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 $\mathbb{R}^3$
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 $k : 1$:
Substituting into the equation:
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:
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:
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 $p$:
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 $\pm\sqrt{A^2 + B^2 + C^2}$:
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:
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 $\vec{n}_2$:
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:
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 $Ax + By + Cz + D_2 = 0$:
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:
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 $\lambda \in \mathbb{R}$:
2. Geometric Determination of the Parameter $\lambda$
The parameter $\lambda$ is uniquely fixed by imposing an extra geometric constraint, such as:
- Passing through a fourth point $P_0(x_0, y_0, z_0)$.
- Being perpendicular to a third plane $\Pi_3: A_3 x + B_3 y + C_3 z + D_3 = 0$ via $(A_1 + \lambda A_2)A_3 + (B_1 + \lambda B_2)B_3 + (C_1 + \lambda C_2)C_3 = 0$.
- Being parallel to a given line with direction ratios $(l, m, n)$.
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.
(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:
Since it passes through $P(2, -1, 3)$:
The equation of the parallel plane is:
Part (b): Distance Between Parallel Planes
Rewrite both planes in standard form: $3x - 4y + 5z - 12 = 0$ ($D_1 = -12$) and $3x - 4y + 5z - 25 = 0$ ($D_2 = -25$).
Plane: $3x - 4y + 5z = 25$; Distance $d = \frac{13\sqrt{2}}{10} \approx 1.838$.
Find the equations of the bisector planes of the dihedral angles between:
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
Multiplying both sides by $6$:
Case 1 ($+$ sign):
Case 2 ($-$ sign):
Step 3: Test for Acute vs Obtuse Bisector
Make both constant terms positive by multiplying $\Pi_1$ by $-1$:
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 $45^\circ$.
Bisector planes: $2x - 8y - 2z + 21 = 0$ and $2x + 2z - 5 = 0$; original planes are orthogonal ($90^\circ$), so both bisect at $45^\circ$.
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:
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:
Since the plane passes through the fixed point $(a, b, c)$:
Step 2: Sphere Passing Through $O, A, B, C$
A general sphere equation in $\mathbb{R}^3$ is:
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:
Step 3: Center of the Sphere and Elimination
The center $(X, Y, Z)$ of this sphere is:
Thus, the intercepts are expressed in terms of the center coordinates as:
Substituting $\alpha, \beta, \gamma$ into Equation 1:
Replacing $(X, Y, Z)$ with current coordinates $(x, y, z)$ establishes the required locus:
Proved: The locus of the sphere center is $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$.