Question

How do you find two unit vectors that make an angle of ${{60}^{\circ }}$ with $v=\left\langle 3,4 \right\rangle $?

Answer 1

Hint: We first have to assume the vectors in the form of $\overrightarrow{u}=\left( x,y \right)$. We find one variable equation from the modulus of the new vector which is1. Then using the angle of ${{60}^{\circ }}$ for two vectors we use the concept of dot product to find another equation. We solve them to find the solution.

Complete step by step solution:
Let us take $\overrightarrow{u}=\left( x,y \right)$ as the required unit vector.
Therefore, the modulus value will be $\left\| \overrightarrow{u} \right\|=\sqrt{{{x}^{2}}+{{y}^{2}}}=1$ which gives ${{x}^{2}}+{{y}^{2}}=1$.
It is given that the angle between $\overrightarrow{u}=\left( x,y \right)$ and $v=\left\langle 3,4 \right\rangle $ is ${{60}^{\circ }}$.
We take the vector dot product of these two vectors and get $\overrightarrow{u}.\overrightarrow{v}=\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\cos \left( 60 \right)$.
The modulus value for $v=\left\langle 3,4 \right\rangle $ will be $\left\| \overrightarrow{v} \right\|=\sqrt{{{3}^{2}}+{{4}^{2}}}=5$.
Putting the values, we get
$\begin{align}
  & \overrightarrow{u}.\overrightarrow{v}=\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\cos \left( 60 \right) \\
 & \Rightarrow \left( x,y \right).\left( 3,4 \right)=1\times 5\times \dfrac{1}{2} \\
\end{align}$
Simplifying we get
$\begin{align}
  & 3x+4y=\dfrac{5}{2} \\
 & \Rightarrow 6x+8y=5 \\
\end{align}$
We have two unknowns x and y and two equations ${{x}^{2}}+{{y}^{2}}=1,6x+8y=5$.
We need to solve them to find the value of the variables.
We can find the value of one variable $y$ with respect to $x$ based on the equation
$6x+8y=5$ where $y=\dfrac{5-6x}{8}$. We replace the value of $y$ in the first equation of
${{x}^{2}}+{{y}^{2}}=1$ and get
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}+{{\left( \dfrac{5-6x}{8} \right)}^{2}}=1 \\
 & \Rightarrow 64{{x}^{2}}+{{\left( 5-6x \right)}^{2}}=64 \\
\end{align}\]
We get the quadratic equation of $x$ and solve
$\begin{align}
  & 64{{x}^{2}}+{{\left( 5-6x \right)}^{2}}=64 \\
 & \Rightarrow 64{{x}^{2}}+36{{x}^{2}}+25-60x=64 \\
 & \Rightarrow 100{{x}^{2}}-60x-39=0 \\
\end{align}$
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $100{{x}^{2}}-60x-39=0$. The values of a, b, c is $100,-60,-39$ respectively.
We put the values and get x as $x=\dfrac{60\pm \sqrt{{{60}^{2}}-4\times 100\times \left( -39 \right)}}{2\times 100}=\dfrac{60\pm \sqrt{19200}}{200}=\dfrac{3\pm 4\sqrt{3}}{10}$.
Putting the value of $x$ we get two values of y as
$y=\dfrac{5-6x}{8}=\dfrac{5-6\left( \dfrac{3\pm 4\sqrt{3}}{10} \right)}{8}=\dfrac{4\mp 3\sqrt{3}}{10}$
Therefore, the vectors are $\overrightarrow{u}=\left( \dfrac{3+4\sqrt{3}}{10},\dfrac{4-3\sqrt{3}}{10} \right);\left( \dfrac{3-4\sqrt{3}}{10},\dfrac{4+3\sqrt{3}}{10} \right)$.

Note: We have to be careful about the choice of the points where we have got two values for the vectors and how the combination of those vectors work. The vector will be calculated with individual values for x.