Rotating a Vector in a Coordinate System: Explaining the Formula

[misogynisticfeminist]

Say, i have a vector a, defined in a coordinate system, x-y-z and i rotate the axes by an angle theta around the z-axis, so i have my z-component invariant in this change of basis. Can someone show me why,

a_x' = a_x cos\theta + a_y sin\theta and

a_x' = -a_x sin\theta + a_y cos\theta

i can't seem to find any place where they show how this is done.

 

[robphy]

Draw a vector and two sets of orthogonal axes.
For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.
Do a little trigononometry.
I'll post a URL with a picture, if I can find one.

Ok here's one:
http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html
(although the vector is not drawn from the origin to the little box).

(Check the subscript on the LHS of your second equation.)

 

[LeonhardEuler]

The new x-axis is a line tilted at some angle, \theta[/tex], to the horizontal. Basically, if you want a'_y[/tex], you want the distance between a point and this line. The formula for this is <br /> |a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}} <br /> where the equation of the line is y=mx+b[/tex]. Since the line passes through the origin, b=0. tan(\theta)=m[/tex], so: <br /> |a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}} <br /> = \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|} <br /> <br /> = |tan(\theta)a_x-a_y|\times|cos(\theta)| <br /> <br /> =|sin(\theta)a_x-cos(\theta)a_y| <br /> <br /> Since, for \theta=0[/tex], we should have a'_y=a_y[/tex], this means the signs will only work out properly if:<br /> a'_y=-sin(\theta)a_x+cos(\theta)a_y <br /> The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.

 

[robphy]

Here's a vector-algebraic proof.

For the first set of axes, we have unit vectors \hat x and \hat y.
For the second set, which is rotated counterclockwise by angle \theta, we have \hat x' and \hat y'.

Note that A_x= \vec A\cdot \hat x and so on.

Now, compute A_{x'}= \vec A\cdot \hat x' where one writes
\vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y.
So, let's plug it in:
\begin{align*}<br /> A_{x&#039;} <br /> &amp;= \vec A\cdot \hat x&#039; \\<br /> &amp;= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x&#039; \\<br /> &amp;=(A_x) \hat x\cdot \hat x&#039; + (A_y) \hat y\cdot \hat x&#039; \\<br /> &amp;=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\<br /> &amp;=(A_x) \cos\theta + (A_y) \sin\theta \\<br /> \end{align}<br />
Do the same for A_{y&#039;}, and note that \hat x\cdot \hat y&#039; =\cos(90^\circ+\theta)= -\sin\theta and \hat y\cdot \hat y&#039; =\cos\theta.

 

[gerben]

The original vector had an x-component and a y-component. I find it easiest to just follow those two components separately.

When the original vector gets rotated the original x-component, let’s call that X, will not be only in the direction of the x-axis anymore, the component of the rotated “original x-component“ that will still be in the direction of the x-axis is:
X\cos(\theta)
but it will now also have a component in the direction of the y-axis, which will be:
-X\sin(\theta)

The same happens with the original y component. The component of the rotated “original y component“ that will still be in the direction of the y-axis is:
Y\cos(\theta)
and the component in the direction of the x-axis will be:
Y\sin(\theta)

To get the x-component of the new vector add the parts in the direction of the x-axis:
X\cos(\theta) + Y\sin(\theta)
To get the y-component of the new vector add the parts in the direction of the y-axis:
-X\sin(\theta) + Y\cos(\theta)

 

[misogynisticfeminist]

hey thanks a lot for the help. Yup the subscript on the LHS should be y. Here's what i have done, a little algebra,

a_x&#039; cos \theta = a_x

a_x&#039; sin \theta= a_y

what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for a_x&#039;

for a_y&#039; i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

Is that right?

edit: i'll look at the rest of the methods too. : )