Lesson 3. Perpendicularity, Planes, and Cross Products
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1 Lesson 3 Perpendicularity, Planes, and Cross Products
2 Example 1: Equation for a Plane Let P (2,3, 1) be a point in space and let V (4, 2,5) be a vector. Find the xyz-equation of the plane containing P that is perpendicular to vector V (that is, V is a normal vector to the plane). The main idea here is that the vector V is perpendicular to the plane at ANY point on the plane. Therefore, any vector with its tail at (2,3,-1) and tip at (x,y,z) on the plane will be perpendicular to V. This vector is: (x, y, z) (2,3, 1) Since (x, y, z) (2,3, 1) and V (4, 2,5) are, their dot product is zero : (x, y, z) (2,3, 1) (4, 2,5) 0 4(x 2) 2(y 3) 5(z 1) 0 4x 2y 5z 3 Created by Christopher Grattoni. All rights reservered.
3 Summary: Equation for a Plane Let P (a,b,c) be a point on the plane and let V=(v,v,v ) be a normal vector to the plane. Then: (x, y,z) P V 0 v (x a) v (y b) v (z c) (We usually like to rewrite in the form Ax+By+Cz=D) Created by Christopher Grattoni. All rights reservered.
4 Summary: Equation for a Plane Given an equation for a plane, you can generate a vector normal to the plane very easily: v (x a) v (y b) v (z c) (v, v, v ) is a normal vector to the plane Ax By Cz D (A,B,C) is a normal vector to the plane Created by Christopher Grattoni. All rights reservered.
5 Example 2: Perpendicular Planes Given the following two planes, show that they are perpendicular: Ideas?? A: 6(x 2) 5(y 11) 2(z 1) 0 B: 3(x 4) 2(y 1) 4(z 8) 0 Yes, their normal vectors should be : (6, 5,2) (3,2, 4) Since the dot product of their normal vectors is 0, Plane A and Plane B are perpendicular. 0 Created by Christopher Grattoni. All rights reservered.
6 Example 3: Parallel Planes Given the following two planes, show that they are parallel: Ideas?? First, show their normal vectors are multiples of each other: B: 3(x 4) 2(y 1) 4(z 8) 0 C: 6(x 3) 4(y 1) 8(z 7) 0 ( 6, 4,8) 2(3,2, 4) Next, make sure they aren't the same plane: (4,1, 8) should not satisfy C : 6(x 3) 4(y 1) 8(z 7) 0: 6(4 3) 4(1 1) 8( 8 7) 170 Therefore Plane B and Plane C are parallel planes. Created by Christopher Grattoni. All rights reservered.
7 Summary: Parallel and Perpendicular Planes Ax By Cz D Given two planes with equations, we Ex Fy Gz H can determine if they are parallel or perpendicular as follows: Perpendicular : (A,B,C) (E,F,G) 0 t. Parallel : k s. (A,B,C) k(e,f,g) and P Plane s.t. P Plane 1 2 Same Plane : k s. (A,B,C) k(e,f,g) t. and P Plane s.t. P Plane 1 2 Created by Christopher Grattoni. All rights reservered.
8 Warning: Lines and Planes Behave Differently! Describing the intersection between two planes is a lot easier than the intersection between two lines in 3 dimensions! Here s the main hiccup to watch out for: two lines in three dimensions do NOT necessarily intersect (they can be skew) Let s look at a few different examples! Created by Christopher Grattoni. All rights reservered.
9 Are the Following Pairs of LINES Perpendicular? L(t) (1, 4,2) t(3,1,1) M(t) (1, 4,2) t(0, 1,1) L(t) and M(t) intersect at (1,4,2) and (3,1,1) (0, 1,1) 0, so these lines are L(t) (2,0, 1) t(3,1,1) M(t) (5,1,0) t(0, 1,1) L(t) (5,1,0) t(3,1,1) M(t) ( 3,1,2) t(0, 1,1) The lines intersect at L(1) M(0) (5,1,0) and (3,1,1) (0, 1,1) 0, so these lines are L(t) and M(t) do not intersect, so these lines are not (they are skew)
10 Are the Following Pairs of LINES Parallel? L(t) (1, 4,2) t(3,1,1) M(t) (5,9,6) t(3,1,1) (1, 4,2) M(t) and the lines have the same generating vector, so they are L(t) (1, 4,2) t(3,1,1) M(t) ( 2,3,1) t(3,1,1) M(1) (1,4,2) L(t) and the lines have the same generating vector, so they are the same line twice. L(t) (1, 4,2) t(3,1,1) M(t) (10,7,5) t(6,2,2) L(3) (10,7,5) M(t) and the generating vectors of the lines are multiples of each other, so they are the same line twice.
