HWQ

• Find the xy trace:

2 2 2( 1) ( 1) 4x y z

22 1 3x y

HWQ

• Find the standard form of the equation of a sphere with endpoints of a diameter

(2,-2,2) and (-1,4,6)

2

2 21 611 4

2 4x y z

Vectors and the Geometry of Space 2015

Section 10.2 Vectors in Space

Vectors in Three-Dimensional Space

Now that we have an understanding of the three-dimensional system, we are ready to discuss vectors in the three-dimensional system. All the information you learned about vectors in Chapter 6 will apply, only now we will add in the third component.

Vectors in component form in three dimensions are written as ordered triples, in other words, now a vector in component form is .

321 ,, vvvv

In three dimensions the zero vector is O = < 0, 0, 0> and the standard unit vectors are . 1,0,0and0,1,0,0,0,1 kji

z

y

x

k

i

j

Each of the unit vectors represents one unit of change in the direction of each of their respective positive axes.

Given the initial point, and the terminal point, , the component form of the vector can be found the same way it was on the Cartesian Plane.

321 ,, pppP 321 ,, qqqQ

321332211 ,,,, vvvpqpqpqvPQ

Be sure to subtract the initial point’s coordinates from the terminal point’s coordinates.

Component form of a vector

The same vector can be written as a combination of the unit vectors.

kvjvivv

321

Standard Unit Vector Notation

We will look at examples using both forms.

More on Vectors in Three-Dimensions

Let and let c be a scalar. 321321 ,,and,, vvvvuuuu

o Vector Equality:

o Magnitude or Length of a Vector:

o Vector Addition:

o Scalar Multiplication:

o Unit Vector in the Direction of :

.and,ifonlyandif 332211 vuvuvuvu

232

22

1 uuuu

332211 ,, vuvuvuvu

321 ,, cucucuuc

u

321 ,,1

uuuuu

u

Note: This is simply the vector multiplied by the reciprocal of its magnitude.

Let’s look at some example problems involving vectors.

Example 1:Sketch the vector with initial point P(2, 1, 0) and terminal point Q(3, 5, 4). Then find the component form of the vector, the standard unit vector form and a unit vector in the same direction.

Solution: First draw a 3D system and plot P and Q. The vector connects P to Q.

P

Q

2 2 21 4 4 1 16 16 33

1 4 4 33 4 33 4 33Unit V

11,4,ector , , , ,

34

33 3 33 3333 33 33

PQ

@@@@@@@@@@@@@@

Second, find the component form of the vector. Do this by subtracting the initial point’s coordinates P(2, 1, 0) from the terminal point’s coordinates Q(3, 5, 4).

Example 1 Continued:

3 2,5 1,4 0 1,4,4

and

4 4

PQ

PQ i j k

@@@@@@@@@@@@@@

@@@@@@@@@@@@@@

Last, find a unit vector in the same direction. Do this by multiplying the vector by the reciprocal of the magnitude.

Note: You can verify it’s a unit vector by finding its magnitude.

113333

3316

3316

331

33

4

33

4

33

1222

Component form

Standard Unit Vector Form

Example 2:Given the vectors find the following:

a. b. c.

6,5,0and4,7,1,3,5,2 zvu

vu

zu

3 zvu

2

Solution:

a.

2,5,3 1,7,4

2

1, 2,

1,

1

5 7, 3 4

u v

b.

3

2,5,3 3 0, 5,6

2 0, 5 15,3

2, 10,

18

21

u z

c.2

2 2,5,3 1,7,4 0, 5,6

4 1 0,10

5,

7

22,

5,

4

6 4 6

u v z

Parallel Vectors

You may recall that a nonzero scalar multiple of a vector has the same direction as the vector (positive scalar) or the opposite direction as the vector (negative scalar). Since this is the case, any nonzero scalar multiple of a vector is considered a parallel vector.

In other words, if two vectors, , are parallel, then there exists some scalar, c such that . The zero vector does not have direction so it cannot be parallel.

vu

anducv

To get the idea, look at these parallel vectors on the Cartesian Plane.

x

y

u

uv

2

uz

2

Example 3:Determine if the vector with initial point, P(3,2,-2) and terminal point, Q(7,5,-3) is parallel to the vector . 3,9,12v

Solution: First find the component form of the vector from P to Q.

1,3,4

23,25,37

PQ

PQ

Second, if the two vectors are parallel, then there exists some scalar, c, such that .,3,43,9,12or cccPQcv

Then –12 = 4c c = -3And -9 = 3c c = -3And 3 = -c c = -3

For the two vectors to be parallel, c would have to be the samefor each coordinate. Since it is, the two vectors are parallel.

Example 4:Determine whether the points A(2,3,-1), B(0,1,3) and C(-3,-2,8) are collinear.

Solution: We need to find two vectors between the three points and determine if they are parallel. If the two vectors are parallel and pass through a common point then the three points must be in the same line.

The vector from A to B is kjikji

422133120

Now we need to find the vector from A to C or B to C.

The vector from A to C is kjikji

955183223

To be parallel: -2 = -5c c = 2/5 -2 = -5c c = 2/5 4 = 9c c = 4/9

Since c is not the same in each case, the vectors are not parallel and the points are not collinear.

Example 4b You Try:

Determine whether the points P(2, -1,4), Q(5,4,6) and R(-4,-11,0) are collinear.

3,5,2PQ @@@@@@@@@@@@@@

6, 10, 4PR @@@@@@@@@@@@@@

2 collinear pointsPR PQ @@@@@@@@@@@@@@@@@@@@@@@@@@@@

Example 5: Find a vector parallel to the vector with magnitude 5.kjiv

23

Solution: Be careful. We might quickly assume that all we need to do is to multiply the vector by 5. This would be fine if we were dealing with a unit vector. Since we are not, we need to multiply by the reciprocal of the magnitude first to get a unit vector and then multiply by 5.

14149123 222 v

kjiv

2314

1ofDirectioninVectorUnit

Vector with magnitude5 in Direction of

1 15 10 55 3 2

14 14 143

12

4

5

14

v

i j kk j ki j i

Solution to Example 5 Continued:

You can verify the new vector is parallel if you look at the form:

Obviously the scalar multiple is .

You can verify the magnitude is 5 by finding the magnitude of the form:

kji

2314

5

14

5

52514

3501425

14100

14225

14

5

14

10

14

15Magnitude

14

5

14

10

14

15

222

kji

More on Vectors in Three-Dimensions

Let and let c be a scalar. 321321 ,,and,, vvvvuuuu

o Dot Product of u and v:

o Angle between 2 vectors u and v :

o Orthogonal vectors have a dot product of 0.

1 1 2 2 3 3u v u v u v u v

cosu v

u v

Example 6: Find the dot product of

Example 7: Find the angle between

0,3, 2 4, 2,3and

1,0,2 3,1,0u and v

12

64.9

Homework:

10.1-10.2 Worksheet (evens)

For extra practice do: 10.2 Pg.719 1-45 odd