Question

3 part question

Answer 1

find the dot product, the magnitudes, and plug them in and solve for theta

Answer 2

-6*13

Answer 3

remember the rules for calculating the dot product

1. multiply the x-coordinates
2. multiply the y-coordinates
3. add the results from steps 1 and 2.

Answer 4

-6+0=-6

Answer 5

good now find the magnitudes of the two original vectors

Answer 6

0

Answer 7

remember that magnitude = sqrt(x^2 + y^2).

Answer 8

6

Answer 9

you must perform the calculation on the original two vectors

Answer 10

sqrt205 and 1

Answer 11

anyway I got to do a quick errand

but once you have the magnitudes plug them in and solve for theta

Answer 12

okay sounds good

Answer 13

so any progress yet?

Answer 14

-6=sqrt205 * 1 (cos x)

Answer 15

good

now solve for x

Answer 16

still there?

cos(x) = -6/sqrt(205)

x = ?

Answer 17

6.28

Answer 18

take the arccos of both sides

x = arccos(-6/sqrt(205)) = ?

Answer 19

114.8

Answer 20

good

so for part II) you will simply plug in |v| and theta into the expression

<|v|cos(theta), |v|sin(theta)>

Answer 21

<|sqrt205|cos(114.8), |1|sin(114.8)>

Answer 22

notice how it says |v| for both so both the |v|'s have to be the same value

Answer 23

like would i add sqrt 205 + 1

Answer 24

how would i get it to be one value

Answer 25

what did you get when you calculated |v| from part I?

Answer 26

sqrt205 and 1

Answer 27

|v| can only have one value

what was the magnititude of vector v?

Answer 28

pay attention to which vector was vector v

Answer 29

oh 1

Answer 30

becase it was (1,0)

Answer 31

that means v = (-6,13) which means |v| is only the magnitude of (6,-13) and nothing else

Answer 32

oh okay, i see what you mean so when you are askiing for v it is (6,-13)

Answer 33

good

so <(sqrt205)cos(114.8), sqrt(205)sin(114.8)> is your vector

Answer 34

then for part III you just need to calculate the dot product and see if it's equal to 0 or not

Answer 35

okay -6+0=-6

Answer 36

you must use the vectors that are given in part III

Answer 37

(-6)+(-42)=-48
13+(-34)=-21

-48*-21=1008

Answer 38

the rule is to multiply first then add

Answer 39

(-6)*(-42)=252
13*(-34)=-442
252+(-442)=-190

Answer 40

good

since the dot product is not 0 then it's not orthogonal

and that's it

Answer 41

could you just check this over to see if i missed anything or not

Answer 42

good but on part II there is an error w/ your parentheses

(sqrt(205)cos(114.8), sqrt(205)cos(114.8)) should be it

Answer 43

well for part I start by finding the modulus (r) and theta

Answer 44

remember that theta = arctan(y/x) and modulus = sqrt(x^2 + y^2).

Answer 45

Part I: Write z1 in polar form. (2 points)

Answer 46

yes. finding theta and the modulus are the correct steps to find the polar form.

Answer 47

x = 1
y = sqrt(3)

find the modulus and the theta

Answer 48

1.65

Answer 49

i did sqrt (1)^2+(sqrt3)^2

Answer 50

actually its 2

Answer 51

the magnitude should be 2 (remember order of operations)

now try the angle

Answer 52

I forgot what the equATIION IS TO FIND theta

Answer 53

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
remember that theta = arctan(y/x) and modulus = sqrt(x^2 + y^2).
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 54

arctan(sqrt3)/1=pi/3

Answer 55

good so your final sol'n is (2,pi/3)

Answer 56

for part I correct =)

Answer 57

yes

for part II: since we want the fourth root you will take the original modulus 2 and raise it to the (1/4) power to get 2^(1/4)

Answer 58

for part III:

you will calculate (theta + 2pi*k)/n

for n = 4 (since we have the fourth roots)
and k = 0,1,2,3

producing a total of four angle values

Answer 59

i dont have to solve 2^(1/4) right just the expression

Answer 60

right

Answer 61

theta + 2pi*0)/4=theta
theta + 2pi*1)/4=theta+pi/2
theta + 2pi*2)/4=theta+pi
theta + 2pi*3)/4=theta+3pi/2

Answer 62

you have to plug in the original theta value

Answer 63

also you need to be very careful with parentheses

(theta + 2pi*k)/n

that beginning parentheses makes a huge difference

Answer 64

is the theta value pi/3

Answer 65

yes

Answer 66

(pi/3 + 2pi*0)/4= pi/12
(pi/3+ 2pi*1)/4=7pi/12
(pi/3+ 2pi*2)/4=13pi/12
(pi/3+ 2pi*3)/4=19pi/12

Answer 67

good

then for part IV you will set up each root

2^(1/4) * [cos(theta) + i*sin(theta)]

and go down the list of your theta values, ending up with four total roots

Answer 68

2^(1/4) * [cos(pi/12) + i*sin(pi/12)]
2^(1/4) * [cos(7pi/12) + i*sin(7pi/12)]
2^(1/4) * [cos(13pi/12) + i*sin(13pi/12)]
2^(1/4) * [cos(19pi/12) + i*sin(19pi/12)]

Answer 69

good that's your sol'n for part IV