PLANESMATH 200 WEEK 2 - FRIDAY

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.

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

MATH 200

▸ 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?

A

B

C

n

MATH 200

▸ 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

PA

B

C

n

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

�x � x1, y � y1, z � z1� · �a, b, c� = 0

▸ We can also multiply this out:

a(x � x1) + b(y � y1) + c(z � z1) = 0

THIS IS CALLED POINT-NORMAL FORM 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

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

MATH 200

▸ Standard form:

▸ ax + by + cz = d

▸ Multiply out point-normal form and combine constant terms on the left-hand side

▸ E.g.

▸ Point-normal using A: -7(x-1) - 8(y-2) - 10(z-1) = 0 ▸ -7x + 7 - 8y + 16 - 10z + 10 = 0 ▸ -7x - 8y - 10z = -33 or 7x + 8y + 10z = 33

▸ Point-normal using C: -7(x+1) - 8y - 10(z-4) = 0 ▸ -7x - 7 - 8y - 10z + 40 = 0 ▸ -7x - 8y - 10z = -33 or 7x + 8y + 10z = 33

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 + 9 - y + 2 + 2z - 10 = 0

▸ 3x - y + 2z = -1

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

A

L(3,1,4)

<1,-1,2>

n = <?,?,?>

WE’LL PUT ALL THE INFORMATION WE HAVE DOWN WITHOUT ANY ATTENTION TO

SCALE OR PROPORTION

MATH 200

▸ 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>

A(1,4,2)

L(3,1,4)

<1,-1,2>

n

�n = �1, �1, 2� � ��2, 3, �2� =

������

i j k1 �1 2

�2 3 �2

������

= �2 � 6, �4 + 2, 3 � 2�= ��4, �2, 1�

MATH 200

▸ Answer:

▸ Point-normal (using A): -4(x-1) - 2(y-4) + (z-2) = 0

▸ Standard: -4x - 2y + z = -10

A(1,4,2)

L(3,1,4)

<1,-1,2>

n = <-4,-2,1>

MATH 200

EXAMPLE 3▸ Find the line of intersection of

the planes P1:x-y+z=4 and P2:2x+4x=10

▸ Start with a diagram:

▸ P1 has normal n1=<1,-1,1>

▸ P2 has normal n2=<2,0,4>