = |w.n|
= | (x0 – x1, y0 – y1, z0– z1). [(A, B, C)/√(A2 + B2 + C2)] |
= |A(x0 – x1) + B(y0 – y1) + C(z0 – z1)|/√(A2 + B2 + C2)
= | Ax0 + By0 + Cz0 – (Ax1 + By1 + Cz1) |/√(A2 + B2 + C2)
= | Ax0 + By0 + Cz0 + D |/√(A2 + B2 + C2) [Because D = – (Ax1 + By1 + Cz1)]
Since the point Q with the coordinates (x1, y1, and z1) is an arbitrary point on the plane in question and D = – (Ax1 + By1 + Cz1), the formula stays the same for any point Q on the plane and, as a result, does not depend on the point Q. To put it another way, the formula for the distance between the point and the plane remains the same regardless of where the point Q lies on the plane. Because of this, the equation for the distance that separates the point P(x0, y0, z0) and the plane : Ax + By + Cz + D = 0 is d = |Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2).
Notable Considerations Regarding the Distance Between a Point and a Plane
•The formula for calculating the distance from a point to a plane is as follows: |Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2)
•If the given point lies on the given plane, then the distance between the point and the plane is equal to zero.
Conclusion
The length of the perpendicular to the plane that passes through the given point can be used to calculate the distance between the point and the plane. To put it another way, we could say that the distance between the point and the plane is equal to the length of the normal vector when it is dropped from the given point onto the given plane. If we want to determine the distance between the point P with the coordinates (x0, y0, z0) and the given plane with the equation Ax + By + Cz = D, then the distance between point P and the given plane is given by the expression |Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2).
