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Question: If\[\left| {A \times B} \right| = \sqrt 3 A \cdot B\], then the value of \[\left| {A \times B} \righ...

If∣A×B∣=3A⋅B\left| {A \times B} \right| = \sqrt 3 A \cdot B, then the value of ∣A×B∣\left| {A \times B} \right|is:
A. (A2+B2+AB)12{\left( {{A^2} + {B^2} + AB} \right)^{\dfrac{1}{2}}}
B. (A2+B2+AB3)12{\left( {{A^2} + {B^2} + \dfrac{{AB}}{{\sqrt 3 }}} \right)^{\dfrac{1}{2}}}
C. A +B
D. (A2+B2+3AB)12{\left( {{A^2} + {B^2} + \sqrt 3 AB} \right)^{\dfrac{1}{2}}}

Explanation

Solution

Hint:

a. You should know vector calculus.
b. You should know vector identities.
c. You should know cosine values for 0,30, 45, 60, and 900,30,{\text{ }}45,{\text{ }}60,{\text{ }}and{\text{ }}90.

Complete step by step solution:

We know that, by vector product,
(A×B)=ABsin⁡θn^\left( {A \times B} \right) = AB\sin \theta \hat n - - - - - - - - - - - - - - - - - - - - (1)

And by Scalar product,
(A⋅B)=ABcos⁡θ\left( {A \cdot B} \right) = AB\cos \theta - - - - - - - - - - - - - - - - - - - - - - (2)
In question, the given quantities are
∣A×B∣=3A⋅B\left| {A \times B} \right| = \sqrt 3 A \cdot B- - - - - - - - - - - - - - - - - - - - - - -(3)
By using
∣A×B∣=∣A∣∣B∣sin⁡θ\left| {A \times B} \right| = \left| A \right|\left| B \right|\sin \theta - - - - - - - - - - - - - - - - - - - - - (4)
We get,
∣A×B∣=ABsin⁡θ\left| {A \times B} \right| = AB\sin \theta - - - - - - - - - - - - - - - - - - - - - - - (5)
Now,
∣A⋅B∣=∣A∣∣B∣cos⁡θ\left| {A \cdot B} \right| = \left| A \right|\left| B \right|\cos \theta
A⋅B=ABcos⁡θA \cdot B = AB\cos \theta - - - - - - - - - - - - - - - - - - - - - - - - - (6)
Substitute equation (6) in the right side of the equation (3) and equation (5) in the left side of the equation (3).
So,
ABsin⁡θ=3ABcos⁡θAB\sin \theta = \sqrt 3 AB\cos \theta
sin⁡θcos⁡θ=3\dfrac{{\sin \theta }}{{\cos \theta }} = \sqrt 3
\tan \theta $$$$ = 3\sqrt 3
⇒θ=60∘\Rightarrow \theta = 60^\circ
Now,
(A+B)2=A2+B2+2A.B{(A + B)^2} = {A^2} + {B^2} + 2A.B
(A+B)2=A2+B2+2ABcos⁡θ{(A + B)^2} = {A^2} + {B^2} + 2AB\cos \theta
(A+B)2=A2+B2+2AB⋅12{(A + B)^2} = {A^2} + {B^2} + 2AB \cdot \dfrac{1}{2}
(A+B)2=A2+B2+AB{(A + B)^2} = {A^2} + {B^2} + AB
Or,
A+B=(A2+B2+AB)12A + B = {\left( {{A^2} + {B^2} + AB} \right)^{\dfrac{1}{2}}}

Hence, Option (A) is correct

Note:
a. Substitution should not be reversed. If it reverses, all calculation will be wasted and time too.
b. Should be aware of the values of cos, sin and tan for 0, 30, 45, 60, and 90.0,{\text{ }}30,{\text{ }}45,{\text{ }}60,{\text{ }}and{\text{ }}90.
c. After getting the final calculation check the options and conveniently rearrange the solution.