linear algebra question

[aande]

Consider two vectors x = {{a},{b}} and y = {{c},{d}} which are both of the same length. Then there is a rotation matrix R(theta) which takes x to y, namely R(theta)x = y. We note that 0 =< theta < 2pi. Use our knowledge of rotation matrices to establish a simple condition on a; b; c; d so that the angle 'theta' satisfies 0 < theta < pi

I have no idea where to start or even what the question is asking really. . . and this question is do in exactly 12 hrs

 

[aande]

I do know that it might have to do with slopes since a and b, and, c and d could be though of as rise/run. for x, the slope is b/a, for y, the slope is d/c. And slopes are related to angles, kinda.

I have found this inequality if the slope of y is greater than slope of x:

d / c > b /a
ad > bc
ad - bc > 0

And I know that ad-bc is the determinant of the 2x2 matrix {x,y} = {{a,b},{c,d}} but I don't know if that helps at all.

Yeah, completely lost.

 

[Vannicke]

Basically what the question is saying, is that if x is some certain place, where is y allowed to be if R(theta)x = y and 0 < theta < pi.

Since 0 < theta < pi, when you think about the circle the x and y are on [centered at the origin] wherever x is on the circle, we can only allow y to be somwhere within the half turn forward [since rotating a maximum of pi is turning a maximum of half the circle]

since x is at {a,b} halfway around the circle is {-a,-b} and so depending on whether the a and b are negative [or in what quadrant of the circle x is in], you can say

c > -a and d > -b if a and b are positive [x is in quadrant I]
c < -a and d > -b if only b is positive [x is in quadrant II]
c > -a and d < -b if only a is positive [x is in quadrant III]
c < -a and d < -b if a and b are negative [x is in quadrant IV]

Basically, this eliminates the parts of the circle that are further than pi from x, by making sure you don't pass -a or -b around the other side of the circle.

Hope it helps, even if its late...

-Van