Plane Equations in Vector and Cartesian Forms

Introduction

Key Concepts

1. Definition of a Plane

A plane in three-dimensional space is a flat, two-dimensional surface that extends infinitely in all directions. It is uniquely determined by three non-collinear points or by a point and a normal vector. Understanding the properties of planes is crucial for various applications in mathematics, physics, and engineering.

2. Cartesian Form of Plane Equations

The Cartesian form of a plane equation is expressed as: $$ ax + by + cz + d = 0 $$ where \( a, b, \) and \( c \) are the coefficients of the variables \( x, y, \) and \( z \) respectively, and \( d \) is the constant term. This form is derived from the general equation of a plane, where \( \mathbf{n} = \langle a, b, c \rangle \) is a normal vector to the plane.

**Example:** Find the Cartesian equation of a plane passing through the point \( (1, 2, 3) \) with a normal vector \( \mathbf{n} = \langle 4, -5, 6 \rangle \).

**Solution:** Using the point-normal form: $$ 4(x - 1) -5(y - 2) +6(z - 3) = 0 \\ 4x - 4 -5y +10 +6z -18 = 0 \\ 4x -5y +6z -12 = 0 $$ Thus, the Cartesian equation is \( 4x -5y +6z -12 = 0 \).

3. Vector Form of Plane Equations

The vector form of a plane equation utilizes vectors to represent all points lying on the plane. It can be expressed as: $$ \mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c} $$ where \( \mathbf{r} \) is the position vector of any point on the plane, \( \mathbf{a} \) is the position vector of a known point on the plane, and \( \mathbf{b} \) and \( \mathbf{c} \) are non-parallel vectors lying on the plane. The parameters \( s \) and \( t \) are scalars.

**Example:** Determine the vector equation of a plane passing through the point \( P(1, 0, -1) \) with direction vectors \( \mathbf{u} = \langle 2, 1, 3 \rangle \) and \( \mathbf{v} = \langle -1, 4, 0 \rangle \).

**Solution:** Using the vector form: $$ \mathbf{r} = \langle 1, 0, -1 \rangle + s\langle 2, 1, 3 \rangle + t\langle -1, 4, 0 \rangle $$

4. Normal Vector and Its Significance

The normal vector \( \mathbf{n} \) of a plane is perpendicular to every vector lying on the plane. It is a crucial component in determining the orientation of the plane in space. The coefficients \( a, b, \) and \( c \) in the Cartesian equation \( ax + by + cz + d = 0 \) directly correspond to the components of the normal vector.

**Example:** For the plane equation \( 3x - 2y + z + 7 = 0 \), the normal vector is \( \mathbf{n} = \langle 3, -2, 1 \rangle \).

5. Distance from a Point to a Plane

The perpendicular distance \( D \) from a point \( P(x_0, y_0, z_0) \) to the plane \( ax + by + cz + d = 0 \) is given by: $$ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} $$ This formula is derived from the projection of the vector from a point on the plane to the external point onto the normal vector.

**Example:** Calculate the distance from the point \( (2, -1, 3) \) to the plane \( x + 2y - z + 4 = 0 \).

**Solution:** $$ D = \frac{|1(2) + 2(-1) -1(3) +4|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|2 -2 -3 +4|}{\sqrt{1 + 4 +1}} = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} $$

6. Intersection of Planes

The intersection of two non-parallel planes is a line. To find this line, one can solve the system of plane equations simultaneously. The direction vector of the intersection line is given by the cross product of the normal vectors of the two planes.

**Example:** Find the line of intersection of the planes \( x + y + z = 6 \) and \( 2x - y + 3z = 14 \).

**Solution:** 1. Find the direction vector \( \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 \): $$ \mathbf{n}_1 = \langle 1, 1, 1 \rangle, \quad \mathbf{n}_2 = \langle 2, -1, 3 \rangle \\ \mathbf{d} = \langle 1, 1, 1 \rangle \times \langle 2, -1, 3 \rangle = \langle (1)(3) - (1)(-1), (1)(2) - (1)(3), (1)(-1) - (1)(2) \rangle = \langle 4, -1, -3 \rangle $$ 2. Find a particular solution by setting one variable (e.g., \( z = 0 \)): $$ x + y = 6 \\ 2x - y = 14 \\ \text{Adding: } 3x = 20 \Rightarrow x = \frac{20}{3}, \quad y = 6 - \frac{20}{3} = -\frac{2}{3} $$ Thus, a point on the line is \( \left( \frac{20}{3}, -\frac{2}{3}, 0 \right) \). 3. The parametric equations of the line are: $$ x = \frac{20}{3} + 4t \\ y = -\frac{2}{3} - t \\ z = -3t $$

7. Plane Parallelism and Perpendicularity

Two planes are parallel if their normal vectors are parallel, i.e., one is a scalar multiple of the other. Planes are perpendicular if their normal vectors are perpendicular, meaning their dot product is zero.

