MCV4U1-UNIT NINE-LESSON TWO
Lesson Two: Intersection of a Line and a Plane
There are three ways a line may intersect a plane.
1.
The line is parallel but not coincident with
the plane.
No solution.
The line intersects the plane in one
point.
One solution.
2.
3.
The line lies right on the plane.
Infinite solutions.
The solutions IS the line.
Method for solving these systems:
1.
2.
2.
Find the parametric equations of the line.
Substitute into the equation of the plane.
Solve for the parameter, (say “t”).
--- if 0t = k, k   , k  0, there is no solution because this is
NEVER true.
---if t=k, k   , there is one solution. Substitute t back into the
parametric equations to find the point of intersection.
---if 0t=0, there are an infinite number of solutions, because this is
ALWAYS true.
MCV4U1-UNIT NINE-LESSON TWO
Example 1. Find the intersection of the line x, y, z   1,3,0  t  1,0,2 and the plane
Solution:
The parametric equations of the line are....
x=1-t
y=3
z=2t
substitute into the plane....
3(1-t) + 3 – 2t + 4=0
3 – 3t + 3 – 2t + 4 = 0
- 5t = - 10

Therefore intersection is a point....
Using the parametric equations again.....
Intersection is a point:
Example 2:
t=2
x =1–t=1–2=-1
y=3
z= 2(2) = 4
x, y, z    1,3,4 
Find the intersection of the line
x2 y4 z

 and the plane
3
2
2
2x - 3y + 5 = 0.
Solution:
Let the parameter be t. Then the parametric equations of the line are....
x=3t+2
y=2t-4
z=2t
substitute into the plane..... 2(3t+2) – 3(2t-4) + 5 = 0
6t + 4 – 6t + 12 + 5 = 0
0t=-21
This is NEVER true, so there is no point of intersection.
Alternate method....
 
 
nd  0  n  d
1.
The line and the plane are parallel/coincident or parallel/distinct.

n

d
3x+y-z+4=0.
MCV4U1-UNIT NINE-LESSON TWO
2.
 
 
nd  0  n  d
The line and the plane intersect at one unique point.

n

d