Wize High School Grade 12 Calculus Textbook > Intersections of Lines & Planes

Intersection of 3 Planes
Given any 3 planes in R3R^3R3, the planes can either 
all intersection -- the system of equations is consistent, or
don't all intersect -- the system of equations is inconsistent

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Consistent: 4 Possible Scenarios
1) The planes intersect at a single point → there is exactly one point of intersection.
2) The planes intersect in a single line → none of the planes are parallel or exactly 2 of the planes are parallel
3) The planes coincide → the intersection is the entire plane itself

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Inconsistent: 4 Possible Scenarios
1) The 3 planes are parallel and at least one is distinct → there are no points of intersection between the 3 planes
2) The planes form an "H" pattern → there are no points of intersection between the 3 planes
4) The planes form a triangular prism ("Δ\orange\DeltaΔ" patter) → there are no points of intersection between the 3 planes

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Finding Points of Intersection
To determine if the 3 planes intersect, we can use substitution or elimination to solve the system of 3 equations.
If we get an exact point (x, y, z)\left(x,\ y,\ z\right)(x, y, z) ➝ the 3 planes intersect at exactly one point
If we get an equation tha makes sense for any values of x, y, or z (ex. 0=00=00=0) ➝ the 3 planes either intersect in a line or a plane
If we get an equation that does make sense for any values of x, y, or z (ex. 3=03=03=0) ➝ the 3 planes don't intersect

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Example: Intersection of 3 Planes
Find the point(s) of intersection (if any) between the following 3 planes:
Π1: x+y+z=3\Pi_1:\ x+y+z=3Π1​: x+y+z=3
Π2: 2x−y−z=0\Pi_2:\ 2x-y-z=0Π2​: 2x−y−z=0
Π3: x−y+4z=4\Pi_3:\ x-y+4z=4Π3​: x−y+4z=4

Check if the planes are parallel
None of n1⃗=[1, 1, 1]\vec{n_1}=\left[1,\ 1,\ 1\right]n1​​=[1, 1, 1], n2⃗=[2, −1, −1]\vec{n_2}=\left[2,\ -1,\ -1\right]n2​​=[2, −1, −1], or n3⃗=[1, −1, 4]\vec{n_3}=\left[1,\ -1,\ 4\right]n3​​=[1, −1, 4] are scalar multiples of one another, so the planes are not parallel.

Solve for one variable in plane 1
Equation 1: x=3−y−zx=3-y-zx=3−y−z

Substitute this into both equations 2 and 3:
Equation 2: 2(3−y−z)−y−z=02\left(3-y-z\right)-y-z=02(3−y−z)−y−z=0 ➝ y+z=2y+z=2y+z=2
Equation 3: (3−y−z)−y+4z=4\left(3-y-z\right)-y+4z=4(3−y−z)−y+4z=4 ➝ −2y+3z=1-2y+3z=1−2y+3z=1

Now we need to solve these two equations
From the new Equation 2: y=2−zy=2-zy=2−z
Substitute this into the new equation 3: −2(2−z)+3z=1-2\left(2-z\right)+3z=1−2(2−z)+3z=1 ➝ 5z=55z=55z=5 ➝ z=1\boxed{z=1}z=1​

Back substitute z=1z=1z=1
−2y+3z=1-2y+3z=1−2y+3z=1
−2y+3(1)=1-2y+3\left(1\right)=1−2y+3(1)=1
y=1\boxed{y=1}y=1​

x=3−y−zx=3-y-zx=3−y−z
x=3−(1)−(1)x=3-(1)-(1)x=3−(1)−(1)
x=1\boxed{x=1}x=1​

Therefore, the planes intersect at an exact point (1,1,1)(1, 1, 1)(1,1,1)

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Example: Intersection of 3 Planes
Solve the following system of equations, then provide a geometric interpretation of the solution(s).
Π1: 2x−2y+2z=−0\Pi_1:\ 2x-2y+2z=-0Π1​: 2x−2y+2z=−0
Π2: x+z=1\Pi_2:\ x+z=1Π2​: x+z=1
Π3: 3x+y+3z=4\Pi_3:\ 3x+y+3z=4Π3​: 3x+y+3z=4

Check if the planes are parallel
None of n1⃗=[2,−2,2]\vec{n_1}=\left[2,-2,2\right]n1​​=[2,−2,2], n2⃗=[1,0,1]\vec{n_2}=\left[1,0,1\right]n2​​=[1,0,1], or n3⃗=[3,1,3]\vec{n_3}=\left[3,1,3\right]n3​​=[3,1,3] are scalar multiples of one another, so the planes are not parallel.

