3.17 Planes (HL)

1. Vector Equation of a Plane

A plane in 3D space is uniquely determined by a known coordinate point $A(a_1, a_2, a_3)$ resting on the plane, and two non-parallel direction vectors $\vec{b}$ and $\vec{c}$ that lie parallel to the plane.

The position vector $\vec{r} = \binom{x}{y}{z}$ mapping any arbitrary point $P(x,y,z)$ on this designated plane is governed by the vector equation:

$\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$

Where $\vec{a}$ is the position vector of point A ($\vec{OA}$), and $\lambda, \mu$ act as two independent scalar parameters ($\lambda, \mu \in \mathbb{R}$).

Derivation: The spatial pathway to point $P$ follows $\vec{OP} = \vec{OA} + \vec{AP}$. Because the vector $\vec{AP}$ is restricted entirely to the plane formed by $\vec{b}$ and $\vec{c}$, it can be expressed universally as a linear combination $\lambda\vec{b} + \mu\vec{c}$.

2. Parametric and Cartesian Forms

Isolating the single vector equation into its foundational axes yields the Parametric Equations:

$x = a_1 + \lambda b_1 + \mu c_1$

$y = a_2 + \lambda b_2 + \mu c_2$

$z = a_3 + \lambda b_3 + \mu c_3$

By systematically eliminating the dual parameters $\lambda$ and $\mu$ using algebraic substitution, the continuous Cartesian Equation is derived:

$Ax + By + Cz = D$

EXAMPLE 1 (Elimination Process)

Define the plane passing through $A(1,2,3)$ running parallel to vectors $\vec{b} = \binom{4}{5}{6}$ and $\vec{c} = \binom{7}{8}{8}$. Formulate the Cartesian boundary.

Vector Form: $\vec{r} = \binom{1}{2}{3} + \lambda\binom{4}{5}{6} + \mu\binom{7}{8}{8}$.

Parametric System:
(1) $x = 1 + 4\lambda + 7\mu$
(2) $y = 2 + 5\lambda + 8\mu$
(3) $z = 3 + 6\lambda + 8\mu$

Algebraic Elimination:
Eliminate $\lambda$ from (1) and (2) using $5 \times (1) - 4 \times (2)$: $5x - 4y = -3 + 3\mu$ (Equation 4).
Eliminate $\lambda$ from (2) and (3) using $6 \times (2) - 5 \times (3)$: $6y - 5z = -3 + 8\mu$ (Equation 5).
Eliminate $\mu$ from (4) and (5) using $8 \times (4) - 3 \times (5)$: $40x - 32y - 18y + 15z = -24 + 9$.
Condensing terms maps to $40x - 50y + 15z = -15$.

Dividing universally by $-5$ yields the definitive Cartesian form: $\mathbf{-8x + 10y - 3z = 3}$.

3. Vector Equation in Normal Form

A plane can be defined more efficiently using a single point $A(a_1, a_2, a_3)$ and a normal vector $\vec{n} = \binom{A}{B}{C}$ that sits perfectly perpendicular to the entire plane surface.

Because any line segment $\vec{AP}$ residing inside the plane is strictly perpendicular to the normal vector $\vec{n}$, their dot product equals zero ($\vec{AP} \cdot \vec{n} = 0$). Substituting position vectors $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ resolves to the Normal Equation Form:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

Notice: This instantly constructs the Cartesian format $Ax + By + Cz = D$, where the coefficients $A, B, C$ are mapped strictly from the normal vector, and the constant $D$ is the evaluated scalar dot product $\vec{a} \cdot \vec{n}$.

EXAMPLE 2 (Utilizing the Normal Vector)

Determine the Cartesian plane equation traversing $A(1,2,3)$ structured strictly perpendicular to normal vector $\vec{n} = \binom{-8}{10}{-3}$.

Utilizing the relation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

The left-hand structure builds immediately to $-8x + 10y - 3z$.

The right-hand constant computes algebraically to $-8(1) + 10(2) - 3(3) = -8 + 20 - 9 = 3$.

The final Cartesian form evaluates identically to Example 1: $\mathbf{-8x + 10y - 3z = 3}$.

Geometric Note: The normal vector required in Example 2 can always be generated independently by cross-multiplying the two parallel vectors from Example 1: $\vec{n} = \vec{b} \times \vec{c} = \binom{4}{5}{6} \times \binom{7}{8}{8} = \binom{-8}{10}{-3}$.

EXAMPLE 3 & 4 (Analyzing Plane Characteristics)

Example 3 (From Cartesian): Analyze the plane $3x - 2y + z = 6$.
The normal vector extracts cleanly from coefficients: $\vec{n} = \binom{3}{-2}{1}$.
Three coordinate points are discovered by forcing axis zeros: $A(2,0,0)$, $B(0,-3,0)$, and $C(0,0,6)$.
Parallel plane vectors formulate geometrically via point differences: $\vec{b} = \vec{AB} = \binom{-2}{-3}{0}$ and $\vec{c} = \vec{AC} = \binom{-2}{0}{6}$.

Example 4 (From Parametric): Analyze $\vec{r} = \binom{3}{1}{2} + \lambda\binom{1}{2}{3} + \mu\binom{5}{0}{2}$.
Two intrinsic parallel vectors exist: $\vec{b} = \binom{1}{2}{3}$ and $\vec{c} = \binom{5}{0}{2}$.
The normal vector evaluates via cross product: $\vec{n} = \vec{b} \times \vec{c} = \binom{4}{13}{-10}$.
Computing the Cartesian structure: $4x + 13y - 10z = 4(3) + 13(1) - 10(2) = \mathbf{5}$.