Is it possible to have the resultant of two vectors be zero? How? What if they are different magnitudes?

< a, b > + < x, y > = < a + x, b + y >

It will only be zero if:
a + x = 0
b + y = 0

Which means that:
a = -x
b = -y

Putting that back into vectors would show that if:
< a, b > + < x, y > = 0
Then:< a, b > = -< x, y >

So it will only be zero if one is the negative multiple if they are the "additive inverses" of each other. That is to say, if you multiply one by -1 and get the other.

Multiplying by -1 doesn't change the magnitude but instead the direction. They point exactly opposite to each other.

... or, intuitively, if you walk straight out from (0,0) along a vector A, the only way you can walk straight back is to follow -A.

hey mate,

I believe that the only way two vectors (v1, v2) can produced a resultant zero vectors is iff v1 = -v2
Thus,
v1 + v2 = 0
Proof,
Consider 2 Real (or Complex) n x 1 vector's
i.e.,
v1 = sum ( a(i)e(i) ; i = 1 to n)
v2 = sum( b(j)e(j) ; j = 1 to n)
where a(i), b(j) are scalar values, and e(k) = kth coordinate vector, i.e.  e(2) for n = 4 = [ 0 , 1 , 0 , 0 ]^T
Then,
v1 + v2 = sum ( a(i)e(i) ; i = 1 to n) + sum( b(j)e(j) ; j = 1 to n)
v1 + v2 = sum( (a(k) + b(k))e(k) ; k = 1 to n)
now for this to be the 0 vector
0 =  sum( (a(k) + b(k))e(k) ; k = 1 to n)
which given the linear independence of e(k) imposes that
a(k) + b(k) = 0 a(k) = -b(k) for all k = 1 to n
as,
v1 = sum( a(i)e(i) ; i = 1 to n) v1 = sum( -b(i)e(i) ; i = 1 to n)
v1 = -sum(b(i)e(i) ; i = 1 to n) = -v2

Whilst I'm sure they made be evidence to the contrary in terms of the existence of two vectors that result in zero, I doubt their validity - but hey who am I to say!

Hope this helps,

David