Vectors

Hi i need to prove the following;

3 distinct points, P,Q are R have position vectors p,q and r respectfully. Show that P,Q and R are collinear (that is, all lie on a straight line) if   p x q + q x r + r x p = 0

ok

OP = p
OQ = q
OR = r

p.q + q.r + r.p = 0
|p||q|cosT + |q||r|cosR + |r||p|cosS = 0

because they are all distinct points, that is, they are non-zero vectors, then the only way that we can obtain a sum of 0 is when:
cosT = 0, cosR = 0 and cosS = 0

because all of the angles = 0, then they are all parallel.
and because they all share a common point O, all three points are collinear.

^ nigger can't trig.

>>3 here
oops. oh well, they share teh same angle pi therefore niggers.

>>5
oh lol shit i meant pi/2

thanks for that anon >>3

>>6
pi/2?
p = (1,0,0)
q = (0,1,0)
r = (0,0,1)
Their dot products are all 0, but they sure as hell aren't collinear.  I think OP means the cross product.

op here, yer i noticed the dot product and figured out u meant cross product.

All 3 terms must add up to 0 and as all are non negative, as lengths are positive and sin(x) cannot be negative between 0-180 degrees then all the angles must be 0 or Pi, so they must lie on a straight line.

>>9
Unlike dot products, cross products are vectors, and that complicates things dramatically.

p = (2,0,0)
q = (1,1,0)
r = (0,2,0)
are clearly collinear, but I don't think any of the 3 cross products are 0 (certainly not p x r)

vectors:
a(a1, a2, a3)
b(b1, b2, b3)
theta is the angle formed by the vectors

2 types of vector multiplication exist:

dot product: results in a constant [cos(theta)*(a1*b1 + a2*b2 + a3*b3)]. The use of the cosine means that vectors with the same direction will turn the cosine factor irrelevant, producing a maximum value, and vectors perpendicular to each other will turn the product null, as cos(pi/2) is 0.

cross product: results in a vector [sin(theta)*c(a2*b3-b2*a3, a3*b1 - a1*b3, a1*b2-a2*b1)]. Notice the subtractions, which mean that axb differs from bxa. To imagine the result of a cross product you can make use of the 'hand rule' (look it up). Opposite of the dot product, the sine in the operation makes it so the maximum value of the vector formed by a cross product is obtained when they are perpendicular i.e. theta=pi/2 (sin(pi/2)=1), and reduces as the angle approaches 0 or pi, effectively becoming null at these values.

to answer your question, which is not exactly well formulated since the sum of 3 vectors a constant does not make (let's just imagine that 0 is a null vector), the cross products of any three collinear vectors result in 0 because sin(0) or sin(pi) is 0, thus multiplying all the values of the final vectors by 0 and creating null vectors (c(0,0,0)).

i'm bored

This is all very fascinating but I don't care.

>>11
0 is an overloaded term; it can be the identity in an additive group, the 0 function, the number 0, the 0 vector...

The vectors need not be collinear; see my example in >>10.  The question is asking for the tips of the vectors to be collinear.

Actually, there's no need to refer to the coordinates.

0 = p x q + q x r + r x p = p x q - r x q - p x r + q x q = p x (q-r) + (q-r) x q = (p-q) x (q-r) = (r-q) x (p-q)

Which is condition for collinearity.