VECTORS

This past week in advanced math we learned about Dot product.
This lesson was pretty simple. 


Dot product= u*v = <x1,y1> * <x2,y2> = x1 x2 + y1 y2


Here are the rules to know when approaching dot product equations:


-If the dot product equals 0, then the vectors are orthogonal (perpendicular).


-If the vectors are multiples of each other, then they are parallel.


-In order to find the angle b/w two vectors use: cos theta = u*v over the magnitude of u and v




Example 1: u(2, -1) v(3,6) w(-5,3)


Find the u*v and v*w. Show that u and v are orthogonal and v and w are parallel


u*v = 2(3) + -1(6) = 0 which means they are perpendicular because it equals 0


v*w = 3(-5) + 6(3) = 3


w/v = 18/-15 = -6/5
-They are multiples of each other therefore they are parallel.


Example 2: To the nearest degree find the angle between the vectors (1,2) and (-3,1)


u*v = 1(-3) + 2(1) = -1


magnitude of u = square root of 5
magnitude of v = the square root of 10


-Now following the formula in point 3 we need to do cos theta
1/ square root of 5 times square root of 10
theata = inverse of cos(-1/ square root of 50) = 81.867
-You find it in the 1st and 3rd quadrents: 98 and 261


wink ;-)

Doing this right before 12>

Dot Product has many types according to the 4 below:
u * v = <x1, y1> * <x2, y2> = x1x2 + y1y2
If it equals 0, then the vectors are perpendicular.
If the vectors are multiples of each other then they're parallel.
Properties of Dot Products:
1. Commutative: u * v = v * u
2. Squared: u* u =! u^2; u * u = |u^2|
3. k(u * v) = ku * v
4. u * u = |u|^2

Ex: u(3, -6) v(4,2) w(-12,-6) -Find the u*v and v*w. Show that u and v are orthogonal and v and w are parallel u*v = 3(4) + -6(2) = 0 which means they are perpendicular because it equals 0 v*w = -12(2) + 2(-6) = -60 w/v = -12/4 = -3 -6/2 = -3 : They are multiples of each other therefore they are parallel

DOT Product: u*v= * =x1x2+y1y2 If the dot product equals zero, then the vectors are orthogonal, which means perpendicular. If the vectors are multiples of each other, then they are parallel. To find the angle between two vectors, use the formula: cos(theta)= u*v/magnitudeU * magnitudeV  Properties of the DOT product: 1.  commutative: u * v = v * u 2.  squared u * u = magnitude of u^2 3.  K(u * v)= Ku * v doesn't distribute 4.  u * (v+w) = u * v + u * w Example 1: Find (2,3) * (4,-5) 2(4)+3(-5) =-7 Example 2: Find (3/5,4/5) * (1/2,-3/2) 3/5(1/2)+4/5(-3/2) =-9/10 Example 3: Find the value of a if the vectors (6,-8) and (4,a) are parallel. a/4=3/6 6a=12 divide both sides by 6 a=2  Example 4:  u(3, -6) v(4,2) w(-12,-6) -Find the u*v and v*w. Show that u and v are orthogonal and v and w are parallel.   u*v = 3(4) + -6(2) = 0 which means they are perpendicular because it equals 0  v*w = -12(2) + 2(-6) = -60 w/v = -12/4 = -3 -6/2 = -3 : They are multiples of each other therefore they are parallel

Dot Product

DOT Product: u*v= * = x1x2+y1y2 -- If the dot product equals zero, then the vectors are orthogonal, which means perpendicular. -- If the vectors are multiples of each other, then they are parallel. -- To find the angle between two vectors, use the formula: cos(theta)= u*v/magnitudeU * magnitudeV -- Properties of the DOT product: 1.  commutative: u * v = v * u 2.  squared u * u does not equal u^2 u * u = magnitude of u^2 3.  K(u * v)= Ku * v doesn't distribute 4.  u * (v+w) = u * v + u * w _____________________________________________________________________ Ex.1: u(3, -6) v(4,2) w(-12,-6) -Find the u*v and v*w. Show that u and v are orthogonal and v and w are parallel u*v = 3(4) + -6(2) = 0 which means they are perpendicular because it equals 0 v*w = -12(2) + 2(-6) = -60 w/v = -12/4 = -3 -6/2 = -3 : They are multiples of each other therefore they are parallel

Dot Product

Dot Product looks like this:
u * v = <x1, y1> * <x2, y2> = x1x2 + y1y2
If it equals 0, then the vectors are perpendicular.
If the vectors are multiples of each other then they're parallel.
Properties of Dot Products:
1. Commutative: u * v = v * u
2. Squared: u* u =! u^2; u * u = |u^2|
3. k(u * v) = ku * v
4. u * u = |u|^2
To find angle b/t two vectors: cos(theta) = u * v / |u| |v|
Examples:
_____________________________________________

1.) Given u = (3, -6), v = (4, 2), w = (-12, -6), find u · v and v · w
and show that u and v are perpendicular and v and w are parallel.

(3)(4) + (-6)(2)
12 + (-12) = 0
-12/4 = -3


(4)(-12) + (2)(-6)
-48 + (-12) = -60
-6/2 = -3

Cool Stuff About Math

A trig chart is a derived out list of trig functions. On there is six trig functions: sine (sin) cosine (cos) secant (sec) cosecant (csc) tangent (tan) and cotangent (cot). You can use the trig chart to evaluate angles in degrees or radians.

