Express a Vector in Component Form

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Transcript of Express a Vector in Component Form

Page 1: Express a Vector in Component Form
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Express a Vector in Component Form

Find the component form of with initial point A(1, –3) and terminal point B(1, 3).

= ⟨x2 – x1 , y2 – y1 ⟩ Component form

= ⟨1 – 1, 3 – (–3)⟩ (x1 , y1 ) = (1, –3) and ( x2 , y2 ) = (1, 3)

= ⟨0, 6⟩

Subtract.

Answer: ⟨0, 6⟩

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Find the component form of given initial point A(–4, –3) and terminal point B(5, 3).

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Find the Magnitude of a Vector

Find the magnitude of with initial point A(1, –3) and terminal point B(1, 3).

Simplify.

(x1 , y1 ) = (1, –3) and ( x2 , y2 ) = (1, 3)

Magnitude formula

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Find the Magnitude of a Vector

Answer: 6

CHECK From Example 1, you know that = ⟨0, 6⟩.

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Find the magnitude of given initial point A(4, –2) and terminal point B(–3, –2).

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Operations with Vectors

A. Find 2w + y for w = ⟨2, –5⟩, y = ⟨2, 0⟩, and z = ⟨–1, –4⟩.

2w + y = 2 ⟨2, –5⟩

+ ⟨2, 0⟩

Substitute.= ⟨4, −10⟩

+ ⟨2, 0⟩

Scalar multiplication

= ⟨4 + 2, –10 + 0⟩

or ⟨6, –10⟩

Vector addition

Answer: ⟨6, –10⟩

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Operations with Vectors

B. Find 3y – 2z for w = ⟨2, –5⟩, y = ⟨2, 0⟩, and z = ⟨–1, –4⟩.

3y – 2z = 3y + (–2z) Rewrite subtraction as addition.

= 3⟨2, 0⟩

+ (–2)⟨–1, –4⟩ Substitute.

= ⟨6, 0⟩

+ ⟨2, 8⟩

Scalar multiplication= ⟨6 + 2, 0 + 8⟩

or ⟨8, 8⟩

Vector addition

Answer: ⟨8, 8⟩

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Find 3v + 2w for v = ⟨4, –1⟩

and w = ⟨–3, 5⟩.

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Vectors are denoted with bold letters

(a, b)

This is the notation for a position vector. This means the point (a, b) is the terminal point and the initial point is the origin.

We use vectors that are only 1 unit long to build position vectors. i is a vector 1 unit long in the x direction and j is a vector 1 unit long in the y direction.

ij

(3, 2)

ij

i i

jjiv 23 +=

v = < a, b

>v = ai + bj

v = < 3, 2 >

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jiji 4352 −++−

If we want to add vectors that are in the form ai + bj, we can just add the i components and then the j components.

jiv 52 +−=

=+wv ji +=

Let's look at this geometrically:

i2−

j5 v

i3

j4−w

ij

When we want to know the magnitude of the vector (remember this is the length) we denote it

v ( ) ( )22 52 +−=Can you see from this picture how to find the length of v?

29=

jiw 43 −=

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A unit vector is a vector with magnitude 1.

jiw 43 −=

If we want to find the unit vector having the same direction as a given vector, we find the magnitude of the vector and divide the vector by that value.

What is ?w

( ) ( )2 23 4= + −w 525 ==

If we want to find the unit vector having the same direction as w we need to divide w by 5.

jiu54

53

−=Let's check this to see if it really is 1 unit long.

2 23 4 25 15 5 25

⎛ ⎞ ⎛ ⎞= + − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

u

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Use this formula to find the a unit vector

Unit vector with the same direction as v.

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Find a Unit Vector with the Same Direction as a Given Vector

Find a unit vector u with the same direction as v = ⟨4, –2⟩.

Unit vector with the same direction as v.

Substitute.

; Simplify.

or

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Find a Unit Vector with the Same Direction as a Given Vector

Rationalize the denominator.

Scalar multiplication

Rationalize denominators.

