Vectors

We know that, A\timesB=ABsinθ Also, A.B=ABcosθ Given, ∣A\timesB∣=\sqrt3A.B Use ∣A\timesB∣=∣A∣∣B∣sinθ Then, ∣A\timesB∣=ABsinθ A.B=∣A∣∣B∣cosθ ∴ ABsinθ=\sqrt3ABcosθ tanθ=\sqrt3 ​⟹θ=60 o Now, (A+B) 2 =A 2 +B 2 +2A.B =A 2 +B 2 +2ABcosθ =A 2 +B 2 +2AB\times =A 2 +B 2 +AB or ∣A+B∣ =(A 2 +B 2 +AB) So, the correct option is (D): (A 2 +B 2 +AB)

Chapter Questions
1 MCQs