Question

a b and c are three non zero vectors such that c is unit vector perpendicular to both a and b if the angle between a and b is pi/6 prove that [a b c]^2=(1÷4)|a|^2|b|^2.

Answer 1

we know that cross product is defined as ,

a x b = absinФ

since c is a unit vector perpendicular to both a and b, hence,

c = (a x b) and |c| = 1

LHS

= [a b c]²

= [c . (a x b)]²

= [c . c]²

= [|c||c|cos0°]²

= 1

RHS

1/4 x |a|² |b|²

= 1/4 x [|a||b|]²

= 1/4 x [axb/sinФ]²      

= 1/4 x [|c|/sinπ/6]²          ( as angle between a and b = π/6)

= 1/4 x [1/1/2]²

= 1/4  x 4

= 1

since LHS = RHS,

hence

[a b c]² = (1/4)|a|²|b|²