Question

x=(4, -5), y=(3, -1) find x*y

Answer 1

I have no idea why this is so hard, I keep getting 60, but that's not one of the answer choices

Answer 2

I'm assuming you want to dot product these vectors?

Answer 3

yes

Answer 4

multiply the corresponding coordinates, then add up the results

x coordinates multiply to 4*3 = 12
y coordinates multiply to -5*(-1) = 5

add up the products of 12 and 5 to get 12+5 = 17

I think the mistake you made was saying 12*5 = 60

Answer 5

oh.. you add at the end. not multiply! haha thank you

Answer 6

would you help with one more vector problem?

Answer 7

yes, if u and v are vectors such that
u = (a,b)
v = (c,d)
a,b,c,d are scalars

then...

u dotproduct v = a*c + b*d

Answer 8

what's your other question?

Answer 9

Determine the angle between c= <5, -2> and d= <1, 3>.

Answer 10

I know it has something to do with finding arctan but that's about it

Answer 11

what do you get when you dot product those vectors?

Answer 12

-1?

Answer 13

good

Answer 14

are you able to find the length of a vector?

Answer 15

is it just 1?

Answer 16

hint:

length of vector c = sqrt(c dot c)

Answer 17

the "c dot c" part is vector c dot product with itself

Answer 18

the length of vector c is not 1

Answer 19

tell me what `c dotproduct c` is equal to

Answer 20

im not sure.. do i just multiply the dot product with both the c coordinates?

Answer 21

Let,
u = <5,-2>
v = <5,-2>
basically assign vector u and vector v to be equal to vector c

now dot product u and v to get what?

Answer 22

oh my gosh I feel so dumb, but I still don't get it. is it 29?

Answer 23

29 is the correct result for this sub-part
but we still have more work to do

Answer 24

so c^2 = 29 which means c = sqrt(29)

the length of vector c is sqrt(29)

hopefully this makes sense?

Answer 25

yes it does

Answer 26

Here's a visual of the vector

Answer 27

you can use the pythagorean theorem to get sqrt(29)

Answer 28

what is the length of vector d?

Answer 29

5.38

Answer 30

sorry I messed up the drawing, it should be

Answer 31

wait nvm, i thought you wanted me to solve for the hyp. sorry

Answer 32

um, do i just find the angle?

Answer 33

we already have that length

Answer 34

what do you get when you dot product vector d by itself?

Answer 35

(vector d) dotproduct (vector d) = ???

Answer 36

<1,3> dotproduct <1,3> = ???

Answer 37

-4?

Answer 38

no

Answer 39

oh, is it 13?

Answer 40

or 7

Answer 41

<1,3> dotproduct <1,3> = 1*1 + 3*3 = 1 + 9 = 10

agreed?

Answer 42

oh my gosh, yes duh I thought we were using the -1 dot product from before, my bad

Answer 43

no we're using vector d now

Answer 44

so

d^2 = 10
d = sqrt(10)

the length of vector d is sqrt(10) units long
this is the length of the hypotenuse

Answer 45

okay makes sense

Answer 46

ok just a few more steps

Answer 47

alright

Answer 48

So the dot product of c and d was found to be -1 (this was the first thing we did for this problem)
That means
\[\Large c \cdot d = -1\]
The lengths of vectors c and d are
\[\Large |c| = \sqrt{29}\]
\[\Large |d| = \sqrt{10}\]

-------------------------------------------------------------

Now plug all this into the formula below and solve for theta

\[\Large \cos(\theta) = \frac{c \cdot d}{|c|*|d|}\]
\[\Large \cos(\theta) = \frac{-1}{\sqrt{29}*\sqrt{10}}\]
\[\Large \cos(\theta) \approx -0.05872202\]
Are you able to determine theta from here?

Answer 49

I understand all of that, but am not sure what to do from there

Answer 50

have you learned about inverse trig functions?

Answer 51

yes, but that was a while ago

Answer 52

you'll apply the arccosine to both sides

Answer 53

I did that and got 2.18, so i dont know what I did wrong

Answer 54

Apply arccosine to both sides to get...
\[\Large \cos(\theta) \approx -0.05872202\]
\[\Large \arccos(\cos(\theta)) \approx \arccos(-0.05872202)\]
\[\Large \theta \approx \arccos(-0.05872202)\]
\[\Large \theta \approx ???\]

Answer 55

you're probably in radian mode

Answer 56

what calculator do you have?

Answer 57

ti-84 plus

Answer 58

do you know how to switch between radian and degree mode?

Answer 59

yes one second