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 $\vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ 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 ($\overrightarrow{OA}$), and $\lambda, \mu$ act as two independent scalar parameters ($\lambda, \mu \in \mathbb{R}$).

Derivation: If $P(x,y,z)$ is any arbitrary point on the plane, the vector $\overrightarrow{AP}$ lies entirely within the plane defined by $\vec{b}$ and $\vec{c}$. Hence, $\overrightarrow{AP}$ can be written as a linear combination of the two direction vectors: $\overrightarrow{AP} = \lambda\vec{b} + \mu\vec{c}$. The position vector of $P$ is then determined by: $\vec{r} = \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.

2. Parametric and Cartesian Forms

Expanding the single vector equation component-by-component 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 from the parametric equations, a single linear 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 $\vec{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$ and $\vec{c} = \begin{pmatrix} 7 \\ 8 \\ 8 \end{pmatrix}$. Formulate the Cartesian boundary.

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

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 $5 \times (1) - 4 \times (2)$ to get Equation 4. Then eliminate $\lambda$ from (2) and (3) using $6 \times (2) - 5 \times (3)$ to get Equation 5. Finally, eliminate $\mu$.

$$\begin{aligned} 5x - 4y &= -3 + 3\mu \quad &\text{--- (Eq. 4)} \\ 6y - 5z &= -3 + 8\mu \quad &\text{--- (Eq. 5)} \\ 8(5x - 4y) - 3(6y - 5z) &= 8(-3 + 3\mu) - 3(-3 + 8\mu) \\ 40x - 32y - 18y + 15z &= -24 + 24\mu + 9 - 24\mu \\ 40x - 50y + 15z &= -15 \\ \mathbf{-8x + 10y - 3z} &\mathbf{= 3} \quad &\text{(Dividing by -5)} \end{aligned}$$

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 $\vec{n} = \begin{pmatrix} A \\ B \\ C \end{pmatrix}$ that sits perfectly perpendicular to the entire plane surface.

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

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

Link to Cartesian Form: This vector dot product directly yields the Cartesian form $Ax + By + Cz = D$. The LHS computes as $\vec{r} \cdot \vec{n} = xA + yB + zC$. The RHS $\vec{a} \cdot \vec{n}$ evaluates strictly to a constant scalar, denoted as $D$.

EXAMPLE 2 (Utilizing the Normal Vector)

Find the equation of the plane passing through $A(1,2,3)$ which is strictly perpendicular to the normal vector $\vec{n} = \begin{pmatrix} -8 \\ 10 \\ -3 \end{pmatrix}$.

Step 1: Utilize the normal relation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} -8 \\ 10 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -8 \\ 10 \\ -3 \end{pmatrix}$

Step 2: The left-hand structure builds immediately to $-8x + 10y - 3z$.

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

Result: 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: $\vec{n} = \vec{b} \times \vec{c} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} \times \begin{pmatrix} 7 \\ 8 \\ 8 \end{pmatrix} = \begin{pmatrix} -8 \\ 10 \\ -3 \end{pmatrix}$.

4. Analyzing Plane Characteristics

EXAMPLE 3 (From Cartesian to Vector Components)

Consider the plane mathematically bounded by $3x - 2y + z = 6$.

(a) Find a normal vector $\vec{n}$:
The coefficients of $x, y, z$ directly provide the normal vector:

$\vec{n} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}$

(b) Find three points on the plane:
For $y = 0, z = 0 \implies 3x = 6 \implies x = 2 \implies \mathbf{A(2, 0, 0)}$.
For $x = 0, z = 0 \implies -2y = 6 \implies y = -3 \implies \mathbf{B(0, -3, 0)}$.
For $x = 0, y = 0 \implies z = 6 \implies \mathbf{C(0, 0, 6)}$.

(c) Find two vectors $\vec{b}$ and $\vec{c}$ parallel to the plane:

$\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 0 - 2 \\ -3 - 0 \\ 0 - 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ 0 \end{pmatrix} \quad \text{and} \quad \vec{c} = \overrightarrow{AC} = \begin{pmatrix} 0 - 2 \\ 0 - 0 \\ 6 - 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ 6 \end{pmatrix}$

(d) Confirm that $\vec{n} \perp \vec{b}$ and $\vec{n} \perp \vec{c}$ :
Verify mathematically utilizing the scalar dot product:

$\vec{n} \cdot \vec{b} = 3(-2) + (-2)(-3) + 1(0) = -6 + 6 + 0 = 0 \quad (\checkmark)$
$\vec{n} \cdot \vec{c} = 3(-2) + (-2)(0) + 1(6) = -6 + 0 + 6 = 0 \quad (\checkmark)$

(e) Write down all forms of equation for this plane:

Vector form: $\vec{r} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ -3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 0 \\ 6 \end{pmatrix}$

Parametric form: $x = 2 - 2\lambda - 2\mu, \quad y = -3\lambda, \quad z = 6\mu$

Normal form: $\vec{r} \cdot \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} = 6$

EXAMPLE 4 (From Parametric to Cartesian)

Analyze the plane structured by $\vec{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}$.

Step 1: Two intrinsic parallel vectors inherently exist: $\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{c} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}$.

Step 2: The plane's normal vector evaluates strictly via cross product:

$\vec{n} = \vec{b} \times \vec{c} = \begin{pmatrix} 4 \\ 13 \\ -10 \end{pmatrix}$

Step 3: Computing the overarching Cartesian structure:

$$\begin{aligned} 4x + 13y - 10z &= 4(3) + 13(1) - 10(2) \\ 4x + 13y - 10z &= 12 + 13 - 20 \\ \mathbf{4x + 13y - 10z} &\mathbf{= 5} \end{aligned}$$