Trig Chart
(bold applies for all six functions)
sin 0 = 0 cos=1
sin 30 (pie/6) = 1/2 =sqaure root3/3
sin 45 (pie/4) =square root2/2 =square root2/2
sin 60 (pie/3) =square root3/3 =1/2
sin 90 (pie/2) =1 =0

scs=undefined sec=1
=2 =2 square root 3/3
=square root 2 =square root 2
=2 square root 3/3 =2
=1 =undefined

tan=0 
cot=undefined=
3 square root3/3 =square root 3
=1 =1
=sqaure root 3 =3 sqare root3/3
=undefined =0

ex:using trig chart evaluate
A. cos 30 B. tan pie/4
square root3/3 1

vectors

Find u*v and v*w.  Show that u and v are orthogonal and v and w are parallel.
u=(3,-6) v=(4,2) w=(-12,-6)
u*v=3(4)+-6(2)=0
v*w=4(-12)+2(-6)=-60
w/v=-12/4=-3
        -6/2=-3
We are given 3 different points:u,v, and w.  First it tells you to find u*v and v*w.  These are called dot products.  To solve a dot product, you have to multiply your x1 and x2 together and your y1 and y2 together.  You get those two products and add them together to get your answer.  If the dot product equals 0, then the vectors are orthogonal.  Orthogonal means perpendicular.  The dot product of u and v equals zero, so that product is orthogonal.  Now it says to show that v and w are parallel.  If the vectors are multiples of each other they are parallel.  When you divide -12 by 4, you get -3, and when you divide -6 by 2, you get -3 also.  -3 are multiples of each other, so v and w are parallel.

Dot Product

Dot Product: u*v = <x1,y1> * <x2,y2> = x1 x2 + y1 y2
-If the dot produect equals 0 thent he vecotrs are orthogonal (perpendicular)
-If the vectors are multiples fo each other they are parallel
-To find the angle b/w two vectors cos theta = u*v over the magnitude of u and v


Example 1: u(3, -6) v(4,2) w(-12,-6)
-Find the u*v and v*w. Show that u and v are orthogonal and v and w are parallel

u*v = 3(4) + -6(2) = 0 which means they are perpendicular because it equals 0
v*w = -12(2) + 2(-6) = -60

w/v = -12/4 = -3
-6/2 = -3 : They are multiples of each other therefore they are parallel

Example 2: To the nearest degree find the angle between the vectors (1,2) and (-3,1)

u*v = 1(-3) + 2(1) = -1

magnitude of u = square root of 5
magnitude of v = the square root of 10

-Now following the formula in point 3 we need to do cos theta
1/ square root of 5 times square root of 10
theata = inverse of cos(-1/ square root of 50) = 81.867
-You find it in the 1st and 3rd quadrents: 98.130 and 261.87

--Daniellee

-The arithmetic mean of number's a and b = a+b/2 -The geometric mean of number's a and b = square root of ab Note: A sequence is a list of numbers. An arithmetic sequence is a sequence that is generated by adding the same number each time. Arithmetic Sequence formula : tn=t1 + (n-1)d t1= first term n= term number d= difference  tn= ___ term A geometric Sequence is a sequence that is generated by multiplying the same number each time. (to divide, we use fractions) formula: tn=t1(r)^(n-1) t1= first term r= what is being multiplied n= term number tn= ___ term Example: 1.) Find the first four terms of the sequence and state whether the sequence is arithmetic, geometric, or neither. tn= 4n+12 n=1 4(1)+12=16 n=2 4(2)+12=20 n=3 4(3)+12=24 n=4 4(4)+12=28 It is arithmetic because you are adding 4 each time! Example: 2.) Find the first four terms of the sequence and state whether the sequence is arithmetic, geometric, or neither. tn=-2n+8 n=1 -2(1)+8=6 n=2 -2(2)+8=4 n=3 -2(3)+8=2 n=4 -2(4)+8=0 It is arithmetic because you are adding -2 each time!

DOT Product!

Hello,I hope everyone had a great weekend. Im currently stressing over the trig exam this week; I'm sure I am not the only one. 
Anyway,

DOT Product:
u*v=<x1, y1> * <x2, y2> = x1x2+y1y2

-- If the dot product equals zero, then the vectors are orthogonal, which means perpendicular.
-- If the vectors are multiples of each other, then they are parallel.
-- To find the angle between two vectors, use the formula:
cos(theta)= u*v/magnitudeU * magnitudeV

-- Properties of the DOT product:
1.  commutative:
u * v = v * u
2.  squared
u * u does not equal u^2
u * u = magnitude of u^2
3.  K(u * v)= Ku * v
doesn't distribute
4.  u * (v+w) = u * v + u * w

Example 1:
Find (2,3) * (4,-5)
2(4)+3(-5)
=-7

Example 2:
Find (3,-5) * (7,4)
3(7)+-5(4)
=1

Example 3:
Find (-3,0) * (5,7)
-3(5)+0(7)
=-15

Example 4:
Find (3/5,4/5) * (1/2,-3/2)
3/5(1/2)+4/5(-3/2)
=-9/10

Example 5:
Find the value of a if the vectors (6,-8) and (4,a) are parallel.
a/4=3/6
6a=12
divide both sides by 6
a=2

