3.18 Intersections Among Lines and Planes (HL)

1. Intersections Between a Line and a Plane

The spatial interaction spanning a 1D vector line and a 2D geometric plane yields three distinct potential outcomes dependent entirely on parallelism and shared spatial coordinates.

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

Spatial Condition Evaluation Methodology
Intersecting Point Substitute the parameterized line coordinates $(x, y, z)$ cleanly into the plane equation $Ax + By + Cz = D$. Solving determines a unique scalar $\lambda$.
Strictly Parallel Verify perpendicularity between the line's direction vector and the plane's normal vector ($\vec{b} \cdot \vec{n} = 0$). If anchor point $\vec{a}$ fails the plane equation, it runs parallel without intersection.
Line Lies on Plane Evaluate identically to the parallel condition ($\vec{b} \cdot \vec{n} = 0$). If the structural anchor point $\vec{a}$ successfully satisfies the plane equation, the entire line rests infinitely on the surface.

Angle Between Line and Plane: Let $\phi$ designate the angle tracing between the direction vector $\vec{b}$ and normal vector $\vec{n}$. The true physical angle $\theta$ between the line and the plane surface acts strictly as the complement ($\theta = 90^\circ - \phi$). Thus, $\sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.

EXAMPLE 1 (Point of Intersection)

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

Parametric components establish points mapping $(1+4\lambda, 2+5\lambda, 3+6\lambda)$.
Insert directly into the targeted plane constraint:
$2(1+4\lambda) + 5(2+5\lambda) - 3(3+6\lambda) = 18$.
Expand structurally: $2 + 8\lambda + 10 + 25\lambda - 9 - 18\lambda = 18$.
Condense variables: $15\lambda + 3 = 18 \Rightarrow 15\lambda = 15 \Rightarrow \lambda = 1$.
Inserting $\lambda=1$ resolves the spatial intersection node: $\mathbf{(5, 7, 9)}$.

EXAMPLE 2 & 3 (Parallel and Incident Cases)

Example 2 (Parallel): Line $L: \vec{r} = \binom{1}{2}{3} + \lambda\binom{4}{5}{6}$ against Plane $\Pi: 2x + 2y - 3z = 1$.
Substituting parametrics yields: $2(1+4\lambda) + 2(2+5\lambda) - 3(3+6\lambda) = 1 \Rightarrow 0\lambda - 1 = 1 \Rightarrow 0\lambda = 2$.
The variable collapses to an impossible mathematical state, proving zero intersection boundaries.
Example 3 (Incident): Line $L: \vec{r} = \binom{1}{2}{3} + \lambda\binom{4}{5}{6}$ against Plane $\Pi: 2x + 2y - 3z = -3$.
Substitution resolves identically to $0\lambda = 0$. This universal truth dictates the geometric line is perfectly embedded within the plane structure.

2. Intersections Between Two Planes

Two distinct infinite geometric planes must either act parallel or intersect continuously to carve an infinite straight 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 structural normal vectors act as scalar multiples ($\vec{n}_1 \parallel \vec{n}_2$).
  • Intersecting Line: Formulates a new vector equation. The direction vector tracing the intersection relies geometrically on the cross product $\vec{b} = \vec{n}_1 \times \vec{n}_2$.
  • Intersection Angle: Equivalent to the interior angle bounded between their individual normal vectors ($\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$).

EXAMPLE 4 (Formulating the Intersection Line)

Evaluate the interaction mapping planes $x + 2y + 3z = 6$ and $4x + 5y + 6z = 15$.

Angle evaluation: Using $\vec{n}_1 = \binom{1}{2}{3}$ and $\vec{n}_2 = \binom{4}{5}{6}$.
$\cos\theta = \frac{4+10+18}{\sqrt{14}\sqrt{77}} = \frac{32}{\sqrt{1078}} \approx 0.974 \Rightarrow \mathbf{\theta \approx 12.9^\circ}$.
Intersection Line mapping (Method A - Two Points):
Force $z=0$ structurally: System evaluates $x+2y=6$ and $4x+5y=15$. Solving yields coordinate $A(0,3,0)$.
Force $z=1$ structurally: System evaluates $x+2y=3$ and $4x+5y=9$. Solving yields coordinate $B(1,1,1)$.
Construct directional path $\vec{AB} = \binom{1}{-2}{1}$. Equation builds exactly to $\mathbf{\vec{r} = \binom{0}{3}{0} + \lambda\binom{1}{-2}{1}}$.
Intersection Line mapping (Method B - Cross Product):
Determine direction utilizing normal cross sections: $\vec{b} = \vec{n}_1 \times \vec{n}_2 = \binom{1}{2}{3} \times \binom{4}{5}{6} = \binom{-3}{6}{-3}$. Because this acts as a perfect scalar multiple of $\binom{1}{-2}{1}$, it validates identical trajectory routing alongside a single mutual anchor point.

3. Intersections Among Three Planes

Introducing a third plane triggers a complex simultaneous $3 \times 3$ algebraic matrix. General states evaluate according to the resultant mathematical validity:

  • Unique Solution $(x,y,z)$: The three rigid planes intersect successfully at exactly one singular coordinate point.
  • Infinitely Many Solutions: The systemic planes intersect seamlessly sharing a continuous infinite line, or act purely coincident.
  • No Solution: The structural planes assemble a hollow triangular geometric prism or evaluate parallel exclusions.

EXAMPLE 7

Evaluate the simultaneous intersection states formatting $2x + 3y + 3z = 3$, $x + y - 2z = 4$, and $5x + 7y + 4z = 10$.

Utilizing standard Gauss elimination or integrated GDC tools outputs a dependent relationship highlighting infinitely many valid configurations driven by a free variable $z$.
Parametric outputs construct: $x = 9 + 9\lambda$, $y = -5 - 7\lambda$, $z = \lambda$.
This inherently maps a complete mathematical intersection line represented safely by $\mathbf{\vec{r} = \binom{9}{-5}{0} + \lambda\binom{9}{-7}{1}}$.