Math 200 Week 3 - Monday Planes

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.

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

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?

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

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

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

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

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 + 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

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 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

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>

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

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>

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>