Question

Show that (1+|a|^2) (1+|b|^2)=|1-a.b|^2+|a+b+a×b|^2 . For any 2 vectors a and b

Answer 1

Note that in the answer below a,b are vectors and |a|, |b| represent their respective 'mods'
To show :
(1+|a|²)(1+|b|²) = (1−a.b)²+|(a+b)+(a×b)|²

RHS = 1 + |a|²|b|²cos²θ − 2|a||b|cosθ+(a+b)² + (a×b)² + 2(a+b)⋅(a×b)
Since, (a×b) is perpendicular to (a+b),
last term, ie, 2(a+b)⋅(a×b) = 0
⇒RHS = 1 + |a|²|b|²cos²θ − 2|a||b|cosθ + (a+b)² + (a×b)²
=1 + |a|²|b|²cos²θ − 2|a||b|cosθ + |a|²+|b|² + 2|a||b|cosθ + |a|²|b|²sin²θ
=1 + |a|²|b|²(cos²θ+sin²θ) + |a|² + |b|²
=1 + |a|²|b|² + |a|² + |b|²
=(1+|a|²)(1+|b|²) = LHS
Hence Proved