Question

If \[\left| {\vec A \times \vec B} \right| = \sqrt 3 \vec A.\vec B\] then the value of \[\left| {\vec A + \vec B} \right|\] is
A. \[{\left[ {{A^2} + {B^2} + \dfrac{{AB}}{{\sqrt 3 }}} \right]^{\dfrac{1}{2}}}\]
B. \[A + B\]
C. \[{\left[ {{A^2} + {B^2} + \sqrt 3 AB} \right]^{\dfrac{1}{2}}}\]
D. \[{\left[ {{A^2} + {B^2} + AB} \right]^{\dfrac{1}{2}}}\]

Answer 1

Hint:From the given data we will find the value of cross product and dot product of the vectors. Then we will equate them as per the given condition. Then we will get the value for the angle between them. After that we will use the formula for the square of the addition of the vectors and then from that we will get the exact answer.

Complete step by step answer:
Given that,
\[\left| {\vec A \times \vec B} \right| = \sqrt 3 \vec A.\vec B\]
We know that, \[\left| {\vec A \times \vec B} \right| = \left| {\vec A} \right|\left| {\vec B} \right|\sin \theta \]
And \[\left| {\vec A.\vec B} \right| = \left| {\vec A} \right|\left| {\vec B} \right|\cos \theta \]
On equating this in the given equation,
\[\left| {\vec A} \right|\left| {\vec B} \right|\sin \theta = \sqrt 3 \left| {\vec A} \right|\left| {\vec B} \right|\cos \theta \]
On cancelling the modulus from both the sides,
\[\sin \theta = \sqrt 3 \cos \theta \]
Shifting the trigonometric functions on one side,
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \sqrt 3 \]
We know that ratio of sin and cos is tan,
\[\tan \theta = \sqrt 3 \]
Thus the angle between them can be obtained as,
\[\theta = {\tan ^{ - 1}}\sqrt 3 \]
Thus the angle will be, \[\theta = \dfrac{\pi }{3}or\theta = \dfrac{{4\pi }}{3}\]
Now as we know that,
\[{\left| {\vec A + \vec B} \right|^2} = {\left| {\vec A} \right|^2} + {\left| {\vec B} \right|^2} + 2\left| {\vec A} \right|\left| {\vec B} \right|\cos \theta \]
Putting the value of the angle we get,
\[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\,or\cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}\]
Then,
\[{\left| {\vec A + \vec B} \right|^2} = {\left| {\vec A} \right|^2} + {\left| {\vec B} \right|^2} + 2\left| {\vec A} \right|\left| {\vec B} \right|\left( { \pm \dfrac{1}{2}} \right)\]
Cancelling the 2,
\[{\left| {\vec A + \vec B} \right|^2} = {\left| {\vec A} \right|^2} + {\left| {\vec B} \right|^2} \pm \left| {\vec A} \right|\left| {\vec B} \right|\]
Taking square root,
\[\left| {\vec A + \vec B} \right| = \sqrt {{{\left| {\vec A} \right|}^2} + {{\left| {\vec B} \right|}^2} \pm \left| {\vec A} \right|\left| {\vec B} \right|} \]
Square root can be written as,
\[\left| {\vec A + \vec B} \right| = {\left( {{{\left| {\vec A} \right|}^2} + {{\left| {\vec B} \right|}^2} \pm \left| {\vec A} \right|\left| {\vec B} \right|} \right)^{\dfrac{1}{2}}}\]
This is the correct answer.
But from options,
\[\left| {\vec A + \vec B} \right| = {\left( {{{\left| {\vec A} \right|}^2} + {{\left| {\vec B} \right|}^2} + \left| {\vec A} \right|\left| {\vec B} \right|} \right)^{\dfrac{1}{2}}}\]

Thus option D is the correct answer.

Note:Note that dot product and cross product of vectors are two different concepts. But both are related to vectors. As they are both used in the equation given we will elaborate them only. We chose option 4 as the answer because it has + sign in it. If the option with – sign is available then we will choose that also.