3.18 Intersections Among Lines and PlanesHL ONLY

1. Intersections Between a Line and a Plane

The interaction between a line and a plane yields three outcomes depending on parallelism and shared coordinates.

Given a line $L: \vec{r} = \vec{a} + \lambda\vec{b}$ and a plane $\Pi: Ax + By + Cz = D$ (with normal vector $\vec{n}$):

Condition	Methodology
Intersecting Point	Substitute the line coordinates $(x, y, z)$ into the plane equation $Ax + By + Cz = D$. Solving determines a unique scalar $\lambda$.
Parallel	Verify perpendicularity between the line's direction vector and the plane's normal vector ($\vec{b} \cdot \vec{n} = 0$). If anchor point $\vec{a}$ does not satisfy the plane equation, the line is parallel and does not intersect.
Line Lies on Plane	Check the parallel condition ($\vec{b} \cdot \vec{n} = 0$). If anchor point $\vec{a}$ satisfies the plane equation, the line lies on the plane.

Angle Between Line and Plane: Let $\phi$ be the acute angle between the direction vector $\vec{b}$ and normal vector $\vec{n}$. The angle $\theta$ between the line and the plane is the complement ($\theta = 90^\circ - \phi$).

$\sin\theta = \cos\phi = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

EXAMPLE 1 (Point of Intersection)

Find the intersection between line $L: \vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$ and plane $\Pi: 2x + 5y - 3z = 18$.

Step 1: The parametric coordinates are $(1+4\lambda, 2+5\lambda, 3+6\lambda)$.

Step 2: Substitute these coordinates into the plane equation:

$$\begin{aligned} 2(1+4\lambda) + 5(2+5\lambda) - 3(3+6\lambda) &= 18 \\ 2 + 8\lambda + 10 + 25\lambda - 9 - 18\lambda &= 18 \\ 15\lambda + 3 &= 18 \\ 15\lambda &= 15 \\ \lambda &= 1 \end{aligned}$$

Step 3: Substitute $\lambda=1$ back into the line equation to find the intersection point $\mathbf{(5, 7, 9)}$.

EXAMPLE 2 (Parallel Case)

Evaluate line $L: \vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$ and Plane $\Pi: 2x + 2y - 3z = 1$.

Substituting the parametric coordinates yields:

$$\begin{aligned} 2(1+4\lambda) + 2(2+5\lambda) - 3(3+6\lambda) &= 1 \\ 2 + 8\lambda + 4 + 10\lambda - 9 - 18\lambda &= 1 \\ (8 + 10 - 18)\lambda + (2 + 4 - 9) &= 1 \\ 0\lambda - 3 &= 1 \\ 0\lambda &= 4 \end{aligned}$$

This equation has no solution. The line is parallel to the plane and does not intersect.

EXAMPLE 3 (Line on Plane)

Evaluate line $L: \vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$ and Plane $\Pi: 2x + 2y - 3z = -3$.

Substitution is identical to Example 2, except for the right side:

$0\lambda - 3 = -3 \implies \mathbf{0\lambda = 0}$

This equation is always true. The line lies on the plane.

2. Intersections Between Two Planes

Two distinct planes either are parallel or intersect to form a line.

Given $\Pi_1: A_1 x + B_1 y + C_1 z = D_1$ and $\Pi_2: A_2 x + B_2 y + C_2 z = D_2$:

Parallel: The normal vectors are scalar multiples ($\vec{n}_1 \parallel \vec{n}_2$).
Intersecting Line: The direction vector of the intersection line is the cross product of the normal vectors: $\mathbf{\vec{b} = \vec{n}_1 \times \vec{n}_2}$.
Intersection Angle: This is the acute angle between their normal vectors ($\cos\theta = \dfrac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$).

EXAMPLE 4 (Formulating the Intersection Line)

Find the intersection of planes $x + 2y + 3z = 6$ and $4x + 5y + 6z = 15$.

Angle: Using normal vectors $\vec{n}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{n}_2 = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$.

$\cos\theta = \dfrac{4+10+18}{\sqrt{14}\sqrt{77}} = \dfrac{32}{\sqrt{1078}} \approx 0.974 \implies \mathbf{\theta \approx 12.9^\circ}$

Intersection Line (Method A - Two Points):
Let $z=0$: The system becomes $x+2y=6$ and $4x+5y=15$.

$$\begin{aligned} x &= 6 - 2y \\ 4(6 - 2y) + 5y &= 15 \\ 24 - 8y + 5y &= 15 \\ -3y &= -9 \implies y = 3 \\ x &= 6 - 2(3) \implies x = 0 \end{aligned}$$

Solving gives point $A(0,3,0)$.

Let $z=1$: The system becomes $x+2y=3$ and $4x+5y=9$.

$$\begin{aligned} x &= 3 - 2y \\ 4(3 - 2y) + 5y &= 9 \\ 12 - 8y + 5y &= 9 \\ -3y &= -3 \implies y = 1 \\ x &= 3 - 2(1) \implies x = 1 \end{aligned}$$

Solving gives point $B(1,1,1)$.
The direction vector is $\overrightarrow{AB} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$. The line equation is $\mathbf{\vec{r} = \begin{pmatrix} 0 \\ 3 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}}$.

Intersection Line (Method B - Cross Product):
Find the direction vector using the cross product of the normal vectors:

$\vec{b} = \vec{n}_1 \times \vec{n}_2 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$

Because this is a scalar multiple of $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$, it represents the same line.

3. Intersections Among Three Planes

Finding the intersection of three planes requires solving a $3 \times 3$ system of equations. The possible outcomes are:

Unique Solution: The planes intersect at one point $(x,y,z)$.
Infinitely Many Solutions: The planes intersect to form a line, or are coincident.
No Solution: The planes form a triangular prism or are parallel.

EXAMPLE 4 (Line Interaction via 3 Planes)

Find the intersection of $2x + 3y + 3z = 3$, $x + y - 2z = 4$, and $5x + 7y + 4z = 10$.

Step 1: Solving the system yields infinitely many solutions with a free variable $z = \lambda$.

Step 2: The parametric equations are:

$x = 9 + 9\lambda$
$y = -5 - 7\lambda$
$z = 0 + 1\lambda$

Step 3: This represents a line, written in vector form as:

$\vec{r} = \begin{pmatrix} 9 \\ -5 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 9 \\ -7 \\ 1 \end{pmatrix}$