Example 6:
If u=(-2,3), find u*u
-2(-2)+3(3)
=13

Example 7:
If u=(5,-3) and v=(3,7), verify that u * v = v * u
5(3)+-3(7)=-6
3(5)+7(-3)=-6
-6=-6

Example 8:
If u=(5,-3) and v=(3,7), verify that 2(u * v) = (2u) * v
2(-6)= -12
10(3)+ -6(7)=-12
-12=-12

The dot product

If the dot product equals zero, the vectors are perpendicular.  If the vectors are multiples of each other, they are parallel.  Formula for dot product:  u · v = · = x1x2 + y1y2 You  use this formula to find the angle between two vectors: cosƟ = u · v/|u||v|   Properties of the dot product 1. Commutative  u · v = v · u u · u ≠ u^2 2. Squared  u · u = |u|^2 3. k(u · v) = ku · v  doesn't distribute 4. u(v + w) = u  · v + u · w Example 1- V=2i+4j   W=i+5j v · w  =  (2)(1) + (4)(5)  =  22 Example 2- Find the angle between  v=  2i + 3j + k and w  =  4i + j + 2k. ||v||=√4+9+1=√14 ||w||=√16+1+4=√21 v . w  =  8 + 3 + 2 = 13 Ɵ=cosƟ^-1((13)/(√14)(√21))

double and half angles O.o oooooo

sin2α= 2sinαcosα

cos2α= cos²α -sin²α

cos2α= 1 -2sin²α

cos2α= 2cos²α -1

tan2α= 2tanα/1 -tan²α

sin(α/2)= ±√((1 - cosα)/2)

cos(α/2)= ± √((1 + cosα)/2)

tan(α/2)= ±√(1 - cosα)/(1+cosα)

tan(α/2)= sinα/(1 + cosα)

tan(α/2)= (1 - cosα)/sinα

the plus or minus things are determined by where the original angle is.

Example 1

Find the exact value of sin120° (Hint... use your trig chart.).

sin120°= sin2(60°)

sin120°= 2sin60°cos60°

sin120°= 2(½)*√(3)/2

sin120° = √(3)/2

Example 2 

Find the exact value of cos15° (Hint... use the trig chart.).

cos15°= cos(30°)/2

cos15°= ± √(1 + cos30°)/2

cos15°= ± √(1+½)/2

cos15°= ± √(3/2)/2

cos15°= ±√3/4

cos15°= ±(√3)/2

cos15°= (√3)/2

Example 3

Simplify as much as possible:

sinΘtanΘ + cos2ΘsecΘ

convert everything to sine and cosine

sinΘ(sinΘ/cosΘ)+cos2Θ(1/cosΘ)

change cos2Θ to cos²Θ -sin²Θ

(sin²Θ/cosΘ) + ((cos²Θ -sin²Θ)/cosΘ))

algebra

(sin²Θ/cosΘ) + cosΘ -(sin²Θ/cosΘ)

cancel

cosΘ 

The Spontaneous Dot Product

 Today's lesson is on the Dot Product.

First here are some key notes you need to know about the dot product

• If the dot product equals zero, then the vectors are orthogonal, which means perpendicular. If the vectors are multiples of each other, then they are parallel. 
• When finding the angle between two vectors, use the formula: cos(theta)= u*v/magnitudeU * magnitudeV

Now how do we find the dot product:

• First you have two points.
• To find the product of something means to multiply.
• u*v= (x1)(x2)+(y1)(y2) is the dot product formula.

Properties of the Dot Product:

• 1.  commutative: u * v = v * u
• 2.  squared: u * u = magnitude of u^2 
• 3.  K(u * v)= Ku * v does not distribute. "I had problems with this on my test LOL."
• 4.  u * (v+w) = u * v + u * w

Examples:

• Find (3,2) * (5,-4): 3(5)+2(-4) =7
• Find (3/4,2/4) * (1/2,-5/2): 3/4(1/2)+2/4/(-5/2) =0 =orthogonal
• Find the value of (a) if the vectors (12,a) and (6,3) are parallel. a/4=6/8 6a=12 divide both sides by 8 a=3

That's the basics of the Dot product.

Sum and Difference Formulas

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
sin(x-y) = sin(x)cos(y) - cos(x)sin(y)


cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
cos(x-y) = cos(x)cos(y) + sin(x)sin(y)


tan(x+y) = [tan(x) + tan(y)] / [1-tan(x)tan(y)]
tan(x-y) = [tan(x) - tan(y)] / [1+tan(x)tan(y)]


These formulas are used to find the exact value of something that isn't on the trig chart, like cos(15°).You will use only angles on the trig chart to plug into the formulas. (30°,45°,60°,90°, and any angle that has a reference angle on the trig chart)


Ex. 1cos(15°)

45-30 = 15

cos(45-30) = cos(45)cos(30) + sin(45)sin(30)

cos(15°) = (√(2)/2)*(√(3)/2) + (√(2)/2)*(1/2)

cos(15°) = (√(6)/4) + (√(2)/4)

cos(15°) = [√(6)+√(2)] /4


Ex. 2tan(75°)

45 + 30 = 75

tan(45+30) = [tan(45) + tan(30)] / [1 - tan(45)tan(30)]

tan(75) = [1 + (√(3)/3)] / [1 - (1)*(√(3)/3)]

tan(75) = [(3+√(3))/3)] / [(3 - √(3))/3)]

tan(75) = [9+√(3)] / [9-√(3)]

tan(75) = [81 + 18√(3) + 3] / [81-3]

tan(75) = [84 + 18√(3)] / [78]

tan(75) = [14 + 3√(3)] / [13]


