Question

i) By definition ( [a, b, c]=a(b times c) )
ii) Also by property, ( mid 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] eq[a, c, b] )
2) Applying the above, ( mid a+b, b+c, c+a]=(a+b){(b+c) times(c+a)} )
( =(a+b) cdot((b times c)+(b times a)+(c times c)+(c times a)) )
[Cross product of vectors is distributive over addition]
( =(a+b) cdot((b times c)+(b times a)+(c times a))[operatorname{sincec} x c=0] )
( =a(b times c)+a(b times a)+a(c times a)+b cdot(b times c)+b cdot(b times a)+b cdot(c times a) )
( =[a, b, c]+[a, b, a]+[a, c, a]+[b, b, c]+[b, c, a] )
( =[a, b, c]+[a, b, c] text { isince }[a, b, a][a, c, a] text { and }[b, b c] e a c h=0 ; text { they form coplanar vectors }) )
( =[a, b, c]+|a, b, c| text { From } 1, text { (ii) above }} )
( =2 mid a, b, c] )
Thus it is proved that ( [a+b, b+c, c+a]=2[a b c] )

# (b) Prove that ſă +5,6 +ē,č + ā] = 2ļābē]

Solution