Solve for one variable in plane 1
Equation 1: x=y−zx=y-zx=y−z

Substitute this into both equations 2 and 3:
Equation 2: (y−z)+z=1\left(y-z\right)+z=1(y−z)+z=1 ➝ y=1\boxed{y=1}y=1​
Equation 3: 3(y−z)+y+3z=43\left(y-z\right)+y+3z=43(y−z)+y+3z=4 ➝ 4y=44y=44y=4 ➝ y=1\boxed{y=1}y=1​

Assign a parameter
Since zzz can take on any value, let's assign z=t\boxed{z=t}z=t​

Substitute the value y=1y=1y=1 and z=tz=tz=t into the equation x=y−zx=y-zx=y−z:
x=1−t\boxed{x=1-t}x=1−t​

Therefore, the planes intersect in a line r⃗=[1,1,0]+t[−1,0,1]\vec r=[1, 1, 0]+t[-1, 0, 1]r=[1,1,0]+t[−1,0,1]

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Example: Intersection of 3 Planes
Solve the following system of equations, then provide a geometric interpretation of the solutions(s).
Π1: x+y−2z=3\Pi_1:\ x+y-2z=3Π1​: x+y−2z=3
Π2: 2x+2y−4z=6\Pi_2:\ 2x+2y-4z=6Π2​: 2x+2y−4z=6
Π3: x+3y−z=0\Pi_3:\ x+3y-z=0Π3​: x+3y−z=0

Check if the planes are parallel
Π1\Pi_1Π1​ and Π2\Pi_2Π2​ are scalar multiples of one another, meaning that the planes are identical. 

Π3\Pi_3Π3​ has a normal vector that is not a scalar multiple of the normal vectors of Π1\Pi_1Π1​ and Π2\Pi_2Π2​, meaning that the three planes must intersect in a line.

Solve for one variable in plane 1
Equation 1: x=3−y+2zx=3-y+2zx=3−y+2z

Substitute this into equation 3:
(3−y+2z)+3y−z=0\left(3-y+2z\right)+3y-z=0(3−y+2z)+3y−z=0
3+2y+z=03+2y+z=03+2y+z=0
z=−3−2yz=-3-2yz=−3−2y

Assign a parameter
Let  y=t\boxed{y=t}y=t​

Substitute this into z=−3−2yz=-3-2yz=−3−2y:
z=−3−2t\boxed{z=-3-2t}z=−3−2t​

Substitute y=ty=ty=t and z=−3−2tz=-3-2tz=−3−2t into x=3−y+2zx=3-y+2zx=3−y+2z:
x=3−t+2(−3−2t)x=3-t+2(-3-2t)x=3−t+2(−3−2t)
x=−3−5t\boxed{x=-3-5t}x=−3−5t​

Therefore, the planes intersect in a line r⃗=[−3,0,−3]+t[−5,1,−2]\vec r=[-3,0,-3]+t[-5,1,-2]r=[−3,0,−3]+t[−5,1,−2]

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Example: Intersection of 3 Planes
Solve the following system of equations, then provide a geometric interpretation of the solutions(s).
Π1: x+y−2z=3\Pi_1:\ x+y-2z=3Π1​: x+y−2z=3
Π2: 2x+2y−4z=6\Pi_2:\ 2x+2y-4z=6Π2​: 2x+2y−4z=6
Π3: 4x+4y−8z=12\Pi_3:\ 4x+4y-8z=12Π3​: 4x+4y−8z=12

Check if the planes are parallel
Π1\Pi_1Π1​, Π2\Pi_2Π2​, and Π3\Pi_3Π3​ are scalar multiples of one another, meaning that the planes are identical. 

Therefore, the 3 planes intersect at infinitely many points. The intersection is the plane itself.

Practice: Intersection of 3 Planes
Solve the following system of equations, then provide a geometric interpretation of the solution(s).

