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# How do you find the angle between the vectors $u = cos\left( {\dfrac{\pi }{3}} \right)i +sin\left( {\dfrac{\pi }{3}} \right)j\;and\;v = cos\left( {\dfrac{{3\pi }}{4}} \right)i + sin\left({\dfrac{{3\pi }}{4}} \right)j$?

Last updated date: 17th Sep 2024
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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
$\theta = \dfrac{{7\pi }}{{12}}$ .

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 .