Question

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

u = <2, -4>, v = <3, -8>

Answer 1

@UnkleRhaukus

Answer 2

\[\cos \theta = \frac{u\cdot v}{\|u\|\cdot\|v\|}\]

Answer 3

\[\|u\|=\sqrt{u_1^2+u_2^2}\]

Answer 4

Alright

Answer 5

so find \(\|u\|\), and \(\|v\|\) first

Answer 6

How I don't understand

Answer 7

The first vector is \[u = \langle2,-4\rangle\]
so its length is \[\|u\|=\sqrt{(2)^2+(-4)^2}\]

Answer 8

So I have to solve that?

Answer 9

yeah simplify this into some irrational number

Answer 10

I just got square root of 2x^2 plus 16

Answer 11

\[\|u\|=\sqrt{(2)^2+(-4)^2}\\
\qquad=\sqrt{4+16}\\
\qquad=\]

Answer 12

2

Answer 13

Wait do you want the decimal or

Answer 14

2 times the square root of 5

Answer 15

keep it as an irrational number

Answer 16

yeah that's right ||u|| = 2√5

Answer 17

now find the length of the other vector

||v|| = √( 3^2 + (-8)^2 )
=

Answer 18

the square root of 73

Answer 19

right,

Answer 20

now what is the dot product of u and v?

Answer 21

Choices:
3.0°
6.0°
-7.0°
16.0°

Answer 22

u * v = 2*3 + (-4)*(-8)

Answer 23

38

Answer 24

so we have
\[\cos \theta = \frac{u\cdot v}{\|u\|\cdot\|v\|}\\
\qquad=\frac{38}{2\sqrt5\cdot\sqrt{73}}\]

Answer 25

19/square root of 365

Answer 26

the angle you are looking for is
\[\theta = \arccos\left(\frac{19}{\sqrt{365}}\right)\]

(make sure your calculator is set to degrees mode)

Answer 27

Is it 6.0 degrees

Answer 28

Can you just check my answer on this last question? I already did it I just want to make sure I'm right

Answer 29

Find a ⋅ b.

a = <2, 4>, b = <2, 5>

Choices:
<4, 9>
24
<4, 20>
16

Answer 30

6 degrees is right,

Answer 31

I chose 24

Answer 32

But I also thought it might be <4,20>

Answer 33

for the dot product just remember to add the products of the corresponding components

a ⋅ b = 2*2 + 4*5
= 4 + 20
= 24

Answer 34

Thanks!

Answer 35

bye!