Question 90 of 379HardMCQ
Let λ∈R,a=λi^+2j^−3k^,b=i^−λj^+2k^.
If ((a+b)×(a×b))×(a−b)=8i^−40j^−24k^,
then ∣λ(a+b)×(a−b)∣2 is equal to : Solutions · 1
Official solution
(a→×(a→×b→)+b→×(a→×b→))×(a→−b→)
=(a→(a→.b→)−b→(a→.a→)+a→(b→.b→)−b→(a→.b→))×(a→−b→)
=(a→.b→)(a→×a→−a→×b→)−(a→.a→)(b→×a→−b→×b→)+(b→.b→)(a→×a→−a→×b→)−(a→.b→)(b→×a→−b→×b→)
=(a→.b→)(b→×a→)−(a→.a→)(b→×a→)+(b→.b→)(b→×a→)−(a→.b→)(b→×a→)
=(b→×a→)(b→.b→−a→.a→)
=(5b→×a→)(5+λ2−13−λ2)
=8(a→×b→)
a→×b→=i^j^k^λ2−31−λ2
=i^(4−3λ)−j^(2λ+3)+k^(−λ2−2)
⇒λ=1
∣a→×(a→−b→)+b→×(a→−b→)∣2
=∣2(a→×b→)∣2=4.35=140 Answer the question to unlock