MCV4U1-UNIT NINE-LESSON THREE
Lesson Three: Intersection of Two Planes
Note: The Greek letter  is often used as a notation for a plane. (i.e.  1 means plane #1 )
Two planes can intersect in one of three ways.
Case 1: Intersection in a line.

n1

n2
Case 2: Intersection in a plane. (Parallel/coincident) – Infinite Solutions

n2

n1
Case 3: No Intersection (Parallel/Non-coincident)

n2
- No Solution

n1
Method for solving these systems:
Examine the normals.
Case 1: The normals are NOT parallel so the planes MUST intersect in a line. Let ANY of the variables equal t (a
parameter). We usually choose z. (i.e. “Let z = t, a parameter”) Solve for x and y in terms of t.
Example 1:
Find the intersection of the planes x + 2y + 7z = 4 and x+3y-3z=1.
Solution:


n1  1,2,7 and n2  1,3,3
r
r
r r
n1  kn2; k   n1 n2 .
Therefore the planes are NOT parallel and MUST intersect in a line.
MCV4U1-UNIT NINE-LESSON THREE
Let z=t, a parameter.......
Then the equations become.....  1 :
x + 2y = - 7t + 4
 2 : x + 3y = 3t + 1
 1   2 :  y  10t  3
 y  10t  3
sub in  1 : x  2(10t  3)  7t  4
 x  27t  10
Therefore the solution is the line
x = - 27t + 10
y = 10t - 3
z=t
r
r
Case 2: The normals are parallel then n1  kn2; k R. Since the planes are coincident, the constants in the equations
“D” will have the same relationship as the normals. That is, D1  kD2; k  .
In fact, it may be easier to just remember that one equation is a scalar multiple of the other equation.
 of the planes 2x-4y+8z=10 and 3x-6y+12z=15.
Example 2: Find the intersection

Solution:
3r
r
r
r
r r
n1  2,4,8 and n 2  3,6,12, so n 2  n1,  n1 n 2
2
3
3
Check for coincidence…. Is D2  D1 ? ....... yes, b /c  15   10
2
2
Therefore the planes are parallel and coincident. We say the intersection is the plane given in parametric form, as
follows…..
Let z=t and y=s, where t and s are parameters. Using EITHER equation (I chose the first one), solve for x in terms of s and

t.
2x  4 y  8z  10
2x  4s  8t  10
2x  4s  8t  10
x  2s  4t  5
x  2s  4t  5
Therefore the solution is the plane y  s
.
zt
r
r
Case 3: The normals are parallel then n1  kn2; k R. Since the planes are non-coincident, the constants in the
equations “D” will NOT have the same relationship as the normals. That is, D1  kD2; k  .

Example 3: Find the intersection of the planes x-y-5z=3 and 2x-2y-10z=4.

Solution:
r
r
r
r
r r
n1  1,1,5 and n2  2,2,10, so n2  2n1,  n1 n2

Check for coincidence….. Is D2  2D1 ? ....... no, b/c  4  2  3
Therefore the planes are parallel/non-coincident and there