Therefore, u = .

Answer: u =

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Find a Unit Vector with the Same Direction as a Given Vector

Check Since u is a scalar multiple of v, it has the same direction as v. Verify that the magnitude of u is 1.

Magnitude Formula

Simplify.

Simplify.

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Find a unit vector u with the same direction as w = ⟨5, –3⟩.

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Write a Unit Vector as a Linear Combination of Unit Vectors

First, find the component form of .

Let be the vector with initial point D(–3, –3) and terminal point E(2, 6). Write as a linear combination of the vectors i and j.

= ⟨x2 – x1 , y2 – y1 ⟩

Component form= ⟨2 – (–3), 6 – (–3)⟩

(x1 , y1 ) = (–3, –3) and ( x2 , y2 ) = (2, 6)

= ⟨5, 9⟩

Subtract.

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Answer: 5i + 9j

Write a Unit Vector as a Linear Combination of Unit Vectors

Then, rewrite the vector as a linear combination of the standard unit vectors.

= ⟨5, 9⟩

Component form

= 5i + 9j ⟨a, b⟩

= ai + bj

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Let be the vector with initial point D(–4, 3) and terminal point E(–1, 5). Write as a linear combination of the vectors i and j.

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Find the component form of the vector v with magnitude 7 and direction angle 60°.

Find Component Form

Component form of v in terms of |v| and θ|v| = 7 and θ

= 60°

Simplify.

Answer:

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Find Component Form

Check Graph v = ≈

⟨3.5, 6.1⟩.

The measure of the angle v makes with the

positive x-axis is about

60° as shown, and

|v| = .

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Find the component form of the vector v with magnitude 12 and direction angle 300°.

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Direction Angles of Vectors

A. Find the direction angle of p = ⟨2, 9⟩

to the nearest tenth of a degree.

Direction angle equation

a = 2 and b = 9

Solve for θ.

Use a calculator.

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Direction Angles of Vectors

Answer: 77.5°

So the direction angle of vector p is about 77.5°, as shown below.

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Direction Angles of Vectors

B. Find the direction angle of r = –7i + 2j to the nearest tenth of a degree.

Use a calculator.

a = –7 and b = 2

Direction angle equation

Solve for θ.

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Direction Angles of Vectors

Answer: 164.1°

Since r lies in Quadrant II as shown below, θ = 180 – 15.9° or 164.1°.

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Find the direction angle of p = ⟨–1, 4⟩

to the nearest tenth of a degree.

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SOCCER A soccer player running forward at 7 meters per second kicks a soccer ball with a velocity of 30 meters per second at an angle of 10° with the horizontal. What is the resultant speed and direction of the kick?

Since the soccer player moves straight forward, the component form of his velocity v1 is ⟨7, 0⟩. Use the magnitude and direction of the soccer ball’s velocity v2to write this vector in component form.

Applied Vector Operations

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v 2 = ⟨| v2 | cos θ, | v2 | sin θ⟩ Component form of v2

= ⟨30 cos 10°, 30 sin 10°⟩ |v2 | = 30 and θ

= 10°

⟨29.5, 5.2⟩

Simplify.

Add the algebraic vectors representing v1 and v2 to find the resultant velocity, r.

r = v1 + v2 Resultant vector= ⟨7, 0⟩

+ ⟨29.5, 5.2⟩

Substitution

= ⟨36.5, 5.2⟩

Vector Addition

Applied Vector Operations

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Applied Vector Operations

The magnitude of the resultant is |r| = or about 36.9. Next find the resultant direction θ.

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Answer: 36.9 m/s; 8.1°

Applied Vector Operations

Therefore, the resultant velocity of the kick is about 36.9 meters per second at an angle of about 8.1° with the horizontal.

⟨a, b⟩

= ⟨36.5, 5.2⟩

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SOCCER A soccer player running forward at 6 meters per second kicks a soccer ball with a velocity of 25 meters per second at an angle of 15° with the horizontal. What is the resultant speed and direction of the kick?