Question

Help with rotation of axes

Answer 1

Find the coordinates (x′, y′) for the given point (x, y), and the angle of rotation θ: (x, y) = (0,3), θ = 90°.

Answer 2

0=x'cos(90)-y'sin(90)
3=x'sin(90)-y'cos(90)

Answer 3

you plugged in the point into the rotation matrix?

Answer 4

$$\Large {
\begin{bmatrix}
\cos \theta &-\sin \theta \\
\sin \theta &\cos \theta \\
\end{bmatrix}
\begin{bmatrix}
x\\ y\\
\end{bmatrix}
= \begin{bmatrix}
x'\\ y'
\end{bmatrix}

}$$

Answer 5

Twas the only formula I know for this :/ I haven't worked with a problem like this before

Answer 6

So we already know x=0 because if it is multiplying then it will just end up 0

Answer 7

i mean x'=0

Answer 8

So the rotation goes to (-3,0)?

Answer 9

correct

Answer 10

$$\Large {
\begin{bmatrix}
\cos \theta &-\sin \theta \\
\sin \theta &\cos \theta \end{bmatrix}\cdot
\begin{bmatrix} x\\ y \end{bmatrix}
= \begin{bmatrix} x'\\ y' \end{bmatrix}
\\~\\\Rightarrow \\~\\
\begin{bmatrix}
\cos 90^0 &-\sin90^0 \\
\sin 90^0 &\cos 90^0 \end{bmatrix}\cdot
\begin{bmatrix} 0\\ 3 \end{bmatrix}
= \begin{bmatrix} x'\\ y' \end{bmatrix}
\\~\\\Rightarrow \\~\\
\begin{bmatrix}
0 &-1 \\
1 &0 \end{bmatrix}\cdot
\begin{bmatrix} 0\\ 3 \end{bmatrix}
= \begin{bmatrix} -3\\ 0 \end{bmatrix}


}$$

Answer 11

Did you cross multiply?

Answer 12

If you want to use your formula, you should have

x'= x cos(90)-ysin(90)
y ' =x sin(90)-y cos(90)

plug in x = 0, y = 3

Answer 13

actually theres a mistake

Answer 14

If you want to use your formula, you should have
x'= x cos(90)-ysin(90)
y ' =x sin(90)+ y cos(90)

plug in x = 0, y = 3

x'= 0 cos(90)- 3 sin(90)
y ' =0 sin(90)+ 3 cos(90)

x'= 0 - 3 *1= -3
y ' =0+ 3*0 = 0