Distances: Point to Plane & Parallel Planes

Welcome to the final lesson in our module on Vectors and 3D Geometry. In the past few lessons, we've built a solid toolkit for working with lines and planes, culminating in our last session where we learned to find the shortest distance between two skew lines. We will now conclude our study of 3D distances by tackling two other fundamental scenarios.

Today's objective is to master the calculation of distances involving planes. Specifically, you will learn how to:

Calculate the perpendicular distance from a given point to a plane.
Calculate the distance between two parallel planes.

These are not just foundational concepts; they are also frequent components of problems in competitive exams like the MHT-CET. We will derive the necessary formulas and also explore some powerful related techniques, like finding the coordinates of the foot of a perpendicular, which are invaluable for problem-solving.

1. Distance of a Point from a Plane

The shortest distance from a point to a plane is always along the line segment that is perpendicular to the plane. The core idea behind finding this distance is to use vector projection. We take a vector from the given point to any known point on the plane and then find the length of its projection onto the plane's normal vector.

This diagram shows the perpendicular distance `D` from a point $(x_1, y_1, z_1)$ to a plane. The formula is derived by projecting a vector onto the plane's normal vector.

The following reading provides a concise derivation of the formula.

Section 9.6—The Distance from a Point to a Plane

This document walks through the derivation of the point-to-plane distance formula using the concept of a scalar projection.

Read the section titled Determining a Formula. Pay close attention to how the projection of the vector \vec{P_0 P_1} onto the normal vector \vec{n} results in the final formula.

As derived, for a point $P (x_{1}, y_{1}, z_{1})$ and a plane with the equation $A x + B y + C z + D = 0$ , the perpendicular distance $d$ is given by the formula:

d = \frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}

Essentially, you substitute the point's coordinates into the plane's equation for the numerator and divide by the magnitude of the normal vector $n = A i^+ B j^+ C k^$ for the denominator.

The following video is specifically geared towards MHT-CET preparation and demonstrates how to apply this formula efficiently.

PLANE in 3D Geometry 🔥 | Most Scoring Topic | LIVE Today 7:30 PM | MHT-CET & JEE

This video from EquationX explains the formula and provides a worked example, highlighting its importance for competitive exams.

First, watch the instructor introduce the distance formula. Then, see how it's applied to a specific problem in the worked example. Notice the direct substitution and calculation, which is the exact process you'll follow.

2. Distance Between Two Parallel Planes

Two planes are parallel if their normal vectors are parallel. This means the coefficients of $x, y,$ and $z$ in their Cartesian equations will be proportional. For example:

Plane 1: $A_{1} x + B_{1} y + C_{1} z + D_{1} = 0$
Plane 2: $A_{2} x + B_{2} y + C_{2} z + D_{2} = 0$
These are parallel if $\frac{A _{1}}{A _{2}} = \frac{B _{1}}{B _{2}} = \frac{C _{1}}{C _{2}}$ .

Two parallel planes have the same normal vector. The distance between them can be found by taking any point on one plane and calculating its distance to the other.

There are two common methods to find the distance between them:

Method 1: Find a Point and Use the Point-to-Plane Formula
This is the most intuitive method.

Choose one of the planes, say $A x + B y + C z + D_{1} = 0$ .
Find the coordinates of any point on this plane. The easiest way is to set two variables to zero (e.g., $x = 0, y = 0$ ) and solve for the third (e.g., $z = - D_{1} / C$ ). This gives you a point $(0, 0, - D_{1} / C)$ .
Calculate the distance from this point to the second plane, $A x + B y + C z + D_{2} = 0$ , using the formula from the previous section.

The following resource provides a clear example of this strategy.

Section 9.6—The Distance from a Point to a Plane

This example shows exactly how to apply the point-finding strategy to calculate the distance between two parallel planes.

Read through Example 2, part a. Note how a simple point (the z-intercept) is chosen on the first plane to simplify the calculation.

Method 2: The Direct Formula
If you have two parallel planes written in the form:

Plane 1: $A x + B y + C z + D_{1} = 0$
Plane 2: $A x + B y + C z + D_{2} = 0$
(Note: The coefficients A, B, and C must be identical)

The distance $d$ between them is given by the shortcut formula:

d = \frac{∣ D _{2} - D _{1} ∣}{A ^{2} + B ^{2} + C ^{2}}

If the coefficients are proportional (e.g., one plane is $x + 2 y + 3 z + 4 = 0$ and the other is $2 x + 4 y + 6 z + 10 = 0$ ), you must first scale one of the equations to make the coefficients of $x, y, z$ identical before applying this formula. In this case, you would rewrite the second plane as $x + 2 y + 3 z + 5 = 0$ .

The next video explains the logic behind this shortcut by considering the planes' distances from the origin.

Vectors: Calculating the distance from a point to a plane or a plane to a plane

This video from Joel Speranza Math provides a nice conceptual explanation for finding the distance between parallel planes and also mentions the point-finding method as an alternative.

Watch the section from this timestamp. Pay attention to how the distances from the origin are used to find the distance between the planes. The sign of the distance indicates whether the planes are on the same or opposite sides of the origin.

3. Foot of the Perpendicular and Mirror Image

For MHT-CET, a very common and important related problem is finding the foot of the perpendicular from a point to a plane, or the mirror image of a point in a plane. The formulas for these are direct applications of the concepts we've discussed.

Foot of the Perpendicular: The point where the perpendicular line from a point $P$ intersects the plane.
Mirror Image: The point $P^{'}$ such that the plane is the perpendicular bisector of the line segment $P P^{'}$ .

The video from EquationX, which we saw earlier, provides an excellent, exam-focused explanation of the formulas and techniques to solve these problems.

PLANE in 3D Geometry 🔥 | Most Scoring Topic | LIVE Today 7:30 PM | MHT-CET & JEE

This video provides direct formulas and worked examples for finding the foot of the perpendicular and the mirror image, which are high-yield topics for your exam.

First, watch the instructor derive the formula for the foot of the perpendicular and mirror image. Notice the only difference is a factor of -2 for the mirror image. Next, see this applied to find the coordinates of the foot in a detailed example. Finally, watch a problem on finding the mirror image of a point. Observe how quickly the answer can be found once you are comfortable with the formula.

The formulas are:
For a point $P (x_{1}, y_{1}, z_{1})$ and a plane $a x + b y + cz + d = 0$ , the coordinates of the foot of the perpendicular $(x, y, z)$ are given by:
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}$
For the mirror image, you simply multiply the right-hand side by 2:
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - 2 (\frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}})$

Conclusion

This lesson concludes our journey through 3D Geometry. You are now equipped to handle a variety of distance-related problems involving points, lines, and planes.

Key Takeaways:

Distance from a Point to a Plane: $d = \frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$ . This is a direct application formula.
Distance Between Parallel Planes: This can be found by either finding a point on one plane and its distance to the other, or by using the direct formula $d = \frac{∣ D _{2} - D _{1} ∣}{A ^{2} + B ^{2} + C ^{2}}$ after ensuring the $A, B, C$ coefficients match.
Foot of Perpendicular & Mirror Image: These are powerful formulas for solving a common type of competitive exam problem, allowing you to find specific coordinates rather than just a distance.

With this, we have completed the Vectors & 3D Geometry module. In our next lesson, we will transition to a new major area of your syllabus: Calculus. We will begin with the foundational topic of Limits, Continuity & Differentiation, starting with how to evaluate limits of functions.

Recommended resources

1. Distance of a Point from a Plane

2. Distance Between Two Parallel Planes

3. Foot of the Perpendicular and Mirror Image

Conclusion