# Let #vec(x)# be a vector, such that #vec(x) = (−1, 1), " and let " R(θ) = [(costheta, -sintheta), (sintheta, costheta )]#, that is Rotation Operator. For #theta=3/4pi# find #vec(y) = R(theta)vec(x)#? Make a sketch showing x, y, and θ?

##### 1 Answer

This turns out to be a counterclockwise rotation. Can you guess by how many degrees?

Let

#T(vecx) = R(theta)vecx,#

#R(theta) = [(costheta,-sintheta),(sintheta,costheta)],#

#vecx = << -1,1 >>.#

Note that this transformation was represented as the **transformation matrix**

What it means is since

#[(costheta,-sintheta),(sintheta,costheta)]xx<< -1,1 >>#

For an **row** dimension and **column** dimension. That is:

#[(y_(11),y_(12), . . . , y_(1n)),(y_(21),y_(22), . . . , y_(2n)),(vdots,vdots, ddots , vdots),(y_(m1),y_(m2), . . . , y_(mn))]#

#= [(R_(11),R_(12), . . . , R_(1k)),(R_(21),R_(22), . . . , R_(2k)),(vdots,vdots, ddots , vdots),(R_(m1),R_(m2), . . . ,R_(mk))]xx[(x_(11),x_(12), . . . , x_(1n)),(x_(21),x_(22), . . . , x_(2n)),(vdots,vdots, ddots , vdots),(x_(k1),x_(k2), . . . , x_(kn))]#

Therefore, for a **column vector**.

Multiplying these two gives:

#[(costheta,-sintheta),(sintheta,costheta)]xx[(-1),(1)]#

#= [(-costheta - sintheta),(-sintheta + costheta)]#

Next, we can plug in

#color(blue)(T(vecx) = R(theta)vecx)#

#= R(theta)[(-1),(1)]#

#= [(-cos((3pi)/4) - sin((3pi)/4)),(-sin((3pi)/4) + cos((3pi)/4))]#

#= [(-cos135^@ - sin135^@),(-sin135^@ + cos135^@)]#

#= [(-(-sqrt2/2) - sqrt2/2),(-sqrt2/2 + (-sqrt2/2))]#

#= color(blue)([(0),(-sqrt2)])#

Now, let's graph this to see what this looks like. I can tell that it's a **counterclockwise rotation**, after determining the transformed vector.

Indeed, a counterclockwise rotation by

*CHALLENGE: Maybe you can consider what happens when the matrix is* *instead. Do you think it will be clockwise?*