Question

Derive an expression for vector triple product

Answer 1

Answer:

Explanation:

Let the 3 vectors be u v and w

u=(u1,u2,u3)

y=(y1,y2,y3)

w=(w1,w2,w3)

The expression for vector triple product is:

u x (v x w) = (u • w) v - (u • v) w

To obtain this lets do this:

v x w = (v2.w3 - v3w2)i + (v3w1 - v1w3)j + (v1.w2 - v2.w3)k

u x (v x w) =(v2.w3 - v3.w2)u  i + (v3w1 - v1w3)u  j + (v1.w2 - v2.w3)u  k

=  (v2.w3 - v3w2)(u1i + u2j + u3k) i + (v3w1 - v1w3) (u1i + u2j + u3k)  j +  (v1.w2 - v2.w3)(u1i + u2j + u3k)  k

= (v2.w3 - v3w2)(-uyk + uzj) + (v3w1 - v1w3) (u1k - u3i) +(v1.w2 - v2.w3) (-u1j+u2i)

= [- u3(v2.w3 - v3w2) + u2 (v2.w3 - v3w2)] i + [u3(v3w1 - v1w3) - u1(v1.w2 - v2.w3)] j + [-u2(v2.w3 - v3w2) + u1(v1.w2 - v2.w3)] k  

Rearranging the terms gives:  

u x (v x w) = (u2.w2 + u3.w3)v1 i - (u2.v2 + u3.v3)w1 i + (u1.w1 + u3.w3)v2 j - (u1.v1 + u3.v3)w2 j + (u1.w1 + u2.w2) v3 k - (u1.v1 + u2.v2)w3 k  

Therefore both adding and substracting u1.v1.w1 i + u2.v2.w2 j +u3.v3.w3 k  from the above equation:

u x (v x w) = (u1.w1 + u2.w2 + u3.w3)v1 i - (u1.v1 + u2.v2 + u3.v3)w1 i + (u1.w1 + u2.w2 + u3.w3)v2 j - (u1.v1 + u2.v2 + u3.v3)w2 j + (u1.w1 + u2.w2 + u3.w3)v3 k - (u1.v1 + u2.v2 + u3.v3)w3 k

Therefore we can write this as:

u x (v x w) = (u1.w1 + u2.w2 + u3.w3)(v1 i + v2 j + v3 k) - (u1v1 + u2v2 + u3v3) (w1 i + w2 j + w3 k)  

= (u • w) v - (u • v) w