Question

Could somebody explain how to do this:
Find the angle between the given vectors to the nearest tenth of a degree.

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

Answer choices:
A. -9.1 Degrees
B. 1.8 Degrees
C. 0.9 Degrees
D. 11.8 Degrees

Not sure how to find an angle between vectors.

Answer 1

Do you know how to find the magnitude or length of a vector?

And also so you know about scalar or dot product of 2 vectors?

Answer 2

I know Scalar and Dot Product. @anikhalder I sort of know magnitude...

Answer 3

That's perfect! So let's go through this:

What is the length of u vector?

Answer 4

I'm not sure how to solve for length...

Answer 5

so its like, suppose i have a vector a = <3, 4>

Then the length of the vector a is :

\[\left| a \right|=\sqrt{3^{2} + {4^{2}}}\]
\[\left| a \right|=\sqrt{25}\]
That is the length of vector a is 5

So using the same way can you solve for the length of vector u?

Answer 6

Oh! Is the magnitude |u| = SqRt. 41 ?

Answer 7

Perfect! You rock!

What about the magnitude (or length) of vector v?

Answer 8

|v| = 5 ?

Answer 9

Awesome! Now do you know the formula for dot (or scalar) product of 2 vectors a and b?

Answer 10

Yes, would the product be: a x b = 32 ?

Answer 11

Yes that's correct! But, I will suggest when writing dot product don't use 'x' sign. You will learn that there is another type of product called the vector or cross product which uses this sign...but we don't need to worry about right now.

Suppose I have a vector a and vector b

Then the dot product of vectors a and b is also defined as:

\[a.b=\left| a \right|\left| b \right|\cos \theta\]
Where theta is the angle between the vectors a and b.

So for our question at hand. Let's replace a by u and b by v.

So, we get:
\[u.v=\left| u\right|\left| v\right|\cos \theta\]

We can replace u.v as 32 and the magnitudes of u and v on the right hand side of the equation to get:

\[\cos \theta=\frac{ u.v }{ \left| u \right|\left| v \right|}\]

Can you solve for costheta now?

Answer 12

Is costheta = .99 ?

Answer 13

@anikhalder

Answer 14

Yes! I find so as well. Now just find the cosine inverse of 0.99 and you'll get it in degrees

Answer 15

I got 8.1 degrees, approximately, but that is not one of my answers.

Answer 16

That's what I get and now I must confess that I am confused as well

Answer 17

Let's verify

Answer 18

Ok

Answer 19

http://onlinemschool.com/math/assistance/vector/angl/

Can you type the values in and see what you get for costheta because I get 0.99 in this website as well

Answer 20

I am getting .99

Answer 21

@anikhalder Can you help me with another problem, and then maybe go back to this one?

Answer 22

oh wait

Answer 23

Express the complex number in trigonometric form.
-3 + 3 square root of three i

Answer 24

http://www.rapidtables.com/calc/math/Arccos_Calculator.htm

Answer 25

And I'm confused as to what that calculator meant.

Answer 26

Check this, we were just using 0.99, we cant do that we have to use the full result which was 0.9995120761

plug this and find the cos inverse. You will see that we get our answer :)

Answer 27

Oh. I see. So the answer would be 1.8 degrees?

Answer 28

Yep! And yes just take 4 minutes and see this video on complex numbers! he explains better than me :)))

Answer 29

What video?

Answer 30

Sorry...my bad i forgot to give the link:

https://www.youtube.com/watch?v=6z6fzPXUbSQ

Answer 31

Thank you so much. (: I watched that video, though, and am still confused on my example. He helped me with a couple others, but the one I posted about is a bit confusing after watching the video.

Answer 32

You mean its like :

\[3+3\sqrt{3}i\]

So,

\[\left| z \right| = \sqrt{3^{2}+(3\sqrt{3})^{2}} = 6\]

and \[\tan \theta = \frac{ 3\sqrt{3} }{ 3}\]

i.e. \[\theta = 60 degrees\]

Now, in polar or trigonometric form,

z = \[\left| z \right|(\cos \theta + i \sin \theta)\]

Just substitute the values and find the answer :)))

Answer 33

What is the Z value?

Answer 34

And is the theta value 120 degrees?

Answer 35

you can say

\[\left| z \right|\] is like the absolute value of the complex number (like the