Question

for
How we can write distributive
write distributive property
À , B and Ở in vector product ?​

Answer 1

Let d=a×(b+c)−a×b−a×c

so it is required to prove that d=0:

d2 = d⋅d

=d⋅(a×(b+c)−a×b−a×c)

=d⋅(a×(b+c))−d⋅(a×b)−d⋅(a×c)

=(d×a)⋅(b+c)−(d×a)⋅b−(d×a)⋅c

=(d×a)⋅(b+c)−(d×a)⋅(b+c)

=0

Therefore d=0, so a×(b+c)=a×b+a×c.

Let a=(a1,a2,a3), b=(b1,b2,b3), c=(c1,c2,c3)∈R3 so

a×(b+c)=∣∣∣∣ijka1a2a3b1+c1b2+c2b3+c3∣∣∣∣=i(a2b3+a2c3−a3b2−a3c2)−j(...)+k(...)

Now try to rearrange the above terms to find the result. See that in the first term we have i(a2b3+a2c3−a3b2−a3c2)=i(a2b3−a3b2)+i(a2c3−a3c2).