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Struggling to understand Arbitary rotation

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I really dont understand how the author comes up with that equation, under the sentence " Now we can see ...." I know that im doing something wrong, but dont know what, here is my thoughts, can someone point out what im doing wrong? Photobucket

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Think of only one plane in that picture: the one that contains v_perp, w and v'_perp. Think of v_perp and w as forming a basis in that plane. Now you simply want to rotate the point (1,0), which leaves you at (cos(alpha),sin(alpha)). Does that make sense?

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Oh yes, that makes sense, the trouble im having is when they introduce the v
ectors, such as

V'T = cos(theta)VT + sin(theta)W

When you say just cos(theta), that makes perfect sense to me, as from my understand its


---->soh, cah, toa

---->cos(theta) = adj/hyp

---->unit circle, so hyp == 1

---->cos(theta) = adj/1

---->cos(theta) * 1 = adj

---->cos(theta) = adj


But when you bring in the vectors such as

cos(theta) * Vec[3] = adj,

i dont understand how they equate to being the hypotenuse, i guess im misunderstanding something here


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Notice please that the basis mentioned by alvaro defines a co-ordinate system. I.e. v_perp and w define a local co-ordinate system in the world space. The co-ordinates cos(theta) and sin(theta) are then referred to the basis vectors v_perp and w, resp.

Assume you have a local co-ordinate system with a x axis
ex := [1 0 0]
and a y axis
ey := [0 1 0]
Both have unit length. Now you specify a point not by its x,y co-ordinates but by a length and an angle.
p := ( L,theta )
and compute the corresponding reactangular x,y co-ordinates as
px := L * cos( theta )
py := L * sin( theta )

To express this as a vector, you apply the co-ordinates to the belonging axes
p := px * ex + py * ey
== L * cos( theta ) * [1 0 0] + L * sin( theta ) * [0 1 0]
Now, you incorporate the L into the axis vectors, i.e. you scale the co-ordinate system:
== cos( theta ) * [L 0 0] + sin( theta ) * [0 L 0]
Do you see already the similarity to the equation in question?

Now, do the same not in the local co-ordinate space but in the global one. The local [L 0 0] and [0 L 0] vectors have a global equivalent, which can be computed by applying the local-to-global transformation. We don't need to do so explicitely, since they are already computed as v_perp and w. Hence, in global co-ordinates we get
p = cos( theta ) * vperp + sin( theta ) * w

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