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### #ActualCruis.In

Posted 03 January 2013 - 01:12 PM

ok to normalize a vector

divide its x,y by its length

the vector in our example is -5,5 (vector towards target from player)

length = sqr( (-5 * 5) + (5 * 5))

length = sqr ( 25 + 25)

length = sqr (50)

length = 7.07

didn't you make a mistake in your magnitude calculation? since the brackets must be added before you find the square root?

normalized = -5/7.07 = -0.7

5/7.07 = 0.7

This is the cosine of the angle between you, and cos(90degrees) = 0, so they are perpendicular to your facing.

now that part I do not get. Could you word that better?

### #2Cruis.In

Posted 03 January 2013 - 12:14 PM

ok to normalize a vector

divide its x,y by its length

the length in our example is -5,5 (vector towards target from player)

length = sqr( (-5 * 5) + (5 * 5))

length = sqr ( 25 + 25)

length = sqr (50)

length = 7.07

didn't you make a mistake in your magnitude calculation? since the brackets must be added before you find the square root?

normalized = -5/7.07 = -0.7

5/7.07 = 0.7

This is the cosine of the angle between you, and cos(90degrees) = 0, so they are perpendicular to your facing.

now that part I do not get. Could you word that better?

### #1Cruis.In

Posted 03 January 2013 - 11:31 AM

If you are facing northeast, your direction vector would be in the direction of (+1, +1), normalised this gives (1/sqrt(2), 1/sqrt(2)).

where does normalize come from and how does (1/sqrt(2) come into it?

when you normalize the vector of enemy from you which is (-5,5) how do you get (-1/sqrt(2), 1/sqrt(2))

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