Ex. 3sin(105°)

45+60 = 105

sin(45+60) = sin(45)cos(60) + cos(45)sin(60)

sin(105) = (√(2)/2)*(1/2) + (√(2)/2)*(√(3)/2)

sin(105) = (√(2)/4) + (√(6)/4)

sin(105) = [√(2)+√(6)] /4

Dot Product

u · v = <x1, y1> · <x2, y2> = x1x2 + y1y2

If the dot product = 0 then the vectors are perpendicular

If the vectors are multiples of each other they are parallel

To find the angle between 2 vectors cosƟ = u · v/|u||v|

Properties of the dot product

1. Commutative 2. Squared 3. k(u · v) = ku · v

u · v = v · u u · u ≠ u^2 doesn't distribute

u · u = |u|^2

4. u(v + w) = u · v + u · w

Ex 1. u = (3, -6) v = (4, 2) w = (-12, -6)

Find u · v and v · w. Show that u and v are perpendicular

and v and w are parallel

(3)(4)+ (-6)(2) (4)(-12) + (2)(-6)

12 + (-12) = 0 -48 + (-12) = -60

-12/4 = -3 -6/2 = -3

Ex 2. To the nearest degree find the angle between the

vectors (1, 2) and (-3, 1)

cosƟ = u · v/|u||v u · v = (1)(-3) + (2)(1) = -1

|u| = sqrt1^2 + 2^2 |v| = sqrt-3^2 + 1^2

= sqrt5 = sqrt10

cosƟ = -1/sqrt5 · sqrt10

Ɵ = cos^-1(-1/sqrt50) = 81.867

81.867 + 180 = 261.8 degrees

180 – 81.867 = 98.130 degrees

Ɵ = 98 degrees, 262 degrees

Reference Angles

Reference Angles are like reduced fractions. They must be between 0degrees and 90degrees.

Steps:
1. Find the original quadrant and sketch.
2. Determine if the angle is positive or negative using unit circle methods.
3. Subtract 360degrees or 180degrees until theta is between 0degrees and 90degrees.

Example 1:
Express each of the following in terms of a reference angle.
a. sin128degrees
subtract 180
=52
sin 128 is positive in the second quadrant
=sin52degrees
b. cos128degrees
subtract 180
=52
cos128 is negative in the second quadrant
=-cos52degrees

Example 2:
Express each of the following in terms of a reference angle.
a. sin310degrees
subtract 360
=50
sin310 is negative in the fourth quadrant
=-sin50degrees
b. sin1000degrees
subtract 360
=640degrees
subtract 360
=sin280
subtract 180
=80
sin280degrees is negative in the fourth quadrant
=-sin80degrees

Example 3:
Express each of the following in terms of a reference angle.
a. cos224.5degrees
subtract 180
=44.5
cos 224.5 is negative in the third quadrant

Inverses

Find the inverse of f(x)=2x-3
x=2y-3
y=x+3/2

To have an inverse that is a function it must pass the horizontal line test. The horizontal line test is when you draw a line across your paper and if your equation that you are given hits the line more than one time, your equation can not have an inverse. If the line only crosses your equation one time, then it can have an inverse and you can solve the equation. To find the inverse you switch the x and y in your equation given and then you solve for y. So for this equation I am given, I used the horizontal line test and since this equation is linear, it will pass the horizontal line test because it only crosses once. To solve this equation, I switched the x and y and then I added 3 to both sides and got 2y=x+3. Then I divided everything by 2 and I got y=x+3/2. So the inverse of this function is y=x+3/2.

Cool Formulas You May Want to Know

Chapter 7
-Convert degrees to radians: degrees = pi over 180
-Convert radians to degrees: rads x 180 over pi = degrees
-k= 1/2r^2(theta)
-k= 1/2rs
-s= r(theta)
-Unit Circle= extremely important!

Chapter 8
-m= tan(alpha)
-csc(theta)=1/sin(theta)
-sec(theta)=1/cos(theta)
-cot(theta)=1/tan(theta)
-sin^2(theta) + cos^2(theta) = 1
-1 + tan^2(theta) = sec^2(theta)
-1 + cot^2(theta) = csc^2(theta)

Chapter 9
-SOHCAHTOA
-sin(theta)= opp/hyp
-cos(theta)= adj/hyp
-tan(theta)= opp/adj
-sin A/a = sinB/b = sinC/c
-Right triangle: 1/2bh
-Non right triangle: 1/2(adj leg)(adj leg)sin(angle below)
-Law of Cosines: leg^2= adj leg^2 + other adj leg^2 - 2(adj leg)(other leg)cos(angle below)

