Prove by writing both sides of the following out in component form that: (a x b) x c)= (a • c)b -(b • c)a?
Prove by writing both sides of the following out in component form that:
(a x b) x c)= (a • c)b -(b • c)a
So I wrote out the right hand side out in terms of x,y, and z for the a,b,c but i'm getting lost on the left hand side.
Let a = (ax,ay,az)
b = (bx,by,bz)
c = (cx,cy,cz)
as it was suggested.
Then, once this is proven, how do i show that multiplication via vector product is non-associative?
Update:In other words, I need help with figuring out the left hand side of this equation plus the "vector product is non-associative" part. Thanks!
More Questions From This User See All
Answers & Comments
Given a basis {i, j, k} define:
a = [m, n, p]
b = [r, s, t]
c = [u, v, w]
we have
a ⨯ b = det([i, j, k; m, n, p; r, s, t])
where inside the square brackets there are the rows of the matrix
i, j, k
m, n, p
r, s, t
determinant is only "formal" as i,j,k are vectors and the other are scalars
a x b = i (n t - p s) + j (p r - m t) + k (m s - n r)
a x b = [n t - p s, p r - m t, m s - n r]
(a x b) x c = det [
i, j, k
n t - p s, p r - m t, m s - n r
u, v, w ]
= i (r (n v + p w) - m (s v + t w)) + j (m s u - n (r u + t w) + p s w) + k (m t u + n t v - p (r u + s v))
(a x b) x c = [r (n v + p w) - m (s v + t w), m s u - n (r u + t w) + p s w, m t u + n t v - p (r u + s v)]
RHS gives
(a·c)b - (b·c)a = (m u + n v + p w) b - (r u + s v + t w) a =
= (m u + n v + p w) [r, s, t] - (r u + s v + t w) [m, n, p] =
= [r (n v + p w) - m (s v + t w), m s u - n (r u + t w) + p s w, m t u + n t v - p (r u + s v)]
comparing the components we see that
(a x b) x c = (a·c)b - (b·c)a