Mathematics / Geometry 3D & Vector Analysis 100% Free Open Access
Chapter 2 โ€ข Theory & Derivations

The Plane in Three Dimensions: Equations, Dihedral Angles & Distance Metrics

ยง2.1 The General Linear Equation & Normal Vectors in Space

1. General First-Degree Equation in $\mathbb{R}^3$

$$\mathbf{\text{Theorem: Every linear equation of the first degree } Ax + By + Cz + D = 0 \text{ (where } A^2 + B^2 + C^2 \ne 0\text{) represents a plane in } \mathbb{R}^3.}$$

Proof: Let $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ be any two distinct points on the surface $Ax + By + Cz + D = 0$. Consider any point $P$ dividing $P_1 P_2$ in ratio $k : 1$:

$$P = \left( \frac{x_1 + k x_2}{1 + k}, \, \frac{y_1 + k y_2}{1 + k}, \, \frac{z_1 + k z_2}{1 + k} \right)$$

Substituting into the equation:

$$A\left(\frac{x_1 + k x_2}{1 + k}\right) + B\left(\frac{y_1 + k y_2}{1 + k}\right) + C\left(\frac{z_1 + k z_2}{1 + k}\right) + D = \frac{(Ax_1 + By_1 + Cz_1 + D) + k(Ax_2 + By_2 + Cz_2 + D)}{1 + k} = \frac{0 + k(0)}{1 + k} = 0$$

Since every point on the line segment $P_1 P_2$ lies entirely on the surface, the surface is a plane. $\blacksquare$

2. Geometric Meaning of Coefficients: The Normal Vector $\vec{n}$

Let $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ lie on the plane. Subtracting their equations:

$$A(x_2 - x_1) + B(y_2 - y_1) + C(z_2 - z_1) = 0 \Longleftrightarrow (A\hat{i} + B\hat{j} + C\hat{k}) \cdot \vec{P_1 P_2} = 0$$

The vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is perpendicular to every displacement vector lying in the plane, establishing that $(A, B, C)$ are the direction ratios of the normal vector to the plane.

3. Intercept Form of the Plane

If the plane intersects the coordinate axes at $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$ ($abc \ne 0$), dividing by $-D$ yields:

$$\mathbf{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$$

ยง2.2 Normal Form & Planes Through Specified Points

1. The Hesse Normal Form: $lx + my + nz = p$

Let $ON$ be the perpendicular drawn from origin $O$ to the plane, of length $p \ge 0$, having direction cosines $(l, m, n)$. For any point $P(x, y, z)$ on the plane, the projection of position vector $\vec{OP} = (x, y, z)$ onto the normal unit vector $\hat{n} = (l, m, n)$ is identically $p$:

$$\mathbf{lx + my + nz = p} \quad (p \ge 0, \quad l^2 + m^2 + n^2 = 1)$$

2. Reduction of General Equation to Normal Form

To convert $Ax + By + Cz + D = 0$ into normal form, transpose $D$ so the RHS is non-negative and divide by $\pm\sqrt{A^2 + B^2 + C^2}$:

$$\frac{-A}{\pm\sqrt{A^2+B^2+C^2}}x + \frac{-B}{\pm\sqrt{A^2+B^2+C^2}}y + \frac{-C}{\pm\sqrt{A^2+B^2+C^2}}z = \frac{D}{\pm\sqrt{A^2+B^2+C^2}}$$

The sign is chosen opposite to $D$ so the constant distance $p$ is positive.

3. Plane Passing Through Three Non-Collinear Points

The plane passing through $P_1(x_1, y_1, z_1)$, $P_2(x_2, y_2, z_2)$, and $P_3(x_3, y_3, z_3)$ is given by the coplanarity determinant:

$$\mathbf{\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0}$$

ยง2.3 Dihedral Angles & Mutual Positions of Two Planes

1. Dihedral Angle Between Two Planes

The angle $\theta$ between two planes $\Pi_1: A_1 x + B_1 y + C_1 z + D_1 = 0$ and $\Pi_2: A_2 x + B_2 y + C_2 z + D_2 = 0$ is equal to the angle between their normal vectors $\vec{n}_1$ and $\vec{n}_2$:

$$\mathbf{\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{\|\vec{n}_1\| \|\vec{n}_2\|} = \frac{|A_1 A_2 + B_1 B_2 + C_1 C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}}$$

2. Special Geometric Orientations

  • Perpendicular Planes ($\Pi_1 \perp \Pi_2$): $$\mathbf{A_1 A_2 + B_1 B_2 + C_1 C_2 = 0}$$
  • Parallel Planes ($\Pi_1 \parallel \Pi_2$): $$\mathbf{\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \ne \frac{D_1}{D_2}}$$
  • Coincident Planes: The ratios of all four coefficients are identical.

