Vector Rotation: How to Rotate a Vector by 90 Degrees using a Rotation Matrix

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To rotate a vector v = x_1e_1 + x_2e_2 by 90 degrees using a rotation matrix, the correct result for a counterclockwise rotation is given by the transformation matrix \(\vec{y} = \begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}\). The initial attempt resulted in a clockwise rotation, indicating a misunderstanding in the application of the rotation matrix. A hint was provided to multiply by -1 to achieve the desired counterclockwise rotation. The discussion highlights the importance of correctly interpreting the transformation matrix to avoid errors in vector rotation. Understanding these concepts is crucial for accurately performing vector rotations in mathematics.
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Homework Statement



Rotate a vector v=x_1e_1+x_2e_2 90 degrees by using the rotation matrix.

The Attempt at a Solution



As you can see in the attached image, I get a 90 degree clockwise rotation. I'm supposed to get a 90 degree counterclockwise rotation. Where do I go wrong?
 

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Hi _Andreas! :smile:
_Andreas said:
Rotate a vector v=x_1e_1+x_2e_2 90 degrees by using the rotation matrix.

As you can see in the attached image, I get a 90 degree clockwise rotation. I'm supposed to get a 90 degree counterclockwise rotation. Where do I go wrong?

erm … not enough sleep? :redface:

Hint: try multiplying by -1 :smile:
 
What you've drawn for your rotated vector is x_1 \hat{e_1} - x_2 \hat{e_2}; which is not the same as what you calculated using the transformation matrix: \vec{y}=\begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}=-x_2 \hat{e_1} + x_1 \hat{e_2}
 
gabbagabbahey said:
What you've drawn for your rotated vector is x_1 \hat{e_1} - x_2 \hat{e_2}; which is not the same as what you calculated using the transformation matrix: \vec{y}=\begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}=-x_2 \hat{e_1} + x_1 \hat{e_2}

Thanks! I did suspect that this was the case, but I wasn't sure.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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