U13L4 The Intersection of Two Planes
The intersection 2 planes has 3 cases:
The planes are coincident.
The planes are parallel and have a point in common.
The equations in scalar form are multiples of each
other.
The planes are parallel and distinct.
The normal vectors are scalar multiples of each
other, but the equations of the planes in scalar
form are not multiples of each other.
The planes intersect in a line.
The normal vectors are not scalar multiples of each other.
Example
Determine the intersection of the planes
3x − 2y + 4z − 17 = 0
(x, y, z) = (5, − 1, 0) + t(2, 1, − 1) + s(−10, − 3, 6), s, t ∈ R
Solution 1
A normal vector of plane 1 is (3, − 2, 4)
A normal vector of plane 2 is (2, 1, − 1) × (−10, − 3, 6) = (3, − 2, 4)
The normal vectors are scalar multiples of each other, so the planes are at
least parallel.
The point (5, − 1, 0) lies on the second plane.
Sub (5, − 1, 0) into the equation of the rst plane.
3(5) − 2(−1) + 4(0) − 17 = 0 so (5, − 1, 0) lies on the second plane.
The two planes are coincident.
fi
U13L4 The Intersection of Two Planes
Page 1 of 3
Determine the intersection of the planes
3x − 2y + 4z − 17 = 0
(x, y, z) = (5, − 1, 0) + t(2, 1, − 1) + s(−10, − 3, 6), s, t ∈ R
Solution 2
The parametric equations for the second plane are:
x = 5 + 2t − 10s
y = − 1 + t − 3s
z = − t + 6s, s, t ∈ R
Sub into the equation of the rst plane:
3(5 + 2t − 10s) − 2(−1 + t − 3s) + 4(−t + 6s) − 17 = 0
15 + 6t − 30s + 2 − 2t + 6s − 4t + 24s − 17 = 0
0=0
This is true for all values of t and s, so the 2 planes are coincident.
Example
Find the intersection of the planes
x − y − z − 12 = 0
3x − 2y − 4z − 8 = 0
(1)
(2)
Solution 1
(1) x -3
−3x + 3y + 3z + 36 = 0
(2)
3x − 2y − 4z − 8 = 0
Add
y − z + 28 = 0
Let z = t, t ∈ R
Then y = t − 28
Sub into plane (1), x = t − 28 + t + 12 = 2t − 16
The intersection of the planes is the line
x = 2t − 16
y = t − 28
z = t, t ∈ R
fi
U13L4 The Intersection of Two Planes
Page 2 of 3
Find the intersection of the planes
x − y − z − 12 = 0
3x − 2y − 4z − 8 = 0
(1)
(2)
Solution 2
By inspection, the normal vectors of the planes are not scalar multiples of
each other, so the planes intersect in a line.
Find a direction vector:
The line of intersection is orthogonal to both normal vectors.
Therefore, a direction vector of the line is
(1, − 1, − 1) × (3, − 2, − 4) = (2, 1, 1)
Find a point:
Let z = 0 in each plane
The equations become
(1)
x − y − 12 = 0
(2)
3x − 2y − 8 = 0
(1) x -3
−3x + 3y + 36 = 0
(2)
3x − 2y − 8 = 0
y + 28 = 0
Add
y = − 28
x + 28 − 12 = 0
Sub into (1)
x = − 16
The equation of the line of intersection is
r ⃗ = (−16, − 28, 0) + t(2,1,1),
U13L4 The Intersection of Two Planes
t∈R
Page 3 of 3