Chapter 10
-cos(alpha plus or minus beta) = cos(alpha)cos(beta) minus or plus sin(alpha)sin(beta)
-sin(alpah plus or minus beta)= sin(alpha)cos(beta) plus or minus cos(alpha)sin(beta)
-tan(alpha+beta) = tan(alpha + beta)/1-tan(alpha)tan(beta)
-tan(alpha-beta) = tan(alpha-beta)/1+tan(alpha)tan(beta)
-sin2(alpha)= 2sin(alpha)cos(alpha)
-cos2(alpha)= cos^2(alpha)-sin^2(alpha)
-cos2(alpha)= 1-2sin^2(alpha)
-cos2(alpha)= 2cos^2(alpha)-1
-tan2(alpha)= 2tan(alpha)/1-tan^2(alpha)
-sin(alpha)/2= plus or minus square root of 1-cos(alpha)/2
-cos(alpha)/2= plus or minus square root of 1+cos(alpha)/2
-tan(alpha)/2= plus or minus square root of 1-cos(alpha)/1+cos(alpha)
-tan(alpha)/2= sin(alpha)/1+cos(alpha)
-tan(alpha)/2= 1-cos(alpha)/sin(alpha)

Some Cool Stuff To Know

tan(α + β) = tanα + tanβ/1 – tanαtanβ – Sum

tan(α - β) = tanα - tanβ/1 + tanαtanβ – Difference

Example 1.

 Suppose tanα = 1/3 and tanβ = ½

a. Find tan(α + β)

tanα + tanβ/1 – tanαtanβ = 1/3 + ½ /1 – (1/3)(1/2)

2/6 + 3/6 / 1 – 1/6 = 5/6 / 5/6 = 1

b. Find tan(α – β)

tanα - tanβ/1 + tanαtanβ = 1/3 – ½ /1 + (1/3)(1/2)

2/6 – 3/6 / 1 + 1/6 = -1/6 / 7/6 = -6/42 = -1/7

Example 2.

 tanα = 2/3 and tanβ = ½

a. tanα + tanβ/1 – tanαtanβ = 2/3 + ½ /1 – (2/3)(1/2)

4/6 + 3/6 / 1 – 2/6 = 7/6 / 2/3 = 21/12 = 7/4

b. tanα - tanβ/1 + tanαtanβ = 2/3 – ½ /1 + (2/3)(1/2)

4/6 – 3/6 / 1 + 2/6 = 1/6 / 4/3 = 3/24 = 1/8

Domain and Range

Domain-the interval of x values where the graph exists.
Rage-the interval of y values where the graph exists.

For zeros x-intercept, root -set = o. Solve for x.
To be a function the graph must pass the vertical line test. For this you draw lines up and down to see if part of the graph crosses the line twice. If if crosses any line more than once it is not a function.
*If given points, the domain is the list of all x values, while range is a list of all y values in a set of points indicated.

1.Polynomials
Domain: 00 or -00 always
Range: odd (-oo or 00). X^2 always (-b/2a, 00) or (-oo, -b/2a)

2.Fractions
Domain: -factor top and bottom
-cancel if possible (mark #)
-set bottom = 0, then solve the x value
-write in interval notation stopping at #s found in { }
Range: is none

Example1.

Find the domain and range of 3+2x^5
Domain = -oo, oo
Range= -oo, 00

Example2.

Is the following a function? (2,-3) (4,-2) (2,2) , (3, -2)
No because when it is drawn out in a graph it does not pass the vertical line test.

Example3.

(x-2) (x-3)/(x-6) (x-3) Find domain and range
-cancel the pay attenction only to top numbers
Domain= (-oo, 3) u (3,6) u (6, 00)
Range= none

Vector Explanations with Rico Gathers

Today we are going to talk about Vectors.

First what's a vector?
 -a vector is a slope. It contains an x-component along with an Y-component.

There are many ways to find vectors. Here's two of the more simple ones. Although all are pretty easy.

1) Vector Addition : V+U= <a,b> + <c,d> = <a+c,b+d>

        -Ex. (1,9) + (4,5)
        
             (1+4,9+5) = <5,14>

2) Vector Subtraction : V-U= <a,b> - <c,d> = <a-c,b-d>


            -Ex. (1,9) - (4,5)
        
             (1-4,9-5) = <-3,4>

That's our lesson for today with Rico Gathers ;-).

Formulas and Vector Equations

I'm bored, so I thought I'd get started a little early :)

Distance: Square root of (x2-x1)^2 + (y2-y1)^2 + (z2+z1)^2

Midpoint: (x1+x2/2  y1+y2/2  z2+z1/2

Sphere: (x-xo)^2 + (y-yo)^2 + (z-zo)^2 = r^2

Vector Equation: (x,y,z) = (xo,yo,zo) + t(a,b,c)

Parametric: x= xo+at  y=yo+bt  z=zo+ct

Magnitude of U = square root of a^2+b^2+c^2

EXAMPLE1: With the give points, find the vector equation and parametric (2,3) (0,1)

(2,3) + (0,1)

= (2,4)

(x,y)= (2,3) + t(2,4) 

x= 2 + 2t

y= 3 + 4t

step 1: Add the points to find your vector

step 2: Plug your first given point and your new vector straight into the formula

step 3: x= the x value of your first given point and the a value of the new vector with t behind it

step 4: y= the y value of your first given point and the b value of the new vector with t behind it 

EXAMPLE2: Find the magnitude of u (3,4)

square root of 3^2 + 4^2

= square root of 9 + 16

= square root of 25

=5

step 1: plug given points into your magnitude formula (square root of x^2 + y^2)

step 2: 3x3 + 4x4

step 3: after you multiply, then add, you are going to get 25

step 4: square root 25 to get your final answer of 5

EXAMPLE 3: Find a vector equation of the line through A(3,4) & B(5,5)

(3,4) + (5,5) = (8,9)

(x,y)= (3,4) + t(8,9)

Step1: Add points to find vector.

