vectors

Precalculus

Vectors• A vector is an object that has a magnitude and a

direction.• Given two points P: & Q: on the plane, a

vector v that connects the points from P to Q is (start with last point)

Notations v = i + j. Or <v1,v2>

The magnitude of v is |v| =The direction is the slopeVectors are equal if they have the same magnitude

and the same direction

),( 11 yx

(x2 −x1)

),( 22 yx

(y2 −y1)

22yx vv

example

• Find the component vector for• r= (3,8) s= (-4,5)

• Find the magnitute of• Find the direction of

rsuv

rsuv

rsuv

Vectors are equal if they have the same magnitude and the same direction

• Find the component vector for• r= (3,8) s= (-4,5)

• Find the magnitute of

The direction of

rsuv

rsuv

rsuv

= −7,−3

rsuv

= 72 + 32 = 58

rsuv

=−3−7

or37

Unit Vectors• Unit vectors are vectors of length 1.• i is the unit vector in the x direction.• j is the unit vector in the y direction.• A unit vector in the direction of v is v/||v||• A vector v can be represented in component form by v = vxi + vyj.• The magnitude of v is ||v|| = • The unit vector is

22yx vv

vxv,vyv

Vector Operations

• Scalar multiplication: A vector can be multiplied by any scalar (or number).

• Example: Let v = 5i + 4j, k = 7. Then kv = 7(5i + 4j) = 35i + 28j.

• Addition/subtraction of vectors: Add/subtract same components.

• Example Let v = 5i + 4j, w = –2i + 3j. • v + w = (5i + 4j) + (–2i + 3j) = (5 – 2)i + (4 + 3)j

= 3i + 7j.

Example:

• 3v – 2w =• ||3v – 2w|| =

v= 5,4 w= −2,3

Example:

• 3v – 2w =• ||3v – 2w|| =

192 +62 = 397

3 5,4 −2 −2,3 = 15,12 + 4,−6 = 19,6

Direction Angles• Given the direction angle of a vector, find the

component form of the vector in the same direction. Use the formula:

Use the formula above for number 29 on pg. 511

v cosθ,sinθ

Direction Angles

• Find the direction angle of the vectors

v= 4,5 w= −2,3

Start by plotting the vector. What trig function will helpfind the angle?

Direction Angles

• Find the direction angle of the vectors

v= 4,5 w= −2,3

θ =tan−1(5

4) ≈ 51.3 θ =tan−1

3

−2

⎛

⎝⎜

⎞

⎠⎟≈ −56.3

But w is in quadrant II, so 180-56.3=123.7

Direction Angles

• Find the direction angle of a vector, then find the component form of the vector with magnitude of 6 in the same direction.

• Ex: −7,9

v cosθ,sinθ

Direction Angles• Given the direction angle of a vector, find the

component form of the vector of magnitude 6 in the same direction. Use the formula:

• Ex:

v cosθ,sinθ

v cosθ,sinθ =

Dot Product

• Dot Product: Multiplication of two vectors.• Let v = vxi + vyj, w = wxi + wyj.

• v · w = vxwx + vywy

• Example: Let v = 5i + 4j, w = –2i + 3j.• v · w = (5)(–2) + (4)(3) = –10 + 12 = 2**

**vectors v and w are orthogonal (perpendicular) iff v · w = 0.

Orthogonal, parallel vectors

• Do now: find cos 90 degrees• If the dot product = 0 the vectors are

orthogonal• If the direction (slope) of the vectors is the

same, the vectors are parallel• examples: Are the following vectors pairs

orthogonal or parallel or neither?u = −5,−1 ,v= 1,−5

w= 3,−5 ,z= −6,10

Orthogonal, parallel vectors• If the dot product = 0 the vectors are orthogonal• If the direction (slope) of the vectors is the same, the vectors are parallel• Examples: • Are the following vectors pairs orthogonal or parallel or neither?

• U and v are orthogonal• W and z are not orthogonal so check direction:• Direction of w = -5/3 and direction of z = -5/3 so

they are parallel!

u = −5,−1 ,v= 1,−5 : u• v=0

w= 3,−5 ,z= −6,10 :w• z=−68

Alternate Dot Product

• Alternate Dot Product formula: • v · w = ||v||||w||cos(θ). • The angle θ is the angle between the two vectors.

θV

W

Angles between 2 vectors

• Using the alternate formula, we solve for θ:• v · w = ||v||||w||cos(θ).

Cosθ =u• vu v

θ =Cos−1u • v

u vThe angle is between 0and 180 degrees

Example

• v = 5i + 4j, w = –2i + 3j. θ =Cos−1u • v

u v

The angle is between 0and 180 degrees

Example

• v = 5i + 4j, w = –2i + 3j. θ =Cos−1u • v

u v

They are neither orthogonal nor parallel so we find the angle…

u • v=2

Example

• v = 5i + 4j, w = –2i + 3j.θ =Cos−1

u • v

u v

θ = cos−12

41 • 13

⎛

⎝⎜

⎞

⎠⎟

θ = cos−12

533

⎛

⎝⎜

⎞

⎠⎟

θ ≈ 85oThe angle is between 0and 180 degrees

Student will be able to solve problems involving velocity and other quantities that can be represented by vectors

• Relation to real life: The Malaysian plane that is lost( in the recent news) could be tracked with calculations of velocity. This is where they started when looking for the wreckage.

• Vocabulary:• Bearing with respect to Navigation-coming out of

the north, measured clockwise• Magnitude: Length • Velocity: (has magnitude and direction)• Speed: the magnitude of velocity is speed.

Real life problem

• Components of a vector (gives east and north speeds)

• Problem: An airplane is flying on a bearing of 170o at 460mph.

• Find the component form of the velocity of the airplane.

Bearing

Wind vectors

• Next problem:• An airplane is flying on a bearing of 340o at

325 mph. A wind is blowing with the bearing of 320o at 40 mph. Find the component form of the velocity of the plane and the wind.

• Then find the actual speed and direction.

Calculating two vectors-1st: velocity vector:

Now Calculate the wind vector

Wind vector:

Wind vector added

• for the wind vector, 320 degrees corresponds to 130 degrees:

• • so the actual velocity =

w=40 cos130,sin130 ≈ −25.71,30.64

v+w= −136.67,336.04