Question

Let u = <-3, 4>, v = <8, 2>. Find u + v

Answer 1

add like parts maybe?

Answer 2

That's what I was thinking.
so
<-3+8 ; 4+2>
<5, 6> ?

Answer 3

perfect .... thought i typed that

Answer 4

I believe that is the answer.
Let u = <3, -1>, v = <-6, -6>. Find 9u + 2v.
With this question, do I multiply the u's by 9 and the v's by 2 then add like parts?

Answer 5

yep

Answer 6

if it helps write the vectors as 9(3x -1y) + 2(-6x-6y)

the resulting ax+by is the vector (a,b)

Answer 7

i just stack them tho, that way the placement stays intact and i dont move things on accident

Answer 8

9<3, -1>
+2<-6, -6>



27, -9
-12, -12
---------
a , b

Answer 9

So it turns into <27 + -12; -9 + -12>
<15, -21>?

Answer 10

yep

Answer 11

Find the angle between the given vectors to the nearest tenth of a degree.

u = <-5, -4>, v = <-4, -3>

Answer 12

you know the dot product equivalence?

Answer 13

|u| |v| cos(a) = u.v

Answer 14

So how would I set this up?
<20, 12> cos a?

Answer 15

you know adding is adding like parts

dot product is multiplying like parts, and adding the results

Answer 16

|u| |v| cos(a) = 20+12

what are the length if u and v?

Answer 17

u is 20 and v is 12

Answer 18

no, the lenghts

if anything sqrt(u.u) and sqrt(v.v)

Answer 19

u = <-5, -4>

|u| = sqrt(u.u) = sqrt(25+16)


v = <-4, -3>
|v| = sqrt(v.v) = sqrt(16+9)

Answer 20

so

5sqrt(41) cos(a) = 20+12

how do we find the value of a?

Answer 21

Divide the other side by 5sqrt41

Answer 22

and do inverse cosine?

Answer 23

yep, and cos^(-1) both sides

a = cos^(-1) (32/(5sqrt(41)))

now you want degrees, so make sure you either convert it, or set your calc on deg mode

Answer 24

Okay, I got 1.78.

Answer 25

Determine whether the vectors u and v are parallel, orthogonal, or neither.

u = <6, -2>, v = <8, 24>
What would I do here to begin solving?