Π1: x−y+z=4\Pi_1:\ x-y+z=4Π1​: x−y+z=4
Π2: 3x+y+z=6\Pi_2:\ 3x+y+z=6Π2​: 3x+y+z=6
Π3: 2x−4y+3z=11\Pi_3:\ 2x-4y+3z=11Π3​: 2x−4y+3z=11

The planes intersect at exactly one point (1,0,3)(1, 0 ,3)(1,0,3)

The planes intersect in the line r⃗=[1,0,3]+t[−1,1,2]\vec r=[1,0,3]+t[-1,1,2]r=[1,0,3]+t[−1,1,2]

The planes intersect in the line r⃗=[−1,1,2]+t[1,0,3]\vec r=[-1,1,2]+t[1,0,3]r=[−1,1,2]+t[1,0,3]

The planes are parallel and distinct, they do no intersect

The planes are identical, and the intersect is the entire plane itself

I don't know

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Example: Intersection of 3 Planes
Find the point(s) of intersection (if any) between the following 3 planes:
Π1: x+y−z=3\Pi_1:\ x+y-z=3Π1​: x+y−z=3
Π2: 2x+2y−2z=0\Pi_2:\ 2x+2y-2z=0Π2​: 2x+2y−2z=0
Π3: −3x−3y+3z=0\Pi_3:\ -3x-3y+3z=0Π3​: −3x−3y+3z=0

Check if the planes are parallel
Planes 1, 2, and 3 all have parallel normal vectors. However, only planes 2 and 3 have Cartesian equations that are scalar multiples of one another.

This manes that planes 2 and 3 are identical, and plane 1 is a distinct parallel plane.

Therefore, there are no points of intersection between the 3 planes.

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Example: Intersection of 3 Planes
Find the point(s) of intersection (if any) between the following 3 planes:
Π1: x+y−z=3\Pi_1:\ x+y-z=3Π1​: x+y−z=3
Π2: 2x+2y−2z=0\Pi_2:\ 2x+2y-2z=0Π2​: 2x+2y−2z=0
Π3: x+2y−z=1\Pi_3:\ x+2y-z=1Π3​: x+2y−z=1

Check if the planes are parallel
Planes 1 and 2 have parallel normal vectors but their Cartesian equations are not scalar multiples of one another. This manes that planes 1 and 2 are parallel but distinct.

Therefore, there are no points of intersection between the 3 planes.

Alternatively
Solve for one variable in plane 1: x=3−y+zx=3-y+zx=3−y+z

Substitute this into both equations 2 and 3:
Equation 2: 2(3−y+z)+2y−2z=02\left(3-y+z\right)+2y-2z=02(3−y+z)+2y−2z=0  ➝ 6=06=06=0
Equation 3: (3−y+z)+2y−z=1\left(3-y+z\right)+2y-z=1(3−y+z)+2y−z=1 ➝ y=−2y=-2y=−2

Since we get the equation 6=06=06=0 which doesn't make sense for any values of x, y, or z, we know that these 3 planes do not intersect.

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Example: Intersection of 3 Planes
Find the point(s) of intersection (if any) between the following 3 planes:
Π1: 2x+y+z=0\Pi_1:\ 2x+y+z=0Π1​: 2x+y+z=0
Π2: x−y−z=1\Pi_2:\ x-y-z=1Π2​: x−y−z=1
Π3: x+y+z=0\Pi_3:\ x+y+z=0Π3​: x+y+z=0

Check if the planes are parallel
None of the planes are parallel

Solve for one variable in plane 1:
Equation 1: y=−2x−zy=-2x-zy=−2x−z

Substitute this into both equations 2 and 3:
Equation 2: x−(−2x−z)−z=1x-\left(-2x-z\right)-z=1x−(−2x−z)−z=1  ➝ 3x=13x=13x=1  ➝  x=13x=\frac{1}{3}x=31​
Equation 3: x+(−2x−z)+z=0x+\left(-2x-z\right)+z=0x+(−2x−z)+z=0 ➝ x=0x=0x=0

Since we get different values of x, there are no points of intersection between the 3 planes.

Practice: Intersection of 3 Planes
Solve the following system of equations, then provide a geometric interpretation of the solution(s).

Π1: x−y+z=0\Pi_1:\ x-y+z=0Π1​: x−y+z=0
Π2: x+y+z=0\Pi_2:\ x+y+z=0Π2​: x+y+z=0
Π3: 2x+y+2z=1\Pi_3:\ 2x+y+2z=1Π3​: 2x+y+2z=1

The planes intersect at exactly one point (1,0,3)(1, 0 ,3)(1,0,3)

The planes intersect in the line r⃗=[1,0,3]+t[−1,1,2]\vec r=[1,0,3]+t[-1,1,2]r=[1,0,3]+t[−1,1,2]

The planes intersect in pairs, but there are no points of intersection between the 3 planes

The planes are parallel and distinct, they do no intersect

The planes are identical, and the intersect is the entire plane itself

I don't know