Anyone good at solving vector question ?

Question

a,b,c are non coplanar,unit vectors,equally inclined to each other at angle N. If (axb)+(bxc)=pa+qb+rc
Find scalars p,q,r in terms of N .

Please tell your working also .

Answer 1

This is a special case of finding the coordinates of a vector v in in a basis a, b, c with known scalar products a.b, b.c, and c.a, when the scalar products v.a, v.b, and v.c are also known. Multiply the eqn v = pa+qb+rc scalarly by a, b and c and write the result in matrix form. Now simply invert the 3x3 matrix G, called the Gram matrix, to express p, q, r in terms of the known scalar products.

Now for the original question. When v = (axb)+(bxc), you get a.v=a.bxc, b.v=0 and c.v=c.axb=a.bxc. The mixed triple product a.bxc is +- the volume of the parallelepiped spanned by a, b and c. Assume that the system a b c (in this order) is right-handed so the sign is +. Also, (a.bxc)^2 equals the determinant of the Gram matrix G. When a b c are unit vectors inclined at angle N, the matrix G has units on the diagonal and cosN elsewhere, and detG = (1+2cosN)(1-cosN)^2.
Multiplying it all out, you end up with
p = r = -sqrt(1+2cosN)
q = 2cos(N)sqrt(1+2cosN)

Answer 2

Let's see.

a x b is perpind to the plane of a-b
b x c is perpind to the plane of b-c

I think that implies that these vectors are both pepind to b. Hence the dot product of either of these with b is zero. That plus a x b = |a||b|sin n should help....

I ran out of steam. Try to sketch what's happening.