ยง2.4 Perpendicular Distance of a Point & Dihedral Bisector Planes

1. Distance from a Point $P_1(x_1, y_1, z_1)$ to a Plane

Let the plane be $\Pi: Ax + By + Cz + D = 0$. The perpendicular distance $d$ from $P_1$ to $\Pi$ is given by:

$$\mathbf{d = \frac{|A x_1 + B y_1 + C z_1 + D|}{\sqrt{A^2 + B^2 + C^2}}}$$

Proof: Translate the coordinate origin to $P_1(x_1, y_1, z_1)$ via $x = x' + x_1, y = y' + y_1, z = z' + z_1$. The equation becomes $Ax' + By' + Cz' + (Ax_1 + By_1 + Cz_1 + D) = 0$. In this frame, the distance from the new origin $(0, 0, 0)$ is the constant term divided by the normal norm. $\blacksquare$

2. Distance Between Parallel Planes

For parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$:

$$\mathbf{d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}}$$

3. Equations of the Dihedral Bisector Planes

The locus of points equidistant from two planes $\Pi_1$ and $\Pi_2$ forms two orthogonal bisector planes:

$$\mathbf{\frac{A_1 x + B_1 y + C_1 z + D_1}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \pm \frac{A_2 x + B_2 y + C_2 z + D_2}{\sqrt{A_2^2 + B_2^2 + C_2^2}}}$$

To distinguish the acute from the obtuse bisector: make $D_1, D_2 > 0$. If $A_1 A_2 + B_1 B_2 + C_1 C_2 > 0$, the positive sign gives the obtuse bisector and the negative sign gives the acute bisector.

ยง2.5 Pencils of Planes & Common Line Intersections

1. Family (Pencil) of Planes Passing Through the Intersection of Two Planes

Let $\Pi_1: A_1 x + B_1 y + C_1 z + D_1 = 0$ and $\Pi_2: A_2 x + B_2 y + C_2 z + D_2 = 0$ be two non-parallel planes. Their intersection is a straight line $L$. Any plane passing through this line $L$ is represented by the linear combination parameter $\lambda \in \mathbb{R}$:

$$\mathbf{\Pi_1 + \lambda \Pi_2 = 0 \Longleftrightarrow (A_1 + \lambda A_2)x + (B_1 + \lambda B_2)y + (C_1 + \lambda C_2)z + (D_1 + \lambda D_2) = 0}$$

2. Geometric Determination of the Parameter $\lambda$

The parameter $\lambda$ is uniquely fixed by imposing an extra geometric constraint, such as:

  1. Passing through a fourth point $P_0(x_0, y_0, z_0)$.
  2. Being perpendicular to a third plane $\Pi_3: A_3 x + B_3 y + C_3 z + D_3 = 0$ via $(A_1 + \lambda A_2)A_3 + (B_1 + \lambda B_2)B_3 + (C_1 + \lambda C_2)C_3 = 0$.
  3. Being parallel to a given line with direction ratios $(l, m, n)$.
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 2.1: Parallel Plane Equation and Inter-Planar Distance

(a) Find the equation of the plane passing through $P(2, -1, 3)$ that is parallel to the plane $3x - 4y + 5z = 12$. (b) Find the perpendicular distance between these two parallel planes.

Part (a): Find the Parallel Plane

Any plane parallel to $3x - 4y + 5z = 12$ has the exact same normal vector $(3, -4, 5)$, so its equation is:

$$3x - 4y + 5z = D$$

Since it passes through $P(2, -1, 3)$:

$$D = 3(2) - 4(-1) + 5(3) = 6 + 4 + 15 = 25$$

The equation of the parallel plane is:

$$\mathbf{3x - 4y + 5z = 25 \quad \Longleftrightarrow \quad 3x - 4y + 5z - 25 = 0}$$

Part (b): Distance Between Parallel Planes

Rewrite both planes in standard form: $3x - 4y + 5z - 12 = 0$ ($D_1 = -12$) and $3x - 4y + 5z - 25 = 0$ ($D_2 = -25$).

$$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|(-12) - (-25)|}{\sqrt{3^2 + (-4)^2 + 5^2}} = \frac{|13|}{\sqrt{9 + 16 + 25}} = \frac{13}{\sqrt{50}} = \mathbf{\frac{13}{5\sqrt{2}} = \frac{13\sqrt{2}}{10}}$$
Final Answer & Physical Insight

Plane: $3x - 4y + 5z = 25$; Distance $d = \frac{13\sqrt{2}}{10} \approx 1.838$.

