Chapter 6a Quiz
When given a magnitude and degree in bearings of two vectors and asked for the combined magnitude and direction angle ...
1. Convert bearing angle to a trig angle to find the component form for both vectors: <mag*cos(trig angle) , mag*sin(trig angle)> 2. Add the the two component forms to find the component form of the resultant vector 3. Find the magnitude of the resultant 4. Find the angle inside the triangle formed by the componets and the resultant 5. Find the bearing for the direction angle
Find a unit vector in the direction of the given vector w = -i - 2j
1. Convert i's and j's to component form w= -1<1,0> - 2<0,1> W = <-1, 0> - <0, 2> W = <-1, -2> 2. Determine the magnitude of the vector |w| = sqrt( (-1)^2 + (-2)^2 ) |w| = sqrt(5) 3.Using the magnitude determine the unit vector unit vector = 1/sqrt(5) * <-1, -2> unit vector = <-1/sqrt(5), -2/sqrt(5)> unit vector = < - sqrt(5) / 5, -2sqrt(5) / 5>
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <2, -7> v = <-4, 14>
1. Determine Slope slope of u = (-7/2) slope of v = (14/-4) --> (-7/2) Same slopes = Parallel Vectors
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <5, -6> v = <-12, -10>
1. Determine Slope slope of u = (-6/5) slope of v = (-10/-12) --> (5/6) 2. Determine if orthognal by finding the dot product u*v = 0 They are Orthogonal Vectors
Let vector U = <-1, 3> and vector V = <2,4> Find the component form of the vector 2u - 4v
1. Determine the component form of vector 2u and vector -4v 2u = 2 * <-1, 3> --> <-2, 6> -4v = -4 * <2, 4> --> <-8, -16> 2. Using vector arithmitic add the first terms and the second terms <-2,6> + <-8, -16> --> <-10, -10> Vector 2u - 4v = <-10, -10>
Find a unit vector in the direction of the given vector u = <-2, 4>
1. Determine the magnitude of the vector |u| = sqrt( (-2)^2 + (4)^2 ) |u| = sqrt(20) |u| = 2sqrt(5) 2. Using the magnitude determine the unit vector unit vector = 1/s2qrt(5) * <-2, 4> unit vector = <-2/2sqrt(5), 4/2sqrt(5)> unit vector = <-1/sqrt(5), 2/sqrt(5)> unit vector = <-sqrt(5)/5, 2sqrt(5)/5>
Prove that the vectors u and v are orthogonal u = <2, 3> v = < 3/2, -1>
1. Find if the dot product of the two vectors is 0 u*v = 2* 3/2 + 3 * -1 u*v = 3 + -3 u*v = 0
Find the magnitude and direction angle of the vector <-1,2>
1. Find magnitude sqrt( (-1)^2 + (2)^2 ) mag: sqrt(5) 2. Find the direction angle Find the angle inside the triangle using trig angles tan -1 ( 2/1) a = 63.4 degrees Identify the Quadrant 2nd Quadrant = theta = 180 - a Calculate Theta = 180 - 63.4 theta = 116.6 degrees 3. State magnitude and direction angle sqrt(5) @ 116.6 degrees
Find the magnitude and direction angle of the vector <3,4>
1. Find magnitude sqrt( (3)^2 + (4)^2 ) mag: 5 2. Find the direction angle <3, 4> is over 3, up 4 so it would be in the first quadrant and have an x value of 3 and a y value of 4 Using this knowledge find the angle of the triangle formed. tan-1 (4/3) = 53.1 degrees 4. 5 @ 53.1 degrees
Find vectorv with the given magnitude and same direction as u |v| = 2 u = <3, -3>
1. Find the angle of the triangle tan -1 (3/3) = 45 degrees 2. Use triangle angle to determine trig angle Quadrant 4: 360 - 45 = 315 3. Find the component form of vector v < 2 * cos(315) , 2* sin(315) > <1.414, -1.414>
Let Q = (3,4) and S = (2, -8) Find the component form and magnitude of 2QS
1. Find the component form of 2QS by finding the distance in the x and distance in the y 2QS = 2 * <-1, -12> 2QS = <-2, -24> 2. Find the magnitude of 2QS using the component form |2QS| = sqrt( (-2)^2 + (-24)^2 ) |2QS| = sqrt(580) |2QS| = 2sqrt(145) Component form of 2QS = <-2, -24> Magnitude of 2QS = 2sqrt(145)
Let P = (-2,2) amd Q = (3,4) Find the component form and magnitude of vector PQ
1. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <5,2> 2. Find the magnitude of PQ using the component form |PQ| = sqrt( (5)^2 + (2)^2 ) |PQ| = sqrt(29) Component form of PQ = <5,2> Magnitude of PQ = sqrt(29)
Prove that vector RS and vector PQ are equivilant by showing that they represent the same vector R=(-4,7) S = (-1,5) P = (0,0) Q=(3, -2)
1. Find the component form of RS by finding the distance in the x and distance in the y RS = <-4,7> --> <-1,5> RS = <3, -2> 2. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <0,0> --> <3,-2> PQ = <3,-2> 3. Compare RS = PQ <3,-2> = <3,-2>
Prove that vector RS and vector PQ are equivilant by showing that they represent the same vector R=(2,1) S = (0,-1) P = (1,4) Q=(-1, 2)
