Given vectors u, v, w, the cross product v × w is normal (perpendicular) to the plane containing v and w, and so any vector
which is orthogonal (perpendicular) to the vector v × w must lie in the plane determined by v and w. Therefore, the triple
cross product u × (v × w), which is orthogonal to both u and v × w, must be of the form Av + Bw for some numbers A and B.
The purpose of the sequel is to prove A = u · w and B = – u · v.
Taking u = u1i + u2j + u3k and v × w = (v2w3 – w2v3)i + (v3w1 – w3v1)j + (v1w2 – w1v2)k,
u × (v × w) = [u2(v1w2 – w1v2) – u3(v3w1 – w3v1)] i
+ [u3(v2w3 – w2v3) – u1(v1w2 – w1v2)] j
+ [u1(v3w1 – w3v1) – u2(v2w3 – w2v3)] k
= (u2w2 + u3w3)v1 i – (u2v2 + u3v3)w1 i
+ (u1w1
+ u3w3)v2 j – (u1v1
+ u3v3)w2 j
+ (u1w1 + u2w2)v3 k – (u1v1 + u2v2)w3 k
= (u1w1 + u2w2 + u3w3)v1 i – (u1v1 + u2v2 + u3v3)w1 i
+ (u1w1
+ u2w2 + u3w3)v2
j – (u1v1 + u2v2
+ u3v3)w2 j
+ (u1w1 + u2w2 + u3w3)v3 k – (u1v1 + u2v2 + u3v3)w3 k
(this last line is obtained from the one above by adding and subtracting u1v1w1i + u2v2w2j + u3v3w3k )
Therefore u × (v × w) = (u1w1 + u2w2 + u3w3)(v1i + v2j + v3k)
– (u1v1 + u2v2 + u3v3)(w1i + w2j + w3k) = (u · w)v – (u · v)w .