1. Math 200
Week 3 - Monday
Planes

2. Main Questions for Today
Math 200
Main Questions for Today
How do we describe planes in space?
Can we find the equation of a plane that satisfies certain conditions?
How do we find parametric equations for the line of intersection of two (non-parallel) planes?
How do we find the (acute) angle of intersection between two planes.

3. Defining a plane B A C A line is uniquely defined by two points
Math 200
Defining a plane
A line is uniquely defined by two points
A plane is uniquely defined by three non-collinear points
Why “non-collinear”?
Suppose we have three points, A, B, and C in space… A B C

4. n B A C We can draw vectors between these points.
Math 200 n A B C
We can draw vectors between these points.
We can find the vector orthogonal to both of these vectors using the cross product.
This is called a normal vector (n = <a,b,c>)
How does this help us describe the set of points that make up the plane?

5. n B A C P Consider a random point P(x,y,z) on the plane.
Math 200 A B C n P
Consider a random point P(x,y,z) on the plane.
If A has components (x1,y1,z1), we can connect A to P with the vector AP = <x - x1, y - y1, z - z1>
The vectors AP and n must be orthogonal!
n • AP = 0
<a, b, c> • <x - x1, y - y1, z - z1> = 0

6. This is called point-normal form for a plane
Math 200
Formula
So, given a point (x1, y1, z1) on a plane and a vector normal to the plane <a, b, c>, every point (x, y, z) on the plane must satisfy the equation
We can also multiply this out:
This is called point-normal form for a plane

7. Find an equation for a plane
Math 200
Find an equation for a plane
Let’s find an equation for the plane that contains the following three points:
A(1, 2, 1); B(3, -1, 2); C(-1, 0, 4)
We need the normal: n
n = AB x AC
AB = <2, -3, 1> and AC = <-2, -2, 3>
n = AB x AC = <-7, -8, -10>
Point-normal using A: -7(x-1) - 8(y-2) - 10(z-1) = 0 A B C n

8. Math 200
Comparing answers
Notice that we had three easy choices for an equation for the plane in the last example:
Point-normal using A: -7(x-1) - 8(y-2) - 10(z-1) = 0
Point-normal using B: -7(x-3) - 8(y+1) - 10(z-2) = 0
Point-normal using C: -7(x+1) - 8y - 10(z-4) = 0
We also could have scaled the normal vector we used:
-14(x-1) - 16(y-2) - 20(z-1) = 0
This makes comparing our answers trickier
To make comparing answers easier, we can put the equations into standard form

9. Point-normal using A: -7(x-1) - 8(y-2) - 10(z-1) = 0
Math 200
Standard form:
ax + by + cz = d
Multiply out point-normal form and combine constant terms on the right-hand side
E.g.
Point-normal using A: -7(x-1) - 8(y-2) - 10(z-1) = 0
-7x y z + 10 = 0
-7x - 8y - 10z = -33 or 7x + 8y + 10z = 33
Point-normal using C: -7(x+1) - 8y - 10(z-4) = 0
-7x y - 10z + 40 = 0

10. Math 200
Example 1
Find an equation for the plane in standard form which contains the point A(-3,2,5) and is perpendicular to the line L(t)=<1,4,2>+t<3,-1,2>
Since the plane is perpendicular to the line L, its direction vector is normal to the plane.
n = <3,-1,2>
Plane: 3(x+3) - (y-2) + 2(z-5) = 0
3x y z - 10 = 0
3x - y + 2z = -1

11. Math 200
Example 2
Find an equation for the plane containing the point A(1,4,2) and the line L(t)=<3,1,4>+t<1,-1,2>
We need two things: a point and a normal vector
Point: we can use A
Normal vector: ???
Draw a diagram to help
n = <?,?,?> L
<1,-1,2> A
(3,1,4)
We’ll put all the information we have down without any attention to scale or proportion

12. Form a second by connecting two points
Math 200
A(1,4,2) L
(3,1,4)
<1,-1,2> n
To get a normal we could take the cross product of two vectors on the plane.
We have one: <1,-1,2>
Form a second by connecting two points
<-2,3,-2>

13. Point-normal (using A): -4(x-1) - 2(y-4) + (z-2) = 0
Math 200
A(1,4,2) L
(3,1,4)
<1,-1,2>
n = <-4,-2,1>
Answer:
Point-normal (using A): -4(x-1) - 2(y-4) + (z-2) = 0
Standard: -4x - 2y + z = -10 or 4x + 2y - z = 10

14. Math 200
Example 3
Find the line of intersection of the planes P1:x-y+z=4 and P2:2x+4z=10
Start with a diagram:
P1 has normal n1=<1,-1,1>
P2 has normal n2=<2,0,4>

15. n = <1,-1,1> x <2,0,4> = <-4, -2, 2> Point:
Math 200
Normal Vector:
A vector parallel to the intersection will be orthogonal to both normal vectors.
n = <1,-1,1> x <2,0,4> = <-4, -2, 2>
Point:
a point in the intersection will satisfy both equations
Because we have three variables and two equations, one variable is free - we can set one equal to a constant.
Let z=0. Then we get
x-y=4 and 2x=10
So x = 5 and y = 1.
Therefore, we have the point (5, 1, 0) common to both planes.
Answer: L: <5, 1, 0> + t<-4, -2, 2>