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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>