Question

Use a metric ruler and a protractor to find 2a--> - 2b-->, if |a-->| = 2cm and |b-->| = 1cm. Then find the magnitude and direction of the resultant.

Answer 1

I don't really get how I would be doing this...

Answer 2

Possible answers:
8 cm, 41°
9 cm, 38°
6 cm, 36°
4 cm, 31°

Answer 3

@jim_thompson5910 @whpalmer4

Answer 4

I think you would use the idea that if vector a = <x,y>

then

x = r*cos(theta)

x = |a|*cos(60)

and

y = r*sin(theta)

y = |a|*sin(60)

Answer 5

Thank you c:

Answer 6

you're welcome

tell me what you get

Answer 7

Alrighty c:

Answer 8

What is r btw?

Answer 9

r is the length of the vector

so r = |a| or r = |b|

Answer 10

ah okai c:

Answer 11

basically if you drew a circle around the tip of the vector, it would be the radius of that circle (the vector is centered at (0,0))

Answer 12

Then for the y part did you mean |b| or |a|?

Answer 13

no it applies to both

Answer 14

r isn't the same for both vectors, it's just in general

Answer 15

in the case of vector a, r = 2

in the case of vector b, r = 1

Answer 16

Oh okai so when I do both those x equations, do I add them together or do I multiply them?

Answer 17

x = r*cos(theta)

x = |a|*cos(60)

x = 2*cos(60)

x = 2*(1/2)

x = 1




y = r*sin(theta)

y = |a|*sin(60)

y = 2*sin(60)

y = 2*(sqrt(3)/2)

y = sqrt(3)


So the vector a has the terminal point at ( 1, sqrt(3) )

Answer 18

So cos(60) = 1/2?

Answer 19

yes

Answer 20

For vector b...


x = r*cos(theta)

x = |b|*cos(135)

x = 1*cos(135)

x = 1*(-sqrt(2)/2)

x = -sqrt(2)/2



y = r*sin(theta)

y = |b|*sin(135)

y = 1*sin(135)

y = 1*(sqrt(2)/2)

y = sqrt(2)/2


and the vector b has the terminal point at ( -sqrt(2)/2 , sqrt(2)/2 )

Answer 21

Oh damn, I was using radians >_<

Answer 22

ah that gets me everytime

Answer 23

once you've found the x,y components, you can add and subtract the vectors


2a - 2b = 2(a-b)


x coordinate of vector a-b: 1 - (-sqrt(2)/2) = 1+sqrt(2)/2

y coordinate of vector a-b: sqrt(3) + sqrt(2)/2 = 1+sqrt(2)/2


Now double everything to find the components for the vector 2(a-b)


x coordinate of vector 2(a-b): 2(1+sqrt(2)/2) = 2+sqrt(2)

y coordinate of vector 2(a-b): 2(sqrt(3) + sqrt(2)/2) = 2*sqrt(3)+sqrt(2)

Answer 24

So effectively, we know that

x = 2+sqrt(2)
y = 2*sqrt(3)+sqrt(2)

the last thing to do is plug them into the formulas

r = sqrt(x^2 + y^2)

and

theta = arctan(y/x)

to find the magnitude and the direction

Answer 25

Okai 1 quick question before I read what you said, you know how you got 2 for 2*cos theta and 1 for 2*cos(60)? Did you minus them to get the final 1 for the x?

Answer 26

yes I did these basic steps

found the x,y component form for vector a
found the x,y component form for vector b

once you have the component form, you can easily add/subtract the vectors

the resultant vector a-b is made up of x,y components as well and the x component is found by subtracting the corresponding x components for vectors 'a' and b (y is found the same way)

Answer 27

Okai good to know, I want to know I can do it later on with another problem that's why I was asking c:

Answer 28

oops made a typo

y coordinate of vector 2(a-b) *should* be 2*sqrt(3)-sqrt(2)

Answer 29

but everything else looks ok

Answer 30

Okai so for r=sqrt (x^2+y^2) I got 4.06 which I rounded to 4cm. Okai so now I know my answer, but how would I get 31degrees?

Answer 31

use

theta = arctan(y/x)

Answer 32

Arctan is just [] right? Or am I wrong?

Answer 33

what do you mean just []

Answer 34

Oh nevermind it's inverse tan right?

Answer 35

yes

Answer 36

arctangent = inverse of the tangent function

Answer 37

ohhh sorry i didn't know

Answer 38

that's ok

Answer 39

I'm not getting the right answer for that part :l I've got 22.9, 55.01, and 1.06 *me trying it different ways*

Answer 40

theta = arctan(y/x)

theta = arctan( (2*sqrt(3)-sqrt(2))/(2+sqrt(2)))

theta = 30.9805340840034

theta = 31 (rounding to the nearest degree)

Answer 41

oh i forgot the extra ( in the front

Answer 42

hmm odd how there wasn't an error

Answer 43

ohh i used + as well instead of -

Answer 44

that's my bad lol

i had + but i edited it to -

Answer 45

not sure how i got + when a-b tells us we're subtracting components

Answer 46

lmao so the first answer is wrong? cx

Answer 47

no, the first answer of 4 is correct

Answer 48

notice how the 4 cm and the 31 degrees match up and it's choice D

Answer 49

oh so the first one i add in the y but the second one i subtract in the y?

Answer 50

no it's all supposed to be subtracting (in the numerator)

Answer 51

before, i mistakenly got this

y = 2*sqrt(3)+sqrt(2)

when it should have been this

y = 2*sqrt(3)-sqrt(2)

Answer 52

But for the first one I used the + and got the correct answer of 4cm

Answer 53

you got close to the correct answer, but rounding made you get there anyway

Answer 54

Well yeah but you know what I mean cx

Answer 55

I got 3.98232284958912 for the magnitude which is much closer

that rounds to 4 as well

Answer 56

Oh okai

Answer 57

I get you c:

Answer 58

ok that's great

Answer 59

Thank you n.n

Answer 60

np