# Trig and Vector Components

Hi -

The problem is below and picture of diagram is attached. Could you tell me what is incorrect about my train of thought? I am suppose to express x and y.

Find the components of the vector with length
C and angle phi as shown, named C vector. Express your answer in terms of C and phi . Write the components in the form (x,y) .

I expressed (x,y) as ... c*cos(phi),c*csc(phi) ... which is incorrect

Thanks for your time.

## Answers and Replies

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robphy
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The component of any vector $\vec V$ along a given axis is
$$V_{axis}=|V|\cos \theta_{\rm between\ \vec V\ and\ axis}$$

In your picture, what is $\theta_{\rm between\ \vec V\ and\ x-axis}$ and $\theta_{\rm between\ \vec V\ and\ y-axis}$?
With the trig identity for $\cos(\phi+\alpha)$, you can express all of your results in terms of $\phi$.

Last edited:
iamhumble said:
Hi -

The problem is below and picture of diagram is attached. Could you tell me what is incorrect about my train of thought? I am suppose to express x and y.

Find the components of the vector with length
C and angle phi as shown, named C vector. Express your answer in terms of C and phi . Write the components in the form (x,y) .

I expressed (x,y) as ... c*cos(phi),c*csc(phi) ... which is incorrect

Thanks for your time.
I'm just wondering, can't you express the answer as
c*sin(phi),c*cos(phi) ?

correct me if im wrong...

Last edited:
robphy
Science Advisor
Homework Helper
Gold Member
Almost... note that c_x < 0.

robphy said:
Almost... note that c_x < 0.
thanks alot, i was just confused because iamhumble made it seem a lot harder...

Vectors

Hello Everyone,
In school I had a test of Vector exercises. The problem is that I have to submit them online and I need the exact value for the questions. Anyway I don't want you to solve them for me. I just need some help where I am stuck.

The first exercise: A boy runs 3.3 m North, 7.1 m NorthEast and 15.6 m West. a)Determine the length of the displacement vector tha goes from the starting point to his final position. b)Determine the direction of the displacement vector.

I was confused by this exercise because there isn't any angle given. So I assumed that "NorthEast" means that he moves 7.1 m with an angle 45 Deg.
Then part a) is easy. For part b) I found the angle formed by the North line and the Resultant vector. Am I right here?

The next exercise is a little fuzzy I think: Let aA + bB + C=0, where
A=(17,-99), B=(-99,28) and C=(4.3,78), where Ax=17 units, Ay=-99 units etc. What is the value of a and b? (All the Upper case letters A,B and C have the vector arrow above).
What I did is: I opened aA + bB + C=0 using unit vectors i, j and k (k=0) and the numbers given above.
Still I have a and b and it's an equation with 2 unknowns.
The final expression I concluded is: (a*17i-b*99i+4.3i)+(b*28j-a*99j+78j)=0...........???

Thank You for your time.

Tide
Science Advisor
Homework Helper
Electro,

Your final equation is really TWO equations since the unit vectors are independent of each other. You should be able to handle it from there.

Thank You Tide,
What I think now, is forming two simultaneous equations.
1) 17a-99b= -4.3
2) 28b-99a= -78
Solving them, I think is the answer for a and b.
Since the i units can't be added to the j units I formed two equations just neglecting i and j. So for i group it will give 0 and for j group too.

Please I need an answer on the previous exercise I posted if I have concepted it right or not.

Thank you
Electro

Thanks all for the much needed assistance. I understand what I did wrong.

:uhh: Anyone can give the answer to my questions?

i think you should start a new post of your own...