**Example:** Determine if the planes \( x - 2y + 3z + 4 = 0 \) and \( 2x -4y +6z +8 = 0 \) are parallel or coincident.

**Solution:** The normal vectors are \( \mathbf{n}_1 = \langle 1, -2, 3 \rangle \) and \( \mathbf{n}_2 = \langle 2, -4, 6 \rangle \). Since \( \mathbf{n}_2 = 2\mathbf{n}_1 \), the planes are parallel. Additionally, since the constant terms are proportional (\( 4 \times 2 = 8 \)), the planes coincide.

8. Parameterization of a Plane

Parameterizing a plane involves expressing the coordinates \( (x, y, z) \) in terms of two parameters, usually denoted as \( s \) and \( t \). This provides a way to describe every point on the plane using these parameters.

**Example:** Parameterize the plane \( 3x + 2y - z = 7 \) using parameters \( s \) and \( t \).

**Solution:** Choose two variables as parameters, say \( y = s \) and \( z = t \). Then: $$ 3x + 2s - t = 7 \Rightarrow x = \frac{7 - 2s + t}{3} $$ Thus, the parameterization is: $$ \begin{cases} x = \frac{7 - 2s + t}{3} \\ y = s \\ z = t \end{cases} $$

9. Angle Between Two Planes

The angle \( \theta \) between two planes can be found using the angle between their normal vectors: $$ \cos\theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} $$

**Example:** Find the angle between the planes \( x + y + z = 3 \) and \( 2x - y + 2z = 4 \).

**Solution:** $$ \mathbf{n}_1 = \langle 1, 1, 1 \rangle, \quad \mathbf{n}_2 = \langle 2, -1, 2 \rangle \\ \mathbf{n}_1 \cdot \mathbf{n}_2 = 1(2) + 1(-1) +1(2) = 3 \\ |\mathbf{n}_1| = \sqrt{1^2 + 1^2 +1^2} = \sqrt{3}, \quad |\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 +2^2} = \sqrt{9} = 3 \\ \cos\theta = \frac{3}{\sqrt{3} \times 3} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \\ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.74^\circ $$

10. Applications of Plane Equations

Plane equations are widely used in various fields such as computer graphics, engineering design, navigation, and physics. They are essential for modeling surfaces, determining spatial relationships, and solving optimization problems.

**Example:** In computer graphics, plane equations are used to represent flat surfaces of 3D models, enabling rendering algorithms to calculate light reflections and shadows accurately.

Advanced Concepts

1. Intersection of Three Planes

The intersection of three planes can result in a single point, a line, or no common intersection depending on the planes' orientations. Solving the system of three plane equations simultaneously determines the nature of their intersection.

**Example:** Find the point of intersection of the planes: $$ \begin{cases} x + y + z = 6 \\ 2x - y + 3z = 14 \\ x - y + z = 2 \end{cases} $$

**Solution:** 1. Subtract the third equation from the first: $$ (x + y + z) - (x - y + z) = 6 - 2 \Rightarrow 2y = 4 \Rightarrow y = 2 $$ 2. Substitute \( y = 2 \) into the third equation: $$ x - 2 + z = 2 \Rightarrow x + z = 4 \Rightarrow x = 4 - z $$ 3. Substitute \( y = 2 \) and \( x = 4 - z \) into the second equation: $$ 2(4 - z) -2 +3z = 14 \Rightarrow 8 -2z -2 +3z = 14 \Rightarrow z = 8 $$ 4. Then, \( x = 4 -8 = -4 \)

Thus, the point of intersection is \( (-4, 2, 8) \).

2. Plane Equations in Parametric Form

While the vector form is common, plane equations can also be expressed parametrically to facilitate certain calculations, such as finding the intersection with lines or other planes.

**Example:** Given a plane with a point \( P(1, -1, 2) \) and two direction vectors \( \mathbf{u} = \langle 3, 4, -2 \rangle \) and \( \mathbf{v} = \langle -1, 2, 5 \rangle \), write its parametric equations.

