Angle of Vector B & X-Axis: Calculate Degrees

[anglum]

Homework Statement

Vector B has x y and z components of 6.7, 4.7, 2.4 and the magnitude of Vector B is equal to 8.599

what is the angle between Vector B and the x-axis. answer in units of degrees

The Attempt at a Solution

i am not sure where to begin on this one... i hate online physics courses BLAH

 

[anglum]

can i figure this out wiht a formula or do i need to draw it out?

 

[Kurdt]

Have you covered direction cosines of vectors?

 

[anglum]

i don't recall covering that

 

[anglum]

would i assume 8.599 squared = 6.7squared + A squared
A = 5.39

then use those as the lenghts of the sides for the triangle to use the cosine of the angle to get my answer?

 

[anglum]

then solving for that angle i get 38.8162364?

 

[BlackWyvern]

Draw a rectangle with the dimensions of the dimensions (lol). Then use pythagoras on the faces to find the angle.

That's THE best way to deal with 3D vectors.

 

[Kurdt]

Ok. Consider a vector $\mathbf{a}=(a_1, a_2, a_3)$ and imagine the angles between that vector and the vectors $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$ are $\alpha$, $\beta$ and $\gamma$ respectively. Knowing that the magnitude of the vector is $a$ we can use pythagorean theroem to say that:

$$cos(\alpha) =\frac{a_1}{a}$$

$$cos(\beta) = \frac{a_2}{a}$$

$$cos(\gamma) = \frac{a_3}{a}$$

Thus we note:

$$\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}=cos(\alpha)\mathbf{\hat{i}}+cos(\beta)\mathbf{\hat{j}}+cos(\gamma)\mathbf{\hat{k}}$$

Now knowing thiscan you find the angle between the x-axis and the vector with the information you have?

 

[anglum]

i am totally lost on ur idea to draw a rectangle of that... once i draw that rectangle how do i use pythagoras on the faces to find the angle?

 

[anglum]

ok KURDT i solved the cos of all 3 of those angles

the first angle is 38.816 degrees
the 2nd angle is 56.867 degrees
the 3rd angle is 73.79337 degrees

correct?

 

[Kurdt]

Yeah they look fine. So its relatively simple once you know about direction cosines which very few establishments teach in my experience.

 

[anglum]

im not sure my next step once i have those 3 values

 

[BlackWyvern]

http://i14.tinypic.com/4y548c8.png

That's what I mean. If you can visualize the vector, you'll be able to solve the problem faster. In an exam, you don't have time to waste.

 

[anglum]

kurdt once i solve for those 3 values what step do i take next??

i understand why i get those 3 angles I am just lost on the last steps i need now

 

[Kurdt]

You have the answer. The first angle is the angle between the vector and the x-axis. The second is the angle between the vector and the y-axis and the 3rd is the angle between the vector and the z-axis.

 

[Dick]

Do you know dot products? That's a much simpler way out of this mess.

 

[anglum]

so from the original problem i posted... .where the angle between Vector B with an arrow drawn over the B and x-axis is 38.826 degrees? becuz when i enter that onto the online classroom it is wrong

 

[BlackWyvern]

I get 35.049. Try that.

 

[anglum]

i tried that black wyvern and it was incorrect

 

[Kurdt]

so from the original problem i posted... .where the angle between Vector B with an arrow drawn over the B and x-axis is 38.826 degrees? becuz when i enter that onto the online classroom it is wrong

Thats because the magnitude you gave seems to be slightly out using the components you gave. Check them and make sure they are correct.

 

[anglum]

my answer of 8.599 for the magnitude of that vector was my answer to part a of the question and that was accepted as a correct answer

so i am not sure what the problem is with my answer of 38.816

 

[BlackWyvern]

Possibly the question is not asking to find the angle in terms of a spherical co-ordinate.

 

[Kurdt]

my answer of 8.599 for the magnitude of that vector was my answer to part a of the question and that was accepted as a correct answer

so i am not sure what the problem is with my answer of 38.816

Given the components from your first post I get 8.529 to 3 d.p. for the magnitude.

 

[anglum]

KURDT you are the man ( not sure if u are a female or not) but u get the idea... for some reason my answer of 8.599 must have been within the range for part a but when using that for part b it threw it way off

thank you sooo much

 

[BlackWyvern]

Me too. I just checked that as well.

 

[Kurdt]

KURDT you are the man ( not sure if u are a female or not) but u get the idea... for some reason my answer of 8.599 must have been within the range for part a but when using that for part b it threw it way off

thank you sooo much

No problem. I am a man (I just have a lot of hair!) and I'm glad I could help. As Dick suggested on the previous page, you may want to brush up on dot products as well. Having learned about direction cosines you should have no problem understanding them.

 

[learningphysics]

EDIT: lol never mind. glad you got the answer anglum. Did the online program accept it?

 

[anglum]

yea learning it did... it kinda pisses me off i lost some pts on it considering my answer of 8.599 was accepted as the answer of the magnitude of vector B however when using that to solve for part b it skewed my answer to where it wasnt accepted... i know its only a few pts difference on the homework but hey every lil but helps

thanks to all who help me wiht my struggles... i cannot begin to describe how difficult these concepts are to gather strictly from an online class

 

[Kurdt]

You'd think they'd scale it so they would accept answers on later questions with max and min answers from previous questions. Never mind, you got there.