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Wize University Linear Algebra Textbook > Equations of Lines and Planes

Intersection of a Line and a Plane

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Intersection of Lines and Planes

Case 1: Infinitely Many POIs

The line lies on the plane   ⟹  \implies all points on the line are on the plane.


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Case 2: No POIs

The line lies on a different parallel plane   ⟹  \implies no points of intersection

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Case 3: One POI

  • The line does not lie on any parallel plane   ⟹  \implies one point of intersection

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Steps to Find Point(s) of Intersection

1. Substitute the line's parametric equations into the standard form of the plane
2. Solve for the parameter:
  • A single valid solution   ⟹  \implies one point of intersection
  • No consistent solution   ⟹  \implies no intersection (line is on a different, parallel plane)
  • Redundant equations   ⟹  \implies infinitely many points of intersection (line lies on the plane)
3. If there is a valid solution, substitute the value of the parameter back into the equation of the line to find the point of intersection.
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Example: Intersection of Line and Plane

Find the point of intersection between the line and the plane:
l: x⃗=⟨1,1,0⟩+t⟨2,−1,3⟩l :\ \vec x = \lang 1,1,0\rang + t\lang 2,−1,3\rang
Π: 2𝑥+3𝑦−𝑧=9\Pi:\ 2𝑥+3𝑦−𝑧=9
Write the line in parametric form:
l: {x=1+2ty=1−tz=3tl:\ \begin{cases} x=1+2t\\ y=1-t\\ z=3t \end{cases}
Substitute these into the equation of the plane:
2(1+2t)+3(1−t)−(3t)=92+4t+3−3t−3t=95−2t=9−2t=4  ⟹  t=−2\begin{aligned} 2(1+2t)+3(1−t)−(3t) &= 9\\ 2+4t+3-3t-3t &= 9\\ 5-2t &= 9\\ -2t &= 4\\ \implies t &= -2 \end{aligned}
Therefore there is a valid solution, and we find the point of intersection by substituting t=−2\colorTwo{t=-2} into the line:
⟨1,1,0⟩+(−2)⟨2,−1,3⟩=⟨−3,3,−6⟩\lang 1,1,0\rang + \colorTwo{(-2)}\lang 2,−1,3\rang = \boxed{\lang -3,3,-6\rang}
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Example: Closest Point on a Plane

Find the point on the plane Π: x−3y=12\Pi:\ x-3y=12 that is closest to the point A(11,−2,3)A(11,-2,3).

Let's call the unknown point PP.
If PP is the point on the plane that is closest to AA, then the vector PA→\overrightarrow{PA} has to be orthogonal to the plane.
We can create a line passing through PP and AA, and use the plane's normal vector n⃗=⟨1,−3,0⟩\vec{n}=\lang 1,-3,0 \rang as its direction vector:
l: ⟨x,y,z⟩=⟨11,−2,3⟩+t⟨1,−3,0⟩l: \ \lang x,y,z\rang = \lang 11,-2,3\rang + t\lang 1,-3,0\rang
Now we simply find the intersection of this line and the given plane!

Intersection of Line & Plane

Converting the line to parametric form:
{x=11+ty=−2−3tz=3\begin{cases} x=11+t\\ y=-2-3t\\ z=3 \end{cases}
Substituting the parametric equations into the equation of the plane:
x−3y=12(11+t)−3(−2−3t)=1211+t+6+9t=1210t=−5  ⟹  t=−12\begin{array}{rcl} x-3y&=&12 \\[0.5em] (11+t)-3(-2-3t)&=&12\\[0.5em] 11+t+6+9t&=&12\\[0.5em] 10t&=&-5\\[0.5em] \implies t&=&-\dfrac{1}{2} \end{array}
Now we find the coordinates of PP, the point on the plane closest to AA, by plugging t=−12t=-\frac{1}{2} into the line's equation:
P⃗=⟨11,−2,3⟩+(−12)⟨1,−3,0⟩=⟨11,−2,3⟩+⟨−12,32,0⟩=⟨212,−12,3⟩\begin{aligned} \vec P &=\lang 11,-2,3\rang + \left(-\textstyle\dfrac{1}{2}\right)\lang 1,-3,0\rang \\[1em] &= \lang 11,-2,3\rang + \left\lang -\textstyle\dfrac{1}{2}, \dfrac{3}{2},0\right\rang \\[1em] &= \boxed{\left\lang \textstyle\dfrac{21}{2},-\dfrac{1}{2},3\right\rang} \end{aligned}

Practice: Intersection of Lines & Planes

Find the point(s) of intersection between the following line and plane:
l:{x=1+ty=1−tz=3+2tl:\begin{cases} x=1+t\\y=1-t\\z=3+2t \end{cases}
Π: x+y+z=3\Pi: \ x+y+z=3

Practice: Intersection of Lines & Planes

Find the point of intersection of the line and the plane:
l: x⃗=⟨1,2,0⟩+t⟨2,1,1⟩l: \ \vec x = \lang 1,2,0\rang + t\lang 2,1,1\rang
Π: x+y−3z=0\Pi: \ x+y−3z=0
Extra Practice
Find the intersection of the line L and the plane P given below
𝐿:(1,1,0)+𝑡(1,0,−2)𝐿:(1,1,0)+𝑡(1,0,−2)
𝑃:𝑥+𝑦+𝑧=3𝑃:𝑥+𝑦+𝑧=3
a) Write the parametric equation of line
L:{x+2y=0x−z=5L: \begin{cases} x+2y=0\\ x-z=5 \end{cases}
b) Find intersection point of the line L with the plane P:3y−3z=2.P:3y-3z=2.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.3
Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2} and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.},
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.1
Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2} and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.},
Practice: Intersection of lines and planes
Practice: Intersection of lines and planes
Find the point of intersection of the line L: [1,2,0]+𝑡[−1,1,1]L:\ \left[1,2,0\right]+𝑡\left[−1,1,1\right] and the plane Π: 𝑥+𝑦−3𝑧=0.\Pi:\ 𝑥+𝑦−3𝑧=0.
Intersection of Lines and Planes
Practice: Intersection of Lines & Planes
Find the point of intersection of the line and the plane:
l: x⃗=⟨1,2,0⟩+t⟨2,1,1⟩l: \ \vec x = \lang 1,2,0\rang + t\lang 2,1,1\rang
Intersection of Lines and Planes
Practice: Intersection of Lines & Planes
Find the point(s) of intersection between the following line and plane:
l:{x=1+ty=1−tz=3+2tl:\begin{cases} x=1+t\\y=1-t\\z=3+2t \end{cases}