Question

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

Answer 1

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 .