Honors Advanced Math Name:_______________________ Points, Lines and Planes in 3D Block: E or G 4 April 2011

3-D dot product and equations of planes

Extending from 2-D to 3-D Here are some key facts we discussed about dot product and perpendicular vectors in two dimensions:

(A) v · w = v1w1 + v2w2.

(B) v · v = | v |2

(C) v · w = | v | | w | cos(θ).

(D) Two vectors v and w are perpendicular if and only if v · w = 0. (E) The vectors perpendicular to <a, b> are k<–b, a> for all reals k.

How do these facts extend to three dimensions? The dot product definition (A) picks up a third term (+v3w3). Facts (B), (C), and (D) hold true without change (the triangle of vectors v, w, and v–w is still planar so the proof shown in class involving Law of Cosines works without any changes). However, fact (E) needs to change significantly, because if you think about it visually, given an arrow in 3-D, there’s a two-dimensional set of perpendicular arrows. If you give all these perpendicular vectors the same starting point, then their ending points form a plane in 3-D space.

Also, a new type of question can be asked in 3-D: how do you find a third vector perpendicular to two given vectors? (Why would this kind of question be posed in 3-D but not in 2-D?) This problem will lead to our defining (in an upcoming lesson) a new 3-D vector operation called the cross-product.

Finding the plane of vectors perpendicular to a 3-D vector Question: Given v = <a, b, c>, what are the vectors perpendicular to v ?

Answer: The perpendicular vectors are any <x, y, z> such that <a, b, c> · <x, y, z> = 0. That is, ax + by + cz = 0. If you think of each vector <x, y, z> as starting at the origin and ending at (x, y, z), then the points (x, y, z) satisfying ax + by + cz = 0 form a plane.

Equations of planes In three-dimensional space, a plane can be uniquely determined by specifying a point of the plane and a vector perpendicular to the plane. Here is how to find the x-y-z equation of this plane.

Suppose <a, b, c> is the given vector perpendicular to the plane. Then the plane’s equation has the form ax + by + cz = d, where the number d can be found by substituting the known point into the equation. Example: Find the equation of the plane through (2, 1, 6) that is perpendicular to <3, 5, 4>.

Solution: The equation has the form 3x + 5y + 4z = d. Substitute the point: 3·2 + 5·1 + 4·6 = d. This gives d = 35, so the plane’s equation is 3x + 5y + 4z = 35.

Homework Brown Adv. Math. p.452/43-45 Brown Adv. Math. p.455/15, 21, 28-31, 33, 34 (Vocabulary note: Cartesian equation means an x-y equation in 2-D or an x-y-z equation in 3-D.)

Points, Lines, and Planes in 3D page 2

Directions: Carefully do each of the following problems. Some of these problems can be done very quickly with formulas proven in problems in the Brown book. Don't use these short-cuts! Instead, use your understanding of the fundamental properties of vectors, lines, and planes. 1. Turn the vector equation ( ) ( )x y z t, , , , , ,= ! +2 3 5 1 3 2 into three parametric equations. 2. Turn the parametric equations x t= +3 4 , y t= !3 , z t= ! +2 3 into the vector equation of a line. 3. Eliminate the parameter to rewrite the equations of the lines from the last two problems as "triple

equations" in x, y, and z. In other words, solve each parametric equation for t and then write the equation of the line as t(x) = t(y) = t(z).

4. Write a new vector equation for your answer to problem 2, where the point is moving twice as fast. 5. Write a vector equation for a line that is parallel to the line in the last problem. 6. a. Find the plane formed by the following three points: ( )1 2 3, , , ( )3 2 1, , , and ( )! !1 2 2, , . b. Find the area of the triangle formed by the three points in part a. c. Do the points ( )5 0 2, , , ( )3 1 4, ,! , and ( )! !1 3 8, , form a plane? Explain why or why not. 7. Consider the following two lines.

lx

y z1

1

24:

!

!= ! = and l

x y z2

2

3

1

4

2

1:

!

!=

!=

!

!

a. Find the intersection of the two lines. b. Find the equation of the plane defined by the two lines. 8. Consider the following equations for a line and a plane.

!

l : x "1

2=y + 3

2( )"1

=z +1

2 and M x y z: 2 2 12! + =

Do the plane and the line intersect? Explain how you can tell. Find the intersection (if one exists) between the line and the plane.

9. Consider a sphere of radius 7, centered at the point

!