Step2: Plug your first given point and your new vector directly into the formula 

This is my blog for this week. Vectors A vector is a slope. vector addition: v+u=+= vector subtraction: v-u= -= vector multiplication: kv=k= To find a vector from two points: p2-p1 Vector equation: (x,y)=(x0, y0)+t(a,b) Parametic equations: x=x0+at y=y0+bt absolute value of v= square root of x^2+y^2= magnitude of a vector Example 1: Give the component form of Ab and find the magnitude of AB when A(1,-2), B(3,-2) (3,-2)-(1,-2)= (3-1,-2+2) = <2,0> vector square root of -4^2+-3^2= square root of 16+9= square root of 25 = 5 magnitude Example 2: Let u= (3,1) and v= (-8,4). Find u+v. (3+-8, 1+4) =(-5,5) Example 3: Let u= (3,1) v= (-8,4) and w= (-6,-2). Calculate each expression. a. u+v (3+-8, 1+4) =<-5,5) b. u-v <3+8, 1-4> =<11,-3> c. 3u+w <9,3> + <-6,-2> =<3,1> Have a lovely day.

vectos

Vector addition - v + u = + =

Vector subtraction - v – u = - =

Vector multiplication – kv = k =

To find a vector from 2 points do P2 – P1

Vector equation - (x, y) = (x˳, y˳) + t(a, b)

Parametric equations:

x = x˳ + at

y = y˳ + bt 

magnitude of a vector

-|u| = sqrt x^2 + y^2     

Example 1: Finding the magnitude/ lenght

http://www.youtube.com/watch?v=6GoMXuE1FOw

Example 2: Addition and subtraction of vectors

http://www.youtube.com/watch?v=3yxYAy7cCBQ&feature=related

Vectors

This past week we learned about vectors.

A vector is the same as a slope. Here are the steps to solving certain vector equations:

1)Vector addition: v+u = + =
2)Vector subtraction: v-u = - =
3)Scalar multiplication: kv = k =
4)To find a vector from two points: P2 - P1
5)Vector equation: (x,y) = (x0,y0) + t(a,b)
6)Parameter equations: x = x0 +at y = y0 + bt
7) /v/ = square root of x^2 + y^2 :this equation finds the magnitude of a vector

Example: Give the component form of AB and find /AB/. A (4,7) B(3,2)

P2 - P1: (3, 2) - (4, 7) = (3-4, 2-7) = <-1, -5>

square root of -1^2 + -5^2
= square root of 1+25
= square root of 26

Example 2: Find vector of OP and give its component form

P(6, 72 degrees)
x/6 = cos50
x = 6 cos 72
x = 1.854

y/6 = sin72
y = 6 sin 72
y = 5.706

<1.854, 5.706>

Example 3: u = (6,4) v = (-1,9)
u + v
<6 + -1, 4 + 9> = <5, 13>

Vectorssss!

Section on Vectors:

A vector=Slope
vector addition: v+u=+= vector subtraction: v-u= -=
 vector multiplication: kv=k= To find a vector from two points:p2-p1 Vector equation:(x,y)=(x0, y0)+t(a,b) Parametic equations:x=x0+aty=y0+bt absolute value of v= square root of x^2+y^2= magnitude of a vector Example 1: Give the component form of Ab and find the magnitude of AB whenA(2,-2), B(4,-2)
(4,-2)-(2,-2)= (4-2,-2+2) = <2,0> vector square root of 2^2+-0^2= square root of 4+0= square root of 4  magnitude Example 2: Let x= (2,6) and y= (-1,4). Find x+y. (2+-1, 6+4)=(1,10) Example 3: Let x= (4,8) y= (1,4) and z= (-5,-2). Calculate each expression. a. x+y(4+1, 8+4)=<5,12> b. x-y<4-1, 8-4>=<3,4> c. 3y+z<3,12> + <-5,-2>=<-2,10>
      

Vector addition and subtraction!:)

Vectors are the same as slope!
Vector addition: u + v. ux+vx, uy+vy
Vector subtration: u - v. ux-vx, uy-vy
To find a vector from 2 points: p2-p1

Examples!
Using u<3,7> v<6,9>.
1.) find u+v
3+6,7+9=<9,16>

2.) find u-v
3-6,7-9=<-3,-2>

3.) find the vector from (-5,6) & (4,-2)
4--5,-2-6=<9,-8>

Vector Addition and Subtraction

Vectors are basically slopes.
Vectors look like this:
<x, y>
Vector addition: v + u = (x + y) + (x0 + y0) = <x + x1, y + y1>
Vector subtraction: same but with minuses (minii?)
Vector equation: (x, y) = (x1, y1) + t(a, b)
Magnitude: |u| = sqrt(x^2 + y^2)

Examples:
Subtract it!
A(2, 3) B(-1, 5)
<-3, 2>

Add it!
A(4, 1) B(3, 2)
<-1, 1>

Vectors!

Hope everyone enjoyed the fair and had a good weekend!

