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

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