3,"4,0( ). a. Write the Cartesian equation for the sphere. b. Describe the intersection of the sphere with the plane

!

z = 4 . Be as specific as possible. c. Line l is given by the equation

!

x,y,z( ) = 3,"4,0( ) + t 2,1,"3 . Find the coordinates of any points where the line intersects the sphere or demonstrate that the line and the sphere have no intersection.

10. Find the angle made by the intersection of the two lines or show that there is no intersection. Give

your answers as decimal approximations in degrees. a. Line 1: x t y z t= + = = ! +4 2 3 1, , and line 2: x s y s z s= + = + = +2 2 2 3 1, , .

b. Line 1: x

y z3

2 1= ! = + and line 2: x

yz!

= + =+

!

1

42

3

3.

c. Line 1: y x= !3 2 and line 2: y x= ! +1

51

Points, Lines, and Planes in 3D page 3

11. Find the distance between the point ( )10 3 2, ,! and the line given by x t= !4 2 , y = 3 , and

z t= ! +1 . 12. Find the distance between the point ( )1 2 3, , and the plane 2 4x y z! + = . 13. Consider the following planes. M x y z

13 4 10: ! + = and M x y z

23 4 6: ! + =

a. Explain how you can tell that these two planes are parallel. b. Find the distance between the planes. 14. Find the angle between the following two planes. Give your answer as a decimal approximation in

degrees. M x y z

13 10: 2 ! + = and M x y z

22 8: ! + =

Answer Key: 1.

!

x = 2 + t ,

!

y = "3+ 3t ,

!

z = 5 + 2t 2.

!

x,y,z( ) = 3,3,"2( ) + t 4,"1,3

3.

!

x " 2 =y + 3

3=z " 5

2

4.

!

x,y,z( ) = 3,3,-2( ) + t 8,"2,6 5.

!

x,y,z( ) = any point not on the last line( ) + t 8,"2,6 Example:

!

x,y,z( ) = 3,3,0( ) + t 4,"1,3 6. a. Using rref gives 4x – 3y + 4z = 10 (or some scaled version of this).

b.

!

AB " AC = AB AC cos# , where

!

" is the angle between the two vectors. Solving that equation

yields

!

" = 98.88o , so the area of the triangle is

!

1

2AB AC sin98.88

o= 6.403

c. No. They are all on the line

!

x,y,z( ) = 5,0,2( ) + t "2,"1,2 . Remember, three points will only determine a plane when they are not collinear.

Points, Lines, and Planes in 3D page 4

7. a. The intersection is

!

"1,5,1( ) . Method: Find parametric equations for both lines, using t as the time variable in one and s as

the time variable in the other. Make a system of equations by setting the x values equal and the y values equal. Solve for s and t. Then check if you get the same z value using either s or t (you should, since the lines intersect). Plug either the s or the t value in and find the coordinates of the point.

b.

!

x,y,z( ) = "1,5,1( ) + t "2,1,1 + s "3,4,"1 (to be discussed further in class) 8. They will intersect because the direction vector for the line

!

1,"1,2 is not perpendicular to the vector

!

2,"2,1 , which is perpendicular to the plane. You can tell they are not perpendicular by taking their dot product. Find parametric equations for the line, then plug your values of x, y, and z into the equation for the plane to find the t value when they intersect. Once you've got the t value,

plug it into the parametric equation to find they intersect at

!

5

3, "

5

3,

4

3

#

$ %

&

' (

9. a.

!

x " 3( )2

+ y + 4( )2

+ z2 = 49

b. The intersection is a circle of radius

!

33 in the plane

!

z = 4 centered at

!

3,"4,4( )

c. There are two points of intersection at

!

3± 27

2, " 4 ±

7

2, m 3

7

2

#

$ %

&

' (

10. a.

!

55.53o. Find vector equations for the two lines, then use the dot product to find the angle

between the two vectors. This method will work in part c as well

b. The two lines do not intersect so it is not meaningful to talk about the angle between them. (I think the best way to determine whether the lines intersect is actually to try to find the point of intersection and see if you get an answer).

c.

!

82.87o

11. The point is on the line so the distance is 0.

12. The distance is

!

1

6. Find the equation of a line that is perpendicular to the plane and goes through

the point ( )1 2 3, , . Find where this line intersects the plane. Then find the distance between the intersection point and ( )1 2 3, , .

13. a. The same vector is perpendicular to both planes.

b. The distance is

!

8

13. Find the equation of any line that is perpendicular to both planes. Find the

points at which this line intersects each plane and find the distance between these points. (And

remember that

!

2

132 "13 =

22" 2 "13

132

=8

13)

14.

!

139.80o The angle is the supplement of

!

40.20o. Use the dot product of the vectors that are

perpendicular to each plane.