-Vector is the same as slope
-Vector addition v+u = <a,b> + <c,d> = <a+b, b+d>
-Vector subtration v-u = <a,b> - <c,d> = <a-c, b-d>
-Scalar multiplication kv = k<a,b> = <ka, kb>
-To find a vecotr from two points so P2 - P1
-Vector equation - (x,y) = (x0,y0) + t(a,b)
-Parameter equations: x = x0 +at     y = y0 + bt
-/v/ = square root of x^2 + y^2 :this equation finds the magnitude of a vector

Example 1: Give the component form of AB and find /AB/

P2 - P1: (3, -2) - (1, -2) = (3-1, -2-2) = <2, -4>

square root of 2^2 + -4^2
= square root of 4 + 6
= square root of 20

Example 2: Find vector of OP and give its component form

P(6, 72 degrees)
x/6 = cos72
x = 6 cos 72
x = 1.854

y/6 = sin72
y = 6 sin 72
y = 5.706

<1.854, 5.706>

Example 3: u = (3,1) v = (-8,4)

u + v
<3 + -8, 1 + 4> = <-5, 5>

--Daniellee

Vectors

A vector is a slope.

vector addition: v+u=+=

vector subtraction: v-u= -=

vector multiplication: kv=k=

To find a vector from two points:

p2-p1

Vector equation:

(x,y)=(x0, y0)+t(a,b)

Parametic equations:

x=x0+at

y=y0+bt

absolute value of v= square root of x^2+y^2= magnitude of a vector

Example 1:

Give the component form of Ab and find the magnitude of AB when A(1,-2), B(3,-2)

(3,-2)-(1,-2)= (3-1,-2+2) = <2,0> vector

square root of -4^2+-3^2= square root of 16+9= square root of 25 = 5 magnitude

Example 2:

Let u= (3,1) and v= (-8,4). Find u+v.

(3+-8, 1+4)

=(-5,5)

Example 3:

Let u= (3,1) v= (-8,4) and w= (-6,-2). Calculate each expression.

a. u+v

(3+-8, 1+4)

=<-5,5)

b. u-v

<3+8, 1-4>

=<11,-3>

c. 3u+w

<9,3> + <-6,-2>

=<3,1>

Vectors!

Hello everyone,
I hope you all had a fantastic weekend!
It's my birthday today, just thought everyone would love to know that, haha :)
Anywayyyy, back to school tomorrow... already...

vector: slope
vector addition: v+u=<a,b>+<c,d>=<a+c,b+d>
vector subtraction: v-u= <a,b>-<c,d>=<a-c,b-d>
vector multiplication: kv=k<a,b>=<ka,kb>

To find a vector from two points:
p2-p1

Vector equation:
(x,y)=(x0, y0)+t(a,b)

Parametic equations:
x=x0+at
y=y0+bt

absolute value of v= square root of x^2+y^2= magnitude of a vector

Example 1:
Give the component form of Ab and find the magnitude of AB.
A(1,-2), B(3,-2)
(3,-2)-(1,-2)= (3-1,-2+2) = <2,0> vector
square root of -4^2+-3^2= square root of 16+9= square root of 25 = 5 magnitude

Example 2:
Polar coordinates of point P are given and O is the origin.  Give its components.
P(6,72degrees)
x=6cos72= 1.854
y=6sin72= 5.706

Example 3:
Let u= (3,1) v= (-8,4) and w= (-6,-2).  Calculate each expression.
a. u+v
(3+-8, 1+4)
=<-5,5)
b. u-v
<3+8, 1-4>
=<11,-3>
c. 3u+w
<9,3> + <-6,-2>
=<3,1>

Vectors

A vector is a slope.

To find a vector from 2 points: P2-P1

Addition:

v + u =< x, y > + < x1, y1 > = < x+x1, y+y1 >

Scalar Multiplication:

kv= k < x, y > = < kx, ky >

Vector Equation:

(x,y) = (xo,yo) + t(a,b)

Parametric Equations:

x=xo+at

y=yo+bt

Magnitude of a Vector:

√(x² + y²)

Component form:

< rcosΘ, rsinΘ >

_________________________________________________________

Ex. 1: Give the component form of Ab and find the magnitude of AB.
A(1,-2), B(3,-2)
(3,-2)-(1,-2)= (3-1,-2+2) = (2,0) vector
square root of -4^2+-3^2= square root of 16+9= square root of 25 = 5 magnitude

Ex.2: Let u= (3,1) and v= (-8,4)

u+v
(3+-8, 1+4)

=(-5,5)

Lovely Vectors

A vector is a slope.

To find a vector from 2 points: P2-P1

Addition:

v + u =< x, y > + < x1, y1 > = < x+x1, y+y1 >

Scalar Multiplication:

kv= k < x, y > = < kx, ky >

Vector Equation: 

(x,y) = (xo,yo) + t(a,b)

Parametric Equations:

x=xo+at 

y=yo+bt

Magnitude of a Vector:

√(x² + y²)

Component form:

< rcosΘ, rsinΘ >

Ex.1: A force of 4 N acts on angle of 30° with the x-axis find in component form

Your force is r.

< 4cos(30°), 4sin(30°) >

=< 2√(3), 2 >

Ex. 2: Find the vector of (2,1) and (4,0). 

a. Find the vector equation.

b. Express in Parametric form.

