Question

How do you find a vector of length 1 in the same direction as $\left( 6,-8 \right)$ ?

Answer 1

Hint: To find a vector of length 1 in the same direction as $\left( 6,-8 \right)$ , that is, the unit vector, let us consider the given vector to be \[\overrightarrow{v}=\left( 6,-8 \right)\] . To find the unit vector in the same direction as \[\overrightarrow{v}\] , we will divide \[\overrightarrow{v}\] by its magnitude. we can find the magnitude of \[\overrightarrow{v}\] by taking the square root of the sum of squares of x and y coordinates. Let us denote the unit vector as $\overrightarrow{u}=\dfrac{\overrightarrow{v}}{\left| \overrightarrow{v} \right|}$ . By substituting the values and solving, we will get the unit vector.

Complete step-by-step solution:
We need to find a vector of length 1 in the same direction as $\left( 6,-8 \right)$ . Vector of length 1 means we have to find the unit vector. Let us consider the given vector to be \[\overrightarrow{v}=\left( 6,-8 \right)\] . To find the unit vector in the same direction as \[\overrightarrow{v}\] , we will divide \[\overrightarrow{v}\] by its magnitude.
So first, we have to find the magnitude of \[\overrightarrow{v}\] .We will denote this as $\left| \overrightarrow{v} \right|$ . We know that for a vector $\overrightarrow{x}=\left( a,b \right)$ , its magnitude is given as $\left| \overrightarrow{x} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ . Therefore, we can find the magnitude of \[\overrightarrow{v}\] as
$\begin{align}
  & \left| \overrightarrow{v} \right|=\sqrt{{{6}^{2}}+{{\left( -8 \right)}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{v} \right|=\sqrt{36+64}=\sqrt{100}=10 \\
\end{align}$
Now, let us find the unit vector in the direction of $\left( 6,-8 \right)$ . We will denote the unit vector as $\overrightarrow{u}$ .
$\overrightarrow{u}=\dfrac{\overrightarrow{v}}{\left| \overrightarrow{v} \right|}$
Let us substitute the values in the above formula.
\[\begin{align}
  & \overrightarrow{u}=\dfrac{\left( 6,-8 \right)}{10} \\
 & \Rightarrow \overrightarrow{u}=\left( \dfrac{6}{10},-\dfrac{8}{10} \right) \\
 & \Rightarrow \overrightarrow{u}=\left( \dfrac{3}{5},-\dfrac{4}{5} \right) \\
\end{align}\]
Hence, the answer is \[\left( \dfrac{3}{5},-\dfrac{4}{5} \right)\].

Note: We can also check whether the vector in the same direction as $\left( 6,-8 \right)$ is of length 1. We will have to take the magnitude of the unit vector. We know that all unit vectors will have magnitude 1.
$\Rightarrow \left| \overrightarrow{u} \right|=\sqrt{{{\left( \dfrac{3}{5} \right)}^{2}}+{{\left( -\dfrac{4}{5} \right)}^{2}}}$
Let us take the squares and simplify.
$\left| \overrightarrow{u} \right|=\sqrt{\dfrac{9}{25}+\dfrac{16}{25}}=\sqrt{\dfrac{25}{25}}=1$
Hence, we have verified the answer.