how to determine whether u, v and w lie in the same plane when positioned so tht their initial points coincide

Question

the vectors
u = (-1, -2, 1)
v = (3, 0, -2)
w = (5, -4, 0)

Answer 1

Calculate their triple product - http://en.wikipedia.org/wiki/Triple_product

If it's zero - they are in the same plane

Answer 2

Calculate the triple product of the vectors. If it is zero the vectors all lie in the same plane.

u • (v X w)

The cross product of v and w will be a vector n that is normal to the plane of v and w. The vector u will therefore be perpendicular to n if it lies in the same plane as v and w. And the dot product of perpendicular vectors is zero.

n = v X w = <3, 0, -2> X <5, -4, 0> = <-8, -10, -12>

u • (v X w) = <-1, -2, 1> • <-8, -10, -12> = 8 + 20 - 12 = -16 ≠ 0

The dot product is not zero. Therefore the vectors do not all lie in the same plane.
___________

Answer 3

Any vector perpendicular to u must also be perpendicular to v and w. If n is a vector perpendicular to u and v, then n = u x v, now if the dot product n.w = 0 then all the vectors u,v and w lie on a same plane.

Lets check,

u x v = (4,1,6) = n

now check n.u = n.v = 0, but
n.w is not = 0 so, u, v, w does not lie in the same plane

Answer 4

ANY three vectors u,v,w will ALWAYS lie in the same plane. (for what it's worth, that plane defined by u,v,w is: (x - u) . [(v-u) × (w -u)] = 0 where x=(x,y,z) is the variable and u,v,w are vector constants. See link below for more.) Anyway, these three particular vectors are however not just COPLANAR but COLLINEAR (all lie in the same line), since as the others points out they are l.d. due to u = v+w