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**Hint:**First of all we need to understand what vectors are . Vectors can be said as physical quantities or objects which have both a magnitude and a direction . If the two vectors are supposed to be $\overrightarrow a $and $\overrightarrow b $. The ‘$\theta $’ is the angle by which the two vectors are separated . Now , to determine that what is the angle between the two vectors we are going to apply the dot product between those two vectors denoted as $\overrightarrow a

.\overrightarrow b $and the dot product is given as $\overrightarrow {a.} \overrightarrow b =

|a||b|\cos \theta $.

**Step by step solution :**

The angle $\theta $ between two vectors $\overrightarrow u $and $\overrightarrow v $as per the question given is related to the modulus ( or magnitude ) and scalar ( or dot ) product of

$\overrightarrow u $and $\overrightarrow u $by the relationship : $\overrightarrow {u.}

\overrightarrow v = |u||v|\cos \theta $

For the question above , The angle between the two vectors $\overrightarrow u $ and

$\overrightarrow v $ will be $\theta $ .

Calculating and simplifying the vectors ,

First , assigning the trigonometric values as the functions given of cosine and sine .

$\overrightarrow u = cos\left( {\dfrac{\pi }{3}} \right)\mathop i\limits^ \wedge + sin\left(

{\dfrac{\pi }{3}} \right)\mathop j\limits^ \wedge \; = \dfrac{1}{2}\mathop i\limits^ \wedge +

\dfrac{{\sqrt 3 }}{2}\mathop j\limits^ \wedge $------- - 1

$\overrightarrow v = cos\left( {\dfrac{{3\pi }}{4}} \right)\mathop i\limits^ \wedge + sin\left(

{\dfrac{{3\pi }}{4}} \right)\mathop j\limits^ \wedge \; = - \dfrac{{\sqrt 2 }}{2}\mathop i\limits^

\wedge + \dfrac{{\sqrt 2 }}{2}\mathop j\limits^ \wedge $----- - 2

Now we will calculate the modulus of $|\overrightarrow u |$= $\left| {\dfrac{1}{2}\mathop

i\limits^ \wedge + \left. {\dfrac{{\sqrt 3 }}{2}\mathop j\limits^ \wedge } \right|} \right.$=

\[\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \]= $\sqrt

{\dfrac{1}{4} + \dfrac{3}{4}} $=$\sqrt 1 $=$1$

$|\overrightarrow v |$= $\left| { - \dfrac{{\sqrt 2 }}{2}\mathop i\limits^ \wedge + \left.

{\dfrac{{\sqrt 2 }}{2}\mathop j\limits^ \wedge } \right|} \right.$= \[\sqrt {{{\left( { - \dfrac{{\sqrt

2 }}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{2}} \right)}^2}} \]= $\sqrt {\dfrac{2}{4} +

\dfrac{2}{4}} $=$\sqrt 1 $=$1$

And now we perform the scalar product :

$\overrightarrow {u.} \overrightarrow v $ = $\left( {\dfrac{1}{2}\mathop i\limits^ \wedge +

\dfrac{{\sqrt 3 }}{2}\mathop j\limits^ \wedge } \right)$. $\left( { - \dfrac{{\sqrt 2 }}{2}\mathop

i\limits^ \wedge + \dfrac{{\sqrt 2 }}{2}\mathop j\limits^ \wedge } \right)$

= $\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{{\sqrt 2 }}{2}} \right) + \left( {\dfrac{{\sqrt 3 }}{2}}

\right)\left( {\dfrac{{\sqrt 2 }}{2}} \right)$

= $

- \dfrac{{\sqrt 2 }}{4} + \dfrac{{\sqrt 2 \sqrt 3 }}{4} \\

\\

$

=$\dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{4}$

Now applying the formula $\overrightarrow {u.} \overrightarrow v = |u||v|\cos \theta $ we get :

We will substitute the values after getting each values of L . H . S . and R . H . S . we calculated before in the above formula to get the angle .

$\dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{4}$= $1.1.\cos \theta $

$\cos \theta $= $\dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{4}$

$\theta = \dfrac{{7\pi }}{{12}}$

**Therefore the angle between the two vectors $\overrightarrow u $ and $\overrightarrow v $ is**

$\theta = \dfrac{{7\pi }}{{12}}$ .

$\theta = \dfrac{{7\pi }}{{12}}$ .

**Note :**The vectors are the objects ( physical quantity ) in real life that are having magnitude and direction . For instance , Force and Velocity .

The modulus actually means $\left| {\overrightarrow r } \right|$= $\sqrt {{a^2} + {b^2}} $

Always remember by scalar product we refer to dot product .

For above question we calculated the L . H . S . and R . H . S . independently using the formula $\overrightarrow {a.} \overrightarrow b = |a||b|\cos \theta $. And then combined and substituted the calculated values .

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