\[\vec {v}=\begin{bmatrix}-3\\4\end{bmatrix}\]
Answer: \(\vec {u}=\begin{bmatrix}\answer {-0.6}\\\answer {0.8}\end{bmatrix}\)
Unit Vector in the Direction of a Given Vector
Recall that a unit vector is a vector of length 1. Given a non-zero vector \(\vec {v}\), we can find a unit vector in the same direction by multiplying \(\vec {v}\) by an appropriate scalar. For example, if \(\vec {v}=\begin{bmatrix}a\\b\end{bmatrix}\) and \(\norm {\vec {v}}=3\), then a unit vector \(\vec {u}\) in the same direction is given by \(\vec {u}=\begin{bmatrix}a/3\\b/3\end{bmatrix}=\begin{bmatrix}a/\norm {\vec {v}}\\b/\norm {\vec {v}}\end{bmatrix}\).
![[Picture]](main.online-7840b1178771ad8aceaee403584f06bd.svg)
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In general, dividing a non-zero vector by its own magnitude produces a unit vector in the same direction. We summarize this observation in a theorem.
Let \(\vec {v}=\begin{bmatrix}v_1\\v_2\\\vdots \\v_n\end{bmatrix}\) be a non-zero vector in \(\mathbb {R}^n\). Vector \(\vec {u}\) given by
\begin{equation*} \vec {u}=\frac {1}{\norm {\vec {v}}}\vec {v}=\begin{bmatrix}v_1/\norm {\vec {v}}\\v_2/\norm {\vec {v}}\\\vdots \\v_n/\norm {\vec {v}}\end{bmatrix} \end{equation*}
is a unit vector in the direction of \(\vec {v}\).
Because \(\vec {u}\) is a positive scalar multiple of \(\vec {v}\), \(\vec {u}\) points in the direction of \(\vec {v}\). We now show that \(\norm {\vec {u}}=1\).
\begin{eqnarray*} \norm {\vec {u}}&=&\sqrt { \Big (\frac {v_1}{\norm {\vec {v}}}\Big )^2+\Big (\frac {v_2}{\norm {\vec {v}}}\Big )^2+\ldots +\Big (\frac {v_n}{\norm {\vec {v}}}\Big )^2}\\ &=&\frac {1}{\norm {\vec {v}}}\sqrt {v_1^2+v_2^2+\ldots +v_n^2}\\ &=&\frac {\norm {\vec {v}}}{\norm {\vec {v}}}=1 \end{eqnarray*}
Find a unit vector in the direction of \(\vec {v}=\begin{bmatrix}2\\-3\\1\\0\\1\end{bmatrix}\).
We first compute \(\norm {\vec {v}}\).
\[\norm {\vec {v}}=\sqrt {4+9+1+1}=\sqrt {15}\]
By Theorem th:unit,
\[\vec {u}=\begin{bmatrix}2/\sqrt {15}\\-3/\sqrt {15}\\1/\sqrt {15}\\0\\1/\sqrt {15}\end{bmatrix}=\frac {1}{\sqrt {15}}\begin{bmatrix}2\\-3\\1\\0\\1\end{bmatrix}\]
Practice Problems
Find a unit vector in the direction of the given vector \(\vec {v}\).
Find a unit vector in the direction of the given vector \(\vec {v}\).
\[\vec {v}=\begin{bmatrix}2\\-3\\6\end{bmatrix}\]
Enter exact answers. No decimal approximations.
Answer: \(\vec {u}=\begin{bmatrix}\answer {2/7}\\\answer {-3/7}\\\answer {6/7}\end{bmatrix}\)
Let \(\vec {v}=\begin{bmatrix}-1\\1\\\sqrt {7}\end{bmatrix}\). Apply the concepts from this section to find a vector \(\vec {w}\) that points in the same
direction as \(\vec {v}\) and whose length is 5.
Enter exact answers. No decimal approximations.
Answer: \(\vec {w}=\begin{bmatrix} \answer {-5/3}\\\answer {5/3}\\\answer {5\sqrt {7}/3} \end{bmatrix}\)