Question

show that (a+b,b+c,c+a)=[abc]^2​

Answer 1

Answer:

i) By definition [a, b, c] = a.(b x c)

ii) Also by property, [a, b, c] = [b, c, a] = [c, a, b], taking in the same clock wise order; however if the order is reversed, then they are not equal; that is [a, b, c] ≠ [a, c, b]

2) Applying the above, [a+b, b+c, c+a] = (a+b).{(b+c) x (c+a)}

= (a+b).{(b x c) + (b x a) + (c x c) + (c x a)}

[Cross product of vectors is distributive over addition]

= (a+b).{(b x c) + (b x a) + (c x a)} [Since c x c = 0]

= a.(b x c) + a.(b x a) + a.(c x a) + b.(b x c) + b.(b x a) + b.(c x a)

= [a, b, c] + [a,b,a] + [a,c,a] + [b,b,c] + [b, c, a]

= [a, b, c] + [a, b, c] {Since [a,b,a] [a,c,a] and [b,bc]each = 0; they form coplanar vectors}

= [a,b,c] + [a,b,c] {From 1.(ii) above}

= 2[a, b, c]

Thus it is proved that [a+b,b+c,c+a]= 2[a b