Vector as a function of theta

Mar 2020
4
0
Sweden
Screenshot 2020-03-22 at 07.58.00.png

Express the vector \(\displaystyle \vec{OA}\) and \(\displaystyle \vec{AB}\) as a function av θ. Suppose that the angle between OA and AB is 90°.
I have begun to define the unitvector: \(\displaystyle \underline{e}_{OA}=\begin{bmatrix}-sin\theta\\ -cos\theta\\0\end{bmatrix}\leadsto \vec{OA}=40\begin{bmatrix}-sin\theta\\ -cos\theta\\0\end{bmatrix}\). I have now a relation for the vector OA as a function of θ. My question is if the overall position of the line segment AB matter at all? I think that AB rotates similarly by an angle theta. However I'm not sure how I can get a relation with the vector AB as a function of θ.

Very thankful for answers!
 
Apr 2015
1,238
359
Somerset, England
hi sandra you are nearly there.

You are correct Ab rotates the same as OA, but from the horizontal, not the vertical.

See if you can carry on from that
 
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Mar 2020
4
0
Sweden
awaw.png
So am I looking for the change in y-direction?
 
Apr 2015
1,238
359
Somerset, England
Keep going you will get there.

vecrot1.jpg
 
Mar 2020
4
0
Sweden
aw.png
To get from A to B I multiplied the unit vector with its magnitude and it is the same vector as in OA?
 
Mar 2020
4
0
Sweden
I'm assuming that the angle theta is the same for the both triangles, how can we know that for sure?
 
Apr 2015
1,238
359
Somerset, England
I'm assuming that the angle theta is the same for the both triangles, how can we know that for sure?
The horizontal and vertical dashed lines are perpendicular, as I noted in the diagram.

I also showed the second angle as (90 - θ) - the angle at B - in the original triangle from the vertical with θ - the angle at O - in it.
So the triangle between AB and the dashed horizontal line contains a right angle and the angle at B or (90 - θ)
So its third angle - the angle at A must be θ.