1. Find the component form of RS by finding the distance in the x and distance in the y RS = <2,1> --> <0,-1> RS = <-2, -2> 2. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <1,4> --> <-1,2> PQ = <-2,-2> 3. Compare RS = PQ <-2,-2> = <-2,-2>
Find the angles between the vectors u = <-4, -3> v = <-1, 5>
1. Find the dot product u*v = -11 2. Find the magnitude of each vector |u| = sqrt( (-4)^2 + (-3)^2 ) |u| = 5 |v| = sqrt( (-1)^2 + (5)^2 ) |v| = sqrt(26) 3. Plug in magnitude and dot product into formula theta = cos -1 (-11 / 5 * sqrt(26) ) theta = 115.6
Find the vector projection of u onto v. Then write u as a sum of two orthogonal vectors one of which is projection of u onto v. u = <-8, 3> v = <-6, -2>
1. Find the dot product u*v = 42 2. Find the magnitude of v |v| = sqrt( (-6)^2 + (-2)^2 ) |v| = sqrt(40) 3. Plug into the formula ProjvU = (42 / sqrt(40)^2) * <-6, -2> ProjvU = <-6.3, -2.1> What will make the dot product 0 and = to u -8 - -6.3 = -1.7 3 - -2.1 = 5.1 u = <-6.3, -2.1> + <-1.7, 5.1>
Find u*v given theta is the angle between u and v theta = 150 |u| = 3 |v| = 8
1. Plug numbers into formula cos 150 = (u*v) / 3 * 8 cos 150 = (u*v) /24 cos 150 * 24 = u*v u*v = -20.785
Find the dot product of u and v u = <4,5> v = <-3, -7>
1. Use dot product formula to find dot product 4 * -3 + 5 * -7 u*v = -47
Find the dot product of u and v u = <5,3> v = <12,4>
1. Use dot product formula to find dot product 5* 12 + 3*4 u*v = 72
Let vector U = <-1, 3> and vector W = <2, -5> Find the compontent form of the vector U - W
1. Using vector arithmitic simply subtract the first terms and the last terms <-1, 3> - <2, -5> --> Vector U-W = <-3, 8> Vector U+V = <-3, 8>
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <15, -12> v = <-4, 5>
1.Determine Slope slope u = (-12/15) --> (-4/5) slope v = (5/-4) 2. Determine if the vectors are orthogonal by finding the dot product U*v = -120 3. The vectors are neither parallel or orthoganal
Find the component form of vector v |v| = 18 and Given angle is 25. Reference image on pg 464 #29
1.Find the x component using cos: |v| * cos(theta) x component = 18 * cos(25) x component = 16.313 2. Find the y component using sin: |v| * sin(theta) y component = 18 * sin(25) y component = 7.607 3. <16.313, 7.607>
Let vector U = <-1, 3> and vector V = <2,4> Find the component form of the vector U+V
1.Using vector arithmitic simply add the first terms together and the last terms together <-1, 3> + <2,4> --> vector U+V = <1, 7> vector U+V = <1, 7>
Given 7(cos 135i + sin135j) State the magnitude and direction angle
7 @ 135 degrees
Component form
< distance in the left/right direction, distance in the up/down direction >
What is vector j in component form?
<0, 1>
What is vector i in component form?
<1, 0>
Formula to Find magnitude given a vectors components
<x,y> sqrt( (x)^2 + (y)^2 )
When given a magnitude and degree in bearings and asked for the component form ...
Convert bearing angle to a trig angle to find the component form: <mag*cos(trig angle) , mag*sin(trig angle)>
An airplane is flying on a bearing of 335 at 530 mph. Find the component form of the velocity of the airplane
Convert bearing angle to a trig angle: Draw the 335 degree angle in bearings Find the angle in triangle - 65 degrees Identify the quadrant - (2nd Qaudrant) Apply appropriate formula - 180 - a Trig angle is 115 Find the component form: <mag*cos(trig angle) , mag*sin(trig angle)> <530 * cos(115) , 530 * sin(115)> <-223.988, 480.343>
Two vectors are parallel ...
If they have the same slope.
For trig angles only When finding the angle of direction for a vector, you need find the angle inside the triangle and then use these formulas to determine the angle of direction. Use a as the angle inside the triangle and theta as the direction angle of the magnitude
When in Quad 1: theta = a When in Quad 2: theta = 180 - a When in Quad 3: theta = 180 + a When in Quad 4: theta = 360 - a
Slope can be determined for vectors through their
component form
The angle between two vectors is
cos-1( u*v / |u| * |v| )
Vectors
describe motion, not location Directed line segment Has direction and magnitutde
Two vectors are orthogonal ...
if and only if their dot product is 0
How is magnitude represented
like absolute value but with an arrow over the variable
A vector is a unit vector when the ...
magnitude of the vector = 1
You find the unit vector of a normal vector by ...
take the vector in component form * 1/magnitude of the vector vector u = 1/|v| * vector v
The dot product of two vectors is
x * x + y * y