Tier 2: Intermediate Exam Example 2.2: Dihedral Bisector Planes & Acute/Obtuse Classification

Find the equations of the bisector planes of the dihedral angles between:

$$\Pi_1: x + 2y + 2z - 9 = 0 \quad \text{and} \quad \Pi_2: 4x - 4y + 2z + 3 = 0$$

Identify which equation bisects the acute angle.

Step 1: Normalizing the Plane Equations

For $\Pi_1$: $\sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.

For $\Pi_2$: $\sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.

Step 2: Formulate the Bisector Plane Equations

$$\frac{x + 2y + 2z - 9}{3} = \pm \frac{4x - 4y + 2z + 3}{6}$$

Multiplying both sides by $6$:

$$2(x + 2y + 2z - 9) = \pm (4x - 4y + 2z + 3)$$

Case 1 ($+$ sign):

$$2x + 4y + 4z - 18 = 4x - 4y + 2z + 3 \implies 2x - 8y - 2z + 21 = 0$$

Case 2 ($-$ sign):

$$2x + 4y + 4z - 18 = -4x + 4y - 2z - 3 \implies 6x + 6z - 15 = 0 \implies 2x + 2z - 5 = 0$$

Step 3: Test for Acute vs Obtuse Bisector

Make both constant terms positive by multiplying $\Pi_1$ by $-1$:

$$\Pi_1: -x - 2y - 2z + 9 = 0, \qquad \Pi_2: 4x - 4y + 2z + 3 = 0$$

Compute the dot product of normals: $A_1 A_2 + B_1 B_2 + C_1 C_2 = (-1)(4) + (-2)(-4) + (-2)(2) = -4 + 8 - 4 = 0$.

Since the normal dot product is identically $0$, the original two planes are mutually perpendicular ($\theta = 90^\circ$)!

Therefore, both bisectors bisect right angles of $45^\circ$.

Final Answer & Physical Insight

Bisector planes: $2x - 8y - 2z + 21 = 0$ and $2x + 2z - 5 = 0$; original planes are orthogonal ($90^\circ$), so both bisect at $45^\circ$.

Tier 3: Honors / Proof Challenge Example 2.3: Variable Intercept Plane & The Circumscribing Sphere Center Locus

A variable plane passes through a fixed point $P(a, b, c)$ ($abc \ne 0$) and intersects the three coordinate axes at points $A, B, C$. Prove that the locus of the center of the sphere circumscribing the tetrahedron $OABC$ is given by:

$$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$$

Step 1: Set up the Variable Plane

Let the intercepts of the variable plane on the coordinate axes be $(\alpha, 0, 0)$, $(0, \beta, 0)$, and $(0, 0, \gamma)$.

The equation of the plane in intercept form is:

$$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$$

Since the plane passes through the fixed point $(a, b, c)$:

$$\mathbf{\frac{a}{\alpha} + \frac{b}{\beta} + \frac{c}{\gamma} = 1} \quad \text{--- (Equation 1)}$$

Step 2: Sphere Passing Through $O, A, B, C$

A general sphere equation in $\mathbb{R}^3$ is:

$$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$

Since the sphere passes through the origin $O(0, 0, 0)$, $d = 0$.

Since it passes through $A(\alpha, 0, 0)$: $\alpha^2 + 2u\alpha = 0 \implies 2u = -\alpha \implies u = -\frac{\alpha}{2}$.

Similarly, passing through $B(0, \beta, 0)$ and $C(0, 0, \gamma)$ gives $v = -\frac{\beta}{2}$ and $w = -\frac{\gamma}{2}$.

The equation of the sphere is:

$$x^2 + y^2 + z^2 - \alpha x - \beta y - \gamma z = 0$$

Step 3: Center of the Sphere and Elimination

The center $(X, Y, Z)$ of this sphere is:

$$X = -u = \frac{\alpha}{2}, \qquad Y = -v = \frac{\beta}{2}, \qquad Z = -w = \frac{\gamma}{2}$$

Thus, the intercepts are expressed in terms of the center coordinates as:

$$\alpha = 2X, \qquad \beta = 2Y, \qquad \gamma = 2Z$$

Substituting $\alpha, \beta, \gamma$ into Equation 1:

$$\frac{a}{2X} + \frac{b}{2Y} + \frac{c}{2Z} = 1 \implies \mathbf{\frac{a}{X} + \frac{b}{Y} + \frac{c}{Z} = 2}$$

Replacing $(X, Y, Z)$ with current coordinates $(x, y, z)$ establishes the required locus:

$$\mathbf{\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2} \quad \blacksquare$$
Final Answer & Physical Insight

Proved: The locus of the sphere center is $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$.