11 Example 4: The Line of Intersection Given the following two planes, find the parametric line describing their intersection: A: 3x 4y z 5 B: 2x 4y 2z 3 Eliminate one variable: 5x z 2 So z 5x 2. We can use this relation to find two points, P and Q on the line of intersection: When x 1, z 5(1) 2 3. Plug back into either 1 1 plane to get that y. So P 1,,3 4 4 When x 1, z 5( 1) 2 7. Plug back into either plane to get that y. So Q 1,, 7 4 4
12 Example 4: The Line of Intersection Given the following two planes, find the parametric line describing their intersection: A: 3x 4y z 5 B: 2x 4y 2z So P 1,,3 and Q 1,, lie on BOTH planes : L(t) P t(q P) 1 L(t) 1,,3 t( 2, 4, 10) t (, ) 4 Created by Christopher Grattoni. All rights reservered.
13 Summary: The Line of Intersection To find the parametric line of intersection between two planes : 1) Eliminate a variable between the two equations to find a relation between 2 variables 2) Find two points on the line of intersection using this relation 3) Use these two points in L(t) P t(q P) Created by Christopher Grattoni. All rights reservered.
14 Example 5: Plane from Three Points Given the following set of three noncollinear points, find the parametric equation of the plane passing through them: P (5,1, 6) Q (3, 1, 4) Parametrically this is easy : R (1,2, 4) For a line, we just have a point P and a generating vector V: L(t) P tv t (, ) For a plane, we just need a point P and two generating vectors, V and W, with different directions (linearly independent vectors): V Q P (3, 1, 4) (5,1, 6) ( 2, 2,10) M(s, t) P tv sw s, t (, ) W R P (1,2, 4) (5,1, 6) ( 4,1,10) s,t (, ) M(s,t) (5,1, 6) t( 2, 2,10) s( 4,1,10)
15 Example 5: Plane from Three Points s,t (, ) M(s,t) (5,1, 6) t( 2, 2,10) s( 4,1,10) Created by Christopher Grattoni. All rights reservered.
16 Example 5: Plane from Three Points Given the following set of three noncollinear points, find the xyz-equation of the plane passing through them: P (5,1, 6) Q (3, 1, 4) R (1,2, 4) You could plug each point into Ax+By+Cz=D and solve a system... but that's no fun! If we could find a normal vector to the surface, we could use: v (x a) v (y b) v (z c) Task : Find a vector perpendicular to both V ( 2, 2,10) and W ( 4,1,10) Created by Christopher Grattoni. All rights reservered.
17 Review: Determinant of a 2 x 2 Matrix Every square (n x n) matrix has a real number associated to it called the determinant a a 1,1 1,2 det a a a a a a 2,1 2,2 Short form: a a a 1,1 1,2 a 2,1 2,2 1,1 2,2 2,1 1,2 a a a a 1,1 2,2 2,1 1,2 Created by Christopher Grattoni. All rights reserved.
18 Review: Determinant of a 3 x 3 Matrix Method: Expansion by Minors a a a 1,1 1,2 1,3 a a a a a a a a a a a a a a a a a a a a a 3,1 3,2 3,3 M 1,1 2,2 2,3 2,1 2,3 2,1 2,2 2,1 2,2 2,3 1,1 1,2 1,3 3,2 3,3 3,1 3,3 3,1 3,2 M M 1,2 1,3 These determinants are the "minors" of the original matrix. Created by Christopher Grattoni. All rights reserved.
19 Example 5: Plane from Three Points Given the following set of three noncollinear points, find the xyz-equation of the plane passing through them: P (5,1, 6) Q (3, 1, 4) R (1,2, 4) You could plug each point into Ax+By+Cz=D and solve a system... but that's no fun! If we could find a normal vector to the surface, we could use: v (x a) v (y b) v (z c) Task : Find a vector perpendicular to both V ( 2, 2,10) and W ( 4,1,10) Created by Christopher Grattoni. All rights reserved.
20 Defining the Cross Product We will introduce a new binary operation called the cross product. The cross product should be a way to take two vectors, a and b, and find a third, a b, that is perpendicular to both. For the right hand rule with a b, the first vector is always your index finger and the second vector is always your middle finger! Created by Christopher Grattoni. All rights reserved.
