3.17 PlanesHL ONLY

1. Vector Equation of a Plane

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

The position vector $\vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ of any point $P(x,y,z)$ on the plane is given 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 $\lambda, \mu$ are scalar parameters ($\lambda, \mu \in \mathbb{R}$).

Derivation: If $P(x,y,z)$ is a point on the plane, the vector $\overrightarrow{AP}$ lies on the plane. Therefore, $\overrightarrow{AP}$ can be written as a linear combination of the direction vectors: $\overrightarrow{AP} = \lambda\vec{b} + \mu\vec{c}$. The position vector of $P$ is: $\vec{r} = \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.

2. Parametric and Cartesian Forms

Expanding the vector equation by components yields the Parametric Equations:

$x = a_1 + \lambda b_1 + \mu c_1$
$y = a_2 + \lambda b_2 + \mu c_2$
$z = a_3 + \lambda b_3 + \mu c_3$

Eliminating the parameters $\lambda$ and $\mu$ from the parametric equations yields the linear Cartesian Equation:

$Ax + By + Cz = D$

EXAMPLE 1 (Elimination Process)

Find the Cartesian equation of the plane passing through $A(1,2,3)$ parallel to vectors $\vec{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$ and $\vec{c} = \begin{pmatrix} 7 \\ 8 \\ 8 \end{pmatrix}$.

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) $z = 3 + 6\lambda + 8\mu$

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

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

3. Vector Equation in Normal Form

A plane can be defined using a point $A(a_1, a_2, a_3)$ and a normal vector $\vec{n} = \begin{pmatrix} A \\ B \\ C \end{pmatrix}$ perpendicular to the plane.

Because any vector $\overrightarrow{AP}$ on the plane is perpendicular to the normal vector $\vec{n}$, their dot product is zero ($\overrightarrow{AP} \cdot \vec{n} = 0$). Substituting position vectors: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \implies \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$. This gives the Normal Equation Form:

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

Link to Cartesian Form: This dot product yields the Cartesian form $Ax + By + Cz = D$. The left side expands to $\vec{r} \cdot \vec{n} = xA + yB + zC$. The right side $\vec{a} \cdot \vec{n}$ is a constant scalar, $D$.

EXAMPLE 2 (Using the Normal Vector)

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

Step 1: Use the 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 side is $-8x + 10y - 3z$.

Step 3: The right side constant is $-8(1) + 10(2) - 3(3) = 3$.

Result: The Cartesian form is $\mathbf{-8x + 10y - 3z = 3}$.

Geometric Note: The normal vector can be found by taking the cross product of the two direction 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 $3x - 2y + z = 6$.

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

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

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

(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 using the 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)

Find the Cartesian equation of the plane $\vec{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}$.

Step 1: Identify the two direction vectors: $\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{c} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}$.

Step 2: Find the normal vector using the cross product:

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

Step 3: Write the Cartesian equation:

$$ 4x + 13y - 10z = 4(3) + 13(1) - 10(2) = 5 $$