The proof is below.
Work Step by Step
We prove that the two results are the same. We will use a two-dimensional vector to prove this. (Ai+Bj)((Ci+Dj)+(Ei+Fj)) We add the two first and then expand: (Ai+Bj)((C+E)i+(D+F)j) We use the definition of the dot product to find: (AC+AE)+(BD+BF) We now use the distributive property to prove this is an equivalent method: (Ai+Bj)((Ci+Dj)+(Ei+Fj)) (Ai)(Ci+Dj)+Ai(Ei+Fj)+Bj(Ci+Dj)+Bj(Ei+Fj) Simplifying gives: AC+AE+BD+BF This is the same result as above.