Is the following vector cross product identity the Associative Law or the Distributive Law of scalar multiplication or neither
p a x q b = p q (a x b),
where p and q are scalars and a and b are non-zero vectors.
Please help, I am really confused!
2007-05-25 04:32:20 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The distributive law would look like a x (b + c) = a x b + a x c
or a.(b x c) = a.b x a.c
The associative law would be (a x b) x c = a x (b x c) which doesn't exist in vector product . Even a.(b x c) = (a.b) x c isn't possible as one is a number and the other is a vector
The second example of the distributive law seems very close to your question
2007-05-25 05:44:44 · answer #1 · answered by fred 5 · 1⤊ 0⤋
Just write out the vectors: u = (u1, u2, ... , un) v = (v1, v2, ... , vn) Then u (dot) v = u1v1 + u2v2 + ... + unvn = v1u1 + v2u2 + ... + vnun = v (dot) u. It's follows from the fact that scalar multiplacation is commutative (that is, 3 * 4 = 4 * 3). You can prove 2) in the same way.
2016-05-17 11:42:48 · answer #2 · answered by mireille 3 · 0⤊ 0⤋
Neither. Why bother with what law it is----just carry out the operation?
p a x q b = pq (a x b) = qp (a x b) if required
Note
pq = qp (scalars)
(a x b) = - (b x a) (vectors)
2007-05-25 23:52:43 · answer #3 · answered by Como 7 · 1⤊ 0⤋
Neither
2007-05-25 04:43:24 · answer #4 · answered by swd 6 · 0⤊ 0⤋
Associative Law
if it were distributive it will be
p a x q b = pqa x pqb,
2007-05-25 04:43:56 · answer #5 · answered by iyiogrenci 6 · 0⤊ 2⤋
distributive
2007-05-25 04:37:39 · answer #6 · answered by uk_wildcat96 2 · 0⤊ 1⤋