**Solution:** Using the parametric form: $$ \begin{cases} x = 1 + 3s - t \\ y = -1 + 4s + 2t \\ z = 2 -2s +5t \end{cases} $$

3. Tangent Planes to Surfaces

In multivariable calculus, the tangent plane to a surface at a given point provides a linear approximation of the surface near that point. It is analogous to the tangent line to a curve in two dimensions.

**Example:** Find the equation of the tangent plane to the surface \( f(x, y) = x^2 + y^2 \) at the point \( (1, 2, 5) \).

**Solution:** 1. The surface equation is \( z = x^2 + y^2 \). 2. The partial derivatives are: $$ f_x = 2x, \quad f_y = 2y $$ At \( (1, 2, 5) \): $$ f_x(1,2) = 2, \quad f_y(1,2) = 4 $$ 3. The tangent plane equation: $$ z - 5 = 2(x - 1) + 4(y - 2) \\ z -5 = 2x -2 +4y -8 \\ 2x +4y -z -5 = 0 $$

4. Dual Forms of Plane Equations

The dual nature of plane equations refers to expressing them in both Cartesian and vector forms, enabling different approaches to problem-solving. This duality is particularly useful in transforming problems from one form to another to simplify calculations.

**Example:** Convert the Cartesian equation \( 3x - 4y + z - 12 = 0 \) to vector form.

**Solution:** 1. Identify a point on the plane by setting two variables to zero: $$ 3(0) -4(0) + z -12 =0 \Rightarrow z=12 \Rightarrow P(0,0,12) $$ 2. Find two direction vectors lying on the plane by setting different pairs of variables: - Let \( x=1, y=0 \): $$ 3(1) + z -12 =0 \Rightarrow z=9 \Rightarrow \mathbf{u} = \langle 1,0,9 \rangle - \langle 0,0,12 \rangle = \langle 1,0,-3 \rangle $$ - Let \( x=0, y=1 \): $$ -4(1) + z -12 =0 \Rightarrow z=16 \Rightarrow \mathbf{v} = \langle 0,1,16 \rangle - \langle 0,0,12 \rangle = \langle 0,1,4 \rangle $$ 3. The vector equation: $$ \mathbf{r} = \langle 0,0,12 \rangle + s\langle 1,0,-3 \rangle + t\langle 0,1,4 \rangle $$

5. Applications in Optimization Problems

Planes are integral in formulating and solving optimization problems, especially those involving linear programming. Constraints are often represented as planes, and feasible regions are determined by their intersections.

**Example:** Maximize \( f(x, y, z) = x + y + z \) subject to the constraints: $$ \begin{cases} x + y + z \leq 10 \\ 2x - y + 3z \leq 20 \\ x \geq 0, y \geq 0, z \geq 0 \end{cases} $$

This problem involves finding the optimal point at the intersection of the constraint planes within the first octant.

6. Plane Equations in Differential Geometry

In differential geometry, planes are studied as tangent planes to surfaces, contributing to the understanding of curvature and the local behavior of surfaces.

**Example:** Given a surface defined by \( F(x, y, z) = x^3 + y^3 + z^3 - 6xyz = 0 \), find the tangent plane at the point \( (1,1,1) \).

**Solution:** 1. Compute the gradient \( \nabla F = \langle 3x^2 -6yz, 3y^2 -6xz, 3z^2 -6xy \rangle \). 2. At \( (1,1,1) \): $$ \nabla F = \langle 3(1)^2 -6(1)(1), 3(1)^2 -6(1)(1), 3(1)^2 -6(1)(1) \rangle = \langle -3, -3, -3 \rangle $$ 3. The tangent plane equation: $$ -3(x -1) -3(y -1) -3(z -1) = 0 \\ -3x +3 -3y +3 -3z +3 =0 \\ x + y + z = 3 $$

7. Line of Intersection Using Vector Methods

Vector methods offer efficient ways to find the line of intersection between two planes by leveraging direction vectors and position vectors.

**Example:** Find the parametric equations of the line of intersection between the planes \( x + 2y + 3z = 9 \) and \( 2x - y + z = 8 \).