21 Defining the Cross Product For the right hand rule with a b, the first vector is always your index finger and the second vector is always your middle finger! Lay two pens on the table and call them a and b. Find ab. How does it relate to ba? ba (ab) Created by Christopher Grattoni. All rights reserved.
22 Defining the Cross Product For the right hand rule with a b, the first vector is always your index finger and the second vector is always your middle finger! Lay two pens on the table in the same spot and call them a and a. Find aa. How does it relate to aa? aa (0,0,0) 0 Created by Christopher Grattoni. All rights reserved.
23 Defining the Cross Product Let i (1,0,0), j (0,1,0), and k (0,0,1). These three vectors are called the standard a set of unit basis vectors for xyz-space. They are unit vectors since i j k 1 They are basis vectors since they point in three different directions (so any vector can be described as sums/multiples of these vectors) Ex. (11, 21,16) can be written as 11i 21j 16k Ex. (5,0,1) can be written as 5i k They are standard basis vectors since they only have x-, y-, or z-components Created by Christopher Grattoni. All rights reserved.
24 Defining the Cross Product i j j k k i k i j j i k j i k k i j i i j j k k Created by Christopher Grattoni. All rights reserved. i j k
25 Calculate V W for V (v,v,v ) and W (w,w,w ) : V W (v i v j v k) (w i w j w k) ij k v w i i v w i j v w i k jk i v w j i v w j j v w j k k i j j i k vwkivwkjv w k k k j i ik j v w (0) v w (k) v w ( j) v w ( k) v w (0) vw( i) v w (j) v w ( i ) v w (0) (v w v w )i (v w v w )j (v w v w )k v v v v v v i j k w w w w w w Created by Christopher Grattoni. All rights reserved. ii 0 j j 0 k k 0 i j k v v v w w w 1 2 3
26 Summary: The Cross Product Calculate V W for V (v,v,v ) and W (w,w,w ) : Remember that V to both V and W! i j k V W v v v Use Mathematica to show w w w V (V W) 0 and W (V W) 0! W generates a vector perpendicular Created by Christopher Grattoni. All rights reserved.
27 Verifying the Cross Product We can verify that the cross product gives us a perpendicular vector to V and W for any choices of V and W. Try this in Mathematica: V={v1,v2,v3}; W={w1,w2,w3}; VxW=Cross[V,W]; Together[V.VxW] Together[W.VxW] Created by Christopher Grattoni. All rights reserved.
28 Example 6: The Cross Product Calculate ( 2, 2,10) ( 4,1,10) : i j k ( 2, 2,10) ( 4,1,10) i j k i 20j 10k ( 30, 20, 10) Created by Christopher Grattoni. All rights reserved.
29 Example 5: Plane from Three Points (CONTINUED) Given the following set of three noncollinear points, find the xyz-equation of the plane passing through them: P (5,1, 6) Now that we have the cross product, we know that V W is a normal vector to the plane! V W ( 2, 2,10) ( 4,1,10) v (x a) v (y b) v (z c) Q (3, 1, 4) R (1,2, 4) ( 30, 20, 10) (from previous slide!) 30(x 5) 20(y 1) 10(z 6) 0 3(x 5) 2(y 1) (z 6) 0 3x 2y z 11 Created by Christopher Grattoni. All rights reserved.
30 Example 6: The Magnitude of the Cross Product Let V (6, 2,5) and W ( 8,1,10). Find V W. You COULD do this the long way: i j k (6, 2,5) ( 8,1,10) i j k i 100 j 10k ( 25, 100, 10) V W ( 25, 100, 10) ( 25, 100, 10) ( 25, 100, 10) Created by Christopher Grattoni. All rights reserved.
31 Example 6 : The Magnitude of the Cross Product Shortcut : V W V W sin( ) where is the angle between V and W. Proof : YOU will prove this in your homework. You MUST know its geometric significance before you take the chapter quiz. Note : Memorize this formula. Study it. Use it. Don't forget it. You will need it along with the analogous formula for dot products: V W V W cos( ) Created by Christopher Grattoni. All rights reserved.
32 Example 6 : The Magnitude of the Cross Product Let V (6, 2,5) and W ( 8,1,10). Find V W. Now the short way: (6, 2,5) ( 8,1,10) So the angle between V and W is V W V W sin( ) sin? Created by Christopher Grattoni. All rights reserved. or 90 Answer Under the Box!
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