Question

Magnitude of vector a is equal to magnitude of vector b is equal to magnitude of their resultant and the value of of all is a then find the value of Vector (A + 2B).(2A-B)​

Answer 1

Answer:

$3( \frac{ {a}^{2} }{2} )$

Explanation:

Given

$| \vec{A}| = | \vec{B}| = | \vec{R}| = a$

To Find

$(\vec{A} + 2\vec{B}). (\vec{2A} +\vec{B})$

And

Ѳ must be 120° and

We know that

$\vec{A}.\vec{A} = {a}^{2} \\ \vec{B}.\vec{B} = {a}^{2} \: and \\\vec{A}. \vec{B} = a \times a \times \cos120 \degree \\\vec{A}. \vec{B} = a \times a \times - \frac{1}{2} \\ \vec{A}. \vec{B} = - \frac{ {a}^{2} }{2}$

Now by using appropriate method

$= > \vec{A}. (2\vec{A} - \vec{B}) + 2\vec{B}.(2\vec{A} - \vec{B})$

Now on Solving brackets

$= > 2(\vec{A}.\vec{A}) - (\vec{A}.\vec{B}) + 4(\vec{B}.\vec{A}) - 2(\vec{B}.\vec{B})$

$= > 2(\vec{A}.\vec{A}) = (\vec{A}.\vec{B}) - 4(\vec{B}.\vec{A}) + 2(\vec{B}.\vec{B})$

Finally on putting values of

$\vec{A}.\vec{A},\vec{B}.\vec{B} \: and \: \vec{A}.\vec{B}$

$= > - 2 {a}^{2} = \frac{ { -a}^{2} }{2} - (\frac{ {-4a}^{2} }{2}) - 2 {a}^{2} \\ = > \cancel{- 2 {a}^{2}} = \frac{ {3a}^{2} }{2} - \cancel{2 {a}^{2}} \\ = > 3( \frac{ {a}^{2} }{2} )$

Answer 2

Hey there

refer to attachment