vectors in space!!! cool stuff in section 8.6b. just as in planes, sets of equivalent directed line...

Vectors in Vectors in Space!!!Space!!!

Cool stuff in Section 8.6bCool stuff in Section 8.6b

Just as in planes, sets of equivalent directed line segments inspace are called vectors.

Denoted with ordered triples:1 2 3, ,v v v v

Zero Vector: 0 0,0,0Standard Unit Vectors:

1,0,0i

0,1,0j

0,0,1k

Vector v can be represented withthese standard unit vectors:

1 2 3, ,v v v v

1 2 3v i v j v k

Let’s see this graphically:

z

y

x

k

ij

(0, 0, 1)

(0, 1, 0)

(1, 0, 0)

v

1 2 3, ,v v v

1v

3v

2v

Vector Relationships in SpaceVector Relationships in SpaceHow do we write the vector v defined by the directed linesegment from P to Q?

, ,P a b c , ,Q x y z

, ,v PQ x a y b z c ��������������

x a i y b j z c k Multiplying a vector by a scalar?

1 2 3 1 2 3, , , ,cv c v v v cv cv cv

Vector Relationships in SpaceVector Relationships in SpaceFor vectors: 1 2 3, ,v v v v

1 1 2 2 3 3, ,v w v w v w • Equality:

1 2 3, ,w w w wv = w if and only if:

1 1 2 2 3 3, ,v w v w v w • Addition: v + w =

1 1 2 2 3 3, ,v w v w v w • Subtraction: v – w =

2 2 21 2 3v v v • Magnitude: |v| =

1 1 2 2 3 3v w v w v w • Dot Product: v w =v

uv

• Unit Vector: , 0v is the unit vector in thedirection of v.

Computing with VectorsComputing with Vectors

3 2,1,41. 6,3,12

0,6, 7 5,5,8 2. 5,11,1

1, 3,4 2, 4,5 3. 3,1, 1

2,0, 64. 2 2 22 0 6 40 2 10

5,3, 1 6,2,3 5. 5 6 3 2 1 3

27

A jet airplane just after takeoff is pointed due east. Its air velocityvector makes an angle of 30 with flat ground with an airspeed of250 mph. If the wind is out of the southeast at 32 mph, calculatea vector that represents the plane’s velocity relative to the pointof takeoff.

Let i point east, j point north, and k point up. A diagram?

Plane’s air velocity:

a = 250cos30 ,0,250sin 30 216.506,0,125 Wind velocity (pointed northwest):

w = 32cos135 ,32sin135 ,0 22.627,22.627,0

A jet airplane just after takeoff is pointed due east. Its air velocityvector makes an angle of 30 with flat ground with an airspeed of250 mph. If the wind is out of the southeast at 32 mph, calculatea vector that represents the plane’s velocity relative to the pointof takeoff.

Let i point east, j point north, and k point up. A diagram?

Velocity relative to the ground is v = a + w:

v = 216.506,0,125 22.627,22.627,0

193.879,22.627,125

= 193.879i + 22.627j + 125k

A rocket soon after takeoff is headed east and is climbing at a80angle relative to flat ground with an airspeed of 12,000 mph.If the wind is out of the southwest at 8 mph, calculate a vector vthat represents the rocket’s velocity relative to the point of takeoff.

Let i point east, j point north, and k point up. A diagram?

Rocket’s velocity relative to the air:

12000cos80 ,0,12000sin80r Air’s velocity relative to the ground:

8cos 45 ,8sin 45 ,0w Rocket’s velocity relative to the ground:

v = 2089.435 i + 5.657 j + 11,817.693 k

Lines in SpaceLines in SpaceFirst-degree equations in three variables graph as…Planes!!!

So how do we get lines?

The pair of equations y = 0 and z = 0 give…the x-axis!!!

For more complicated lines, we can use:

• one vector equation, or

• a set of three parametric equations.

Suppose L is a line through the point

for some real number t.

The vector v is a directionvector for line L.

0 0 0 0, ,P x y z

and in the direction of a nonzero vector v = , ,a b c

Then for any point on L, , ,P x y z0P P tv��������������

x

y

z

0 0 0 0, ,P x y z

, ,P x y z

L

, ,v a b c

Let:

x

y

z

0 0 0 0, ,P x y z

, ,P x y z

L

, ,v a b c, ,r OP x y z

��������������

0 0 0 0 0, ,r OP x y z ��������������

0P P tv��������������

Then 0r r tv

The vector equation of line L: 0r r tv In component form: 0 0 0, , , , , ,x y z x y z t a b c

0 0 0, ,x at y bt z ct

Let:

x

y

z

0 0 0 0, ,P x y z

, ,P x y z

L

, ,v a b c, ,r OP x y z

��������������

0 0 0 0 0, ,r OP x y z ��������������

0P P tv��������������

Then 0r r tv

This can be expressed as the parametric equations:

0x x at

0 0 0, , , ,x y z x at y bt z ct

0y y bt 0z z ct

If L is a line through the point

0r r tv , ,r x y z

0 0 0 0, ,r x y z

Equations for a Line in SpaceEquations for a Line in Space

0 0 0 0, ,P x y z in the direction

, ,a b cof a nonzero vector v = , then a point , ,P x y zis on L if and only if

• Vector Form: where

OR

• Parametric Form:

0x x at 0y y bt 0z z ct where t is a real number.

The line through

Practice ProblemsPractice Problems

0 4,3, 1P 2,2,7v

with direction vector

can be written:

4,3, 1 2,2,7r t • in vector form as

• or in parametric form as 4 2x t 3 2y t 1 7z t

How can we “see” this graphically???How can we “see” this graphically???

Using the standard unit vectors i, j, and k, write a vector equation for the line containing the points A(3, 0, –2) and B(–1, 2, –5), and compare it to the parametric equations for the line.

Practice ProblemsPractice Problems

v AB��������������The line is in the direction of

So using , the vector equation of the line becomes

0r r tv

1 3,2 0, 5 2

0r OA��������������

4,2, 3

, , 3,0, 2 4,2, 3x y z t

Using the standard unit vectors i, j, and k, write a vector equation for the line containing the points A(3, 0, –2) and B(–1, 2, –5), and compare it to the parametric equations for the line.

Practice ProblemsPractice Problems

, , 3,0, 2 4,2, 3x y z t , , 3 4 ,2 , 2 3x y z t t t

3 4 2 2 3xi yj zk t i t j t k The parametric equations are the three component equations:

3 4x t 2y t 2 3z t