a,b are vectors. If you know that |a|=2, |b|=3 and |a+b|=3, how wolud you find the coordinates of vector a and vector b?

You can''t. You can only find the angle between the two. It''s easy to find with the cosine law.

c² = a² + b² + 2ab * cos(angle)

Where c = |a+b|

Cédric

Basically, the information you have is the length of three sides of a triangle.

Exactly what I''m looking for...

c² = a² + b² + 2ab * cos(angle):

if c=|a+b|, are a and b the same as |a| and |b|?

Yes, they are.

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I have another problem dude:

two vectors a and b:

(2*a-b)*(a+b)=45, |a|=6 and a*b=-2

I have to find the length of b(|b|) ( +the angle between a and b, I''m sure I know how to find the angle... )

Why do you need that?

Cédric, the subtle poster

just develop your first equation:
(2*a-b)*(a+b)=2*a*a+2*a*b-b*a-b*b=45
ie 2*a²+a*b-b²=45
so 2*6²+(-2)-b²=45
b²=25
|b|=5

for the angle, you know that a*b=|a|*|b|*cos(angle)
so cos(angle)=-2/(6*5)=-0.0666666666666666666666
so angle=ArcCos(-0.06666666666666)=1.637 radians

CQFD (too easy!)

That was a homework question (the OP never bothered to tell otherwise). Please read the forum FAQ.

quote:

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Original post by mikamikaze
CQFD (too easy!)

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CQFD is a french expression, you know. It doesn''t make sense at all to the English speakers here (I don''t remember the English equivalent).

Cédric

quote:

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Original post by cedricl
CQFD is a french expression, you know. It doesn''t make sense at all to the English speakers here (I don''t remember the English equivalent).

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It ought to, at least to the erudite ones. CQFD stands for "Ce qu''il fallait démontrer." This is the functional equivalent of the Latin expression QED ("quod erat demonstrandem", IIRC). Both mean, "Which was to be proved," and are used to show that the desired conclusion of a particular set of problem-solving steps has been reached, and the answer provided.

Peace,
ZE.

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