vector problem

eoins23456

two vectors where the modulus of a equals the modulus of b which equals the modulus of a-b...find the angle between a and b.slightly confused with this one.anyone got any ideas?

straight_As

I may have over simplified this completely, so don't laugh if I'm wrong

|a| = |b| = |(a-b)|

|a| = |b| = |ba| .... ba = a-b

Apply the triangle law => equilateral triangle => all angles = 60'

Hope I'm right :pac:

Michaelrsh

straight_As wrote: »

I may have over simplified this completely, so don't laugh if I'm wrong

|a| = |b| = |(a-b)|

|a| = |b| = |ba| .... ba = a-b

Apply the triangle law => equilateral triangle => all angles = 60'

Hope I'm right :pac:

a.b = !a! !b! cos(thetos)

! = modules

zam

cosx = (a.b) over (|a|.|b|)

?

Michaelrsh

zam wrote: »

cosx = (a.b) over (|a|.|b|)

?

Oui!!

Michaelrsh

zam wrote: »

cosx = (a.b) over (|a|.|b|)

?

Remember that a.b is the dot product of a and b

Fremen

[latex] |a-b| = \sqrt{|a|^2 + |b|^2 - 2 a.b}[/latex]

Any help to you?

Fremen

Straight as, you're right, but I think you need to justify it a little better than that. What did you mean by |ab|, for example?

straight_As

(The vector (ba)) = (the vector a) - (the vector b)
=> |(the vector (ba))| = |(the vector a) - (the vector b)|

Universal vector law maybe? Don't really know what its proper title is tbh.

eoins23456

rightio that make sense i suppose.thanks for the quick responses!