Question

prove that c(uxv) = cuxcv

Answer 1

There seems to be mistake here. We have c(uxv) = cu x v = u x cv and not cu x cv.

Let u = a₁i + a₂j + a₃k, and v = b₁i + b₂j + b₃k

Then uxv =

i	j	k
a1	a2	a3
b1	b2	b3

= i(a₂ b₃-a₃ b₂) - j(a₁b₃-a₃*b₁) + k(a₁b₂-a₂b₁)

so that c(uxv) = c[i(a₂ b₃-a₃ b₂) - j(a₁b₃-a₃b₁) + k(a₁b₂-a₂b₁)]

Also cu x v =

i	j	k
ca1	ca2	ca3
b1	b2	b3

= i(ca₂b₃-ca₃ b₂) - j(ca₁b₃-ca₃b₁) + k(ca₁b₂-ca₂b₁) = c[i(a₂ b₃-a₃ b₂) - j(a₁b₃-a₃b₁) + k(a₁b₂-a₂b₁)] = c(u x v).

Also, u x cv =

i	j	k
a1	a2	a3
cb1	cb2	cb3

= i(ca₂b₃-ca₃ b₂) - j(ca₁b₃-ca₃b₁) + k(ca₁b₂-ca₂b₁) = c[i(a₂ b₃-a₃ b₂) - j(a₁b₃-a₃b₁) + k(a₁b₂-a₂b₁)] = c(u x v).