if A=sinui+cosuj+uk, B=cosui-sinuj-3k and C=2i+3j-k, evaluate d/du(Ax(BxC))

Question

Vector triple product

Answer 1

if A=sinui+cosuj+uk, B=cosui-sinuj-3k and C=2i+3j-k, evaluate d/du(Ax(BxC))

 

BxC = (cosu, -sinu, -3)x(2, 3, -1) = (u1, u2, u3)(v1, v2, v3)

BxC = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)

BXC = (-sinu.(-1) - (-3).3, -3.2 - cosu.(-1), cosu.3 - (-sinu).2)

BxC = (sinu + 9, -6 + cosu, 3.cosu + 2.sinu)

Ax(BxC) = (sinu, cosu, u)x(sinu + 9, -6 + cosu, 3.cosu + 2.sinu)

Ax(BxC) = (cosu.( 3.cosu + 2.sinu) – u.( -6 + cosu), u.(sinu + 9) – sinu.( 3.cosu + 2.sinu), sinu.( -6 + cosu) – cosu.(sinu + 9))

 

d(Ax(BxC))/du = (sinu(u – 6.cosu) + 4(1 + cos^2u) – cosu,

                            sinu(1 – 4.cosu) + 6(1 + sin*2u) + u.cosu,

                            -6.cosu – 9.sinu)