Prove parts (a) and (d), of Theorem 6.1.1, justifying each step with the name of a vector

Question:

Prove parts (a) and (d), of Theorem 6.1.1, justifying each step with the name of a vector space axiom or by referring to previously established results.
Theorem 6.1.1
If u, v and w are vectors in a real inner product space, and k is any scalar, then
(a) (0, v) = (v,0) = 0
(d) (u - v, w ) = (u,w) + (v,w)