Find vector P2-P1 :

(4-2, 0-1) = <2,-1>

a. Using the first point (2,1) and the vector <2,-1>, we can write the equation in the form (x,y) = (xo,yo) + t(a,b)

(x,y) = (2,1) + t(2,-1)

b. Parametric form:

x = 2 + 2t

y = 1 - t

Ex. 3<4,2> - 2<1,-2>

<12,6> - <2,-4>

<10, 10>

vectors

Vectors are slopes

to add vectors: v + u =< x, y > + < x1, y1 > = < x+x1, y+y1 >

Scalar Multiplication kv= k < x, y > = < kx, ky >

To find a vector from 2 points: P2-P1

The vector equation is: (x,y) = (xo,yo) + (a,b)

Parametric Equations: x=xo+at ; y=yo+bt

|v| = √ x² + y² = the magnitude of the vector

Component form < rcosΘ, rsinΘ >

a force can be expressed as a vector

Example 1: A force of 2 N acts on angle of 45° with the x-axis find in component form

< 2cos(45°), 2sin(45°) >

=< √2, √2 >

the only step) plug into the formula r is 2 and Θ is 45°

Dot Product

u•v = < X1, Y1> • < X2,Y2 > = X1 X2 + Y1 Y2

-If the dot product equal to 0, then the vectors are orthogonal which is just the really over complicated way of saying perpendicular

-If the vectors are multiples of each other they are parallel

-To find the angle between 2 vectors use the formula: cosΘ= (u*v) / (|u||v|)

-Properties of the dot products:

Commutative: u*v = v*u

Squared: u*u is not u² u*u = |u²|

k(u*v) = ku*v (does not distribute)

u*(v + w) = u*v + u*w

Example 1: u = (2,-4) v = (6,3)

u•v = 3(4) + (-6)(2) = o

*it is perpendicular

Step1) Plug into formula (see above)

Step2) Multiply and add or plug into calculator if you are lazy

Vectors

Vector addition - v + u = + =

Vector subtraction - v – u = - =

Vector multiplication – kv = k =

To find a vector from 2 points do P2 – P1

Vector equation - (x, y) = (x˳, y˳) + t(a, b)

Parametric equations:

x = x˳ + at

y = y˳ + bt

-|u| = sqrt x^2 + y^2 ---> magnitude of a vector

Ex 1. Given A(4, 2) and B(9, -1), express AB in a compund

form and find |AB|

(9, -1) – (4, 2) = <5, -3>

|AB| = sqrt 5^2 + (-3)^2 = sqrt 25 + 9 = sqrt 34

Ex 2. A force of 2N is represented by an angle of 145 degrees

with the x axis. Find its components.

x/2 = cos145 y/2 = cos145

x = 2cos145 = -1.638 y = 2cos145 = 1.147

<-1.638, 1.147>

On the unit circle:  0 degrees 90 degrees 180 degrees 270 degrees 360 degrees Or: pie/2 pie 3pie/2 2pie There are no steps when using the unit circle.  Sin Theta = y/r Cos Theta = x/r Tan Theta = y/x Csc Theta = r/y Sec Theta = r/x Cot Theta = x/y   Example 1 Show that the point P is on the unit circle. Solution: We need to show that this point satisfies the equation of the unit circle, that is, x2+y2=1. Therefore P is on the unit circle.

vectors

A(1,-2)  B(3,-2)
P2-P1=(3,-2)-(1,-2)=<2,0>
component AB= square root of 2^2+0^2= square root of 4=2

The directions tell you to express AB in component form.  We are given two points.  We have point A at (1,-2) and point B at (3,-2).  Now you have to find a vector from the two points you have.  To do this, you use the formula you have to know which is P2-P1.  That means point 2 minus point 1.  So now you do (3,-2)-(1,-2) and you just subtract it like normal subtraction and you get <2,0> as your vector.  Now you find the component of AB.  To do this, you use almost the same exact formula as the pathagorean thereom.  Next you use you vector to find the component of AB.  This formula is the square root of x^2+y^2.  So we do the square root of 2^2+0^2 and we got the square root of 4.  That reduces to 2, and that is your answer.

Vectors & Dot Product

• Vector - slope
• Vector addition v + u = <a,b> + <c,d> = <a+c,b+d>
• Scalor Multiplication kv= k<a,b> = <ka,kb>
• To find a vector from 2 points do p2-p1
• Vector equation - (x,y) = (xo,yo) + t(a,b)
• Parametric Equations: x=xo+at ; y=yo+bt
• Negative absolute value of v = square root of x^2+y^2 which = the magnitude of the vector
• Component form < r cos theta, r sin theta >
• FORCE = VECTOR

Example 1: A force of 10 N acts on angle of 130 degrees with the x-axis find in component form

< 10 cos 130 degrees, 10 sin 130 degrees >

= <-6.428, 7.66>

Step1) Draw out your triangle

Step2) Plug givens into your component form formula

Step3) Your 10 is r and 130 is your theta

Dot Product 

U x V = < X1 x Y1 > x < X2 x Y2 > = X1 X2 + Y1 Y2

-If the dot product = o, then the vectors are orthogonal (perpendicular) 

-If the vectors are multiples of each other they are parallel 

-To find the angle between 2 vectors Cos Theta = U x V/ Abs value of U x Abs value of V

-Properties of the dot products: 

1. Communtive .....  U x V = V x U
2. Squared ......  U x U not equal to U^2 ; U x U = Absolute Value of U^2
3. K (U x V) = KU x V (doesn't distribute)
4. U x (V + W) = U x V + U x W

Example 1: U = (3,6) V = (4,2). Find the Dot product of U V

(3,6) (4,2)

12+12 = 24

Step1: Multiply the x's together
Step2: Multiply the y's together
Step3: Add the values together