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The mystery of normalizations with dotproducts

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Ok, this is not really a mystery, per se, but definitely one for me. I have 2 questions. 1) If I get the dotproduct between 2 vectors, I theoretically get the distance one vector would have on the other (like a shadow)....this is according to a game programming book I have. If this is true, what is a cheap way to find the angle between 2 vectors? 2) Does doing a dotproduct on 2 unnormalized vectors yield the same answer as if one or both of those vectors were normalized? 3) ~ I''''m a wannabe programmer ~

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2 ways to make a dot product:

1) Either use the vectors directly

2) use the angle between the two and the norm of both vector.

Technically speaking, computing a dot product is simpler with the vector directly since it dont require the use of computing a cosine.

1) v_1 (DOT) v_2 = x_1*x_2 + y_1*y_2 + z_1*z_2 = a number

Remember that the dot product is a NUMBER not a vector.

2) v_1 (DOT) v_2 = |v_1|*|v_2|*cos(angle between the vectors)

So basically, to answer your first question, equating 1) with 2) will give you a way to get the angle. It is numerically consuming to do so, but still, it is possible.

Your notion of the dot product is a little off. If one of the vector is NORMALIZED, THEN v_1 (DOT) n_1 will yield the projection of v_1 on n_1 like the shadow you talked about.

To answer you second question, you should see by looking at 1) and/or 2) (especially 2)) that the answers are dependant on the lenght of the vectors so... The answer to your question is NO.

Edited by - Genre on December 25, 2001 1:49:40 PM

Edited by - Genre on December 25, 2001 1:50:12 PM

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1) Just to simplify the above formula:

v = acos( a . b / |a| |b| )

However, if both a and b are normalized:

v = acos( a . b )

because |a| |b| = 1

2) No

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Yes... Much simpler. I assume that people want to know all the steps between the question and the answer... Do they?

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