- #1

sherlockjones

- 31

- 0

In the xy-plane let us find a point of reflection [tex] Q [/tex] of a general point [tex] P = (a,b) [/tex] in the line [tex] y = x [/tex]. The line [tex] x-y = 0 [/tex] has a normal vector [tex] <-1, 1> [/tex] and so a vector equation of the straight line through [tex] P [/tex] and perpendicular to [tex] y = x [/tex] is given by [tex] <x,y> = t<1,-1> + <a,b> = <a+t,b-t> [/tex]. The point of intersection [tex] M = (x,y) [/tex] of the two lines is obtained from [tex] a+t = b-t [/tex] or [tex] t = \frac{1}{2}(b-a) [/tex]. Thus [tex] M = (\frac{1}{2}(a+b), \frac{1}{2}(a+b)) [/tex]. The point [tex] Q = (b,a) [/tex] because [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]. Thus [tex] w = Av [/tex] where [tex] A [/tex] is the identity matrix.

I do not get how the normal vector of [tex] x-y = 0 [/tex] is [tex] <1,-1> [/tex]. Isn't that saying that the x-component is 1 and the y-component is -1? Also how did they get the vector equation [tex] <x,y> = t<1,-1> + <a,b> [/tex]? Finally, why does [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]?

Thanks