Use matrix multiplication to find the image of v=(-2,1,2) when it is rotated through an angle of θ=-π/4 radians about the y-axis
i know it has something to do with trig .... i just cant recall it at the moment
class is starting, linear algebra :) ill try to chk this back in abt 2 hours.
The matrix rotation is \[A=\left( \begin{array}{ccc} \cos(\theta)& 0& \sin(\theta) \\ 0 &1 &0 \\ - \sin(\theta) & 0 & \cos(\theta) \\ \end{array} \right) \] Use \[\theta= -\frac \pi 4 \] and Multiply A by the transpose of of (-2,1,2)
This a evaluated at -\pi/4 \[ \left( \begin{array}{ccc} \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \end{array} \right) \] Multiply it by \[ \left( \begin{array}{c} -2 \\ 1 \\ 2 \\ \end{array} \right) \] to get \[\left( \begin{array}{c} -2 \sqrt{2} \\ 1 \\ 0 \\ \end{array} \right) \]
thanks alot
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