Question:

Suppose a vector $\vec{V}$ makes an angle $\phi$ with respect to the y axis. What could be the x and y components of the vector $\vec{V}$?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

$v_x = v\sin{\phi}, v_y = v\cos{\phi}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. Here's vector*V*at an angle

*φ*with respect to the

*y*-axis. It's gonna have a

*x*-component in the

*x*-direction like this that's

*v*subscript

*x*and this is the opposite leg of the right triangle and so when you hear the word opposite, you should be thinking about

*sin*because the

*sin*of

*φ*is the length of the opposite

*v x*divided by the length of the hypotenuse

*v*and so we can find the

*x*-component by multiplying both sides by

*v*and we get

*v x*is

*v sin φ*— oops not

*Θ*,

*φ*— and that's the answer there although it's slightly ambiguous the way they ask the question because it's possible that the angle

*φ*could be on the other side of the

*y*-axis in which case, the vector would be like this and

*φ*would be in here and in this part (b), you would have to say that

*v x*is the negative of

*v sin φ*because it's going to the left there's

*v x*. Okay. But keeping that in mind, the

*y*-component is not ambiguous because it's with respect to the

*y*-axis which you understand to be the positive

*y*-axis, didn't say negative

*y*-axis so we have here

*v y*. The

*cos*of—let's put this in a different color I guess— the

*cos*of

*φ*is adjacent over hypotenuse that's

*v y*over

*v*and multiply both sides by

*v*and you get

*v y*is

*v*times

*cos φ*and there's the

*y*-component and here's the

*x*-component. So when you are thinking about components, don't memorize the way to get it in terms of

*cos*or

*sin*because as you can see, it depends on how the angle is measured. So the

*x*-component is typically found using the

*cos*because it's more common to have your angle with respect to the

*x*-axis but you can't always count on that being true and this question demonstrates that where sometimes the angle is with respect to the

*y*-axis in which case, the

*x*-component is found using

*sin*; always draw a picture.