Question

If |A×B| = 3 A.B then the value of |A + B| is

• A. $(A^2+B^2+\frac{AB}{\sqrt 3})^{\frac{1}{2}}$
• B. A + B
• C. (A2+B2+$\sqrt 3$AB)$^{\frac{1}{2}}$
• D. (A2+B2+AB)$^{\frac{1}{2}}$

Answer 1

Answer: (D)

(A2+B2+AB)$^{\frac{1}{2}}$

Answer 2

We know that,
A×B=ABsinθ Also, A.B=ABcosθ Given, ∣A×B∣=√3A.B
Use ∣A×B∣=∣A∣∣B∣sinθ Then, ∣A×B∣=ABsinθ A.B=∣A∣∣B∣cosθ
∴ ABsinθ=√3ABcosθ tanθ=√3 ​⟹θ=60o
Now, (A+B)2=A2+B2+2A.B =A2+B2+2ABcosθ
=A2+B2+2AB×$\frac{1}{2}$
=A2+B2+AB or ∣A+B∣
=(A2+B2+AB)$^{\frac{1}{2}}$
So, the correct option is (D): (A2+B2+AB)$^{\frac{1}{2}}$