**Solution:** 1. Find the direction vector by computing \( \mathbf{n}_1 \times \mathbf{n}_2 \): $$ \mathbf{n}_1 = \langle 1, 2, 3 \rangle, \quad \mathbf{n}_2 = \langle 2, -1, 1 \rangle \\ \mathbf{d} = \langle 1,2,3 \rangle \times \langle 2,-1,1 \rangle = \langle (2)(1) - (3)(-1), (3)(2) - (1)(1), (1)(-1) - (2)(2) \rangle = \langle 5, 5, -5 \rangle = 5\langle 1,1,-1 \rangle $$ 2. Simplify direction vector: \( \mathbf{d} = \langle 1,1,-1 \rangle \) 3. Find a particular point by setting \( z = 0 \): $$ x + 2y = 9 \\ 2x - y = 8 \\ \text{Multiply first equation by 2: } 2x +4y = 18 \\ \text{Subtract second equation: } 5y = 10 \Rightarrow y = 2 \\ x = \frac{9 - 2y}{1} = 9 -4 =5 \\ \text{Point: } (5, 2, 0) $$ 4. Parametric equations: $$ x = 5 + t \\ y = 2 + t \\ z = 0 - t = -t $$

8. Intersection with Coordinate Axes

Determining where a plane intersects the coordinate axes involves setting two variables to zero and solving for the third. These intercepts can aid in graphing the plane.

**Example:** Find the intercepts of the plane \( 2x - 3y + z = 12 \).

**Solution:** - **x-intercept:** Set \( y = 0, z = 0 \): $$ 2x = 12 \Rightarrow x = 6 \Rightarrow (6, 0, 0) $$ - **y-intercept:** Set \( x = 0, z = 0 \): $$ -3y = 12 \Rightarrow y = -4 \Rightarrow (0, -4, 0) $$ - **z-intercept:** Set \( x = 0, y = 0 \): $$ z = 12 \Rightarrow (0, 0, 12) $$

9. Angle Between a Line and a Plane

The angle between a line and a plane is complementary to the angle between the line's direction vector and the plane's normal vector.

**Example:** Find the angle between the line \( \mathbf{r} = \langle 1,2,3 \rangle + t\langle 4,5,6 \rangle \) and the plane \( 7x - 8y + 9z = 10 \).

**Solution:** 1. Direction vector of the line: \( \mathbf{d} = \langle 4,5,6 \rangle \) 2. Normal vector of the plane: \( \mathbf{n} = \langle 7, -8, 9 \rangle \) 3. Compute the angle \( \phi \) between \( \mathbf{d} \) and \( \mathbf{n} \): $$ \cos\phi = \frac{\mathbf{d} \cdot \mathbf{n}}{|\mathbf{d}||\mathbf{n}|} = \frac{4(7) +5(-8)+6(9)}{\sqrt{4^2+5^2+6^2} \sqrt{7^2+(-8)^2+9^2}} = \frac{28 -40 +54}{\sqrt{77}\sqrt{194}} = \frac{42}{\sqrt{77}\sqrt{194}} \approx 0.367 $$ 4. The angle between the line and the plane \( \theta \) is: $$ \theta = 90^\circ - \phi \approx 90^\circ - \cos^{-1}(0.367) \approx 90^\circ -67.9^\circ =22.1^\circ $$

10. Intersection of a Plane with a Sphere

Intersecting a plane with a sphere results in a circle, a point, or no intersection, depending on the distance between the sphere's center and the plane.

**Example:** Find the intersection of the plane \( x + y + z = 6 \) with the sphere \( x^2 + y^2 + z^2 - 4x - 6y - 2z +12 =0 \).

**Solution:** 1. Rewrite the sphere equation: $$ x^2 -4x + y^2 -6y + z^2 -2z = -12 \\ (x^2 -4x +4) + (y^2 -6y +9) + (z^2 -2z +1) = -12 +4 +9 +1 \\ (x -2)^2 + (y -3)^2 + (z -1)^2 = 2 $$ Center: \( (2,3,1) \), Radius: \( \sqrt{2} \) 2. Distance \( D \) from center to plane: $$ D = \frac{|1(2) +1(3) +1(1) -6|}{\sqrt{1^2 +1^2 +1^2}} = \frac{|6 -6|}{\sqrt{3}} = 0 $$ 3. Since \( D = 0 < \) Radius, intersection is a circle. 4. Radius of the circle \( r = \sqrt{R^2 - D^2} = \sqrt{2 -0} = \sqrt{2} \)

Comparison Table

Summary and Key Takeaways