Unit Vector in the Direction of a Given Vector

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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 $\vec {u}=\begin {bmatrix}a/3\\b/3\end {bmatrix}=\begin {bmatrix}a/\norm {\vec {v}}\\b/\norm {\vec {v}}\end {bmatrix}$ .

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 $\vec {v}$ .

Proof: Because $\vec {u}$ is a positive scalar multiple of $\vec {v}$ , $\vec {u}$ points in the direction of $\vec {v}$ . We now show that $\norm {\vec {u}}=1$ .
$\begin{eqnarray*} \norm{\vec{u}}&=&\sqrt{ \Big(\frac{v_1}{\norm{\vec{v}}}\Big)^2+\Big(\frac{v_2}{\norm{\vec{v}}}\Big)^2+\ldots +\Big(\frac{v_n}{\norm{\vec{v}}}\Big)^2}\\ &=&\frac{1}{\norm{\vec{v}}}\sqrt{v_1^2+v_2^2+\ldots +v_n^2}\\ &=&\frac{\norm{\vec{v}}}{\norm{\vec{v}}}=1 \end{eqnarray*}$
$\blacksquare$

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 $\vec {v}$ .

$\vec {v}=\begin {bmatrix}-3\\4\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {-0.6}\\\answer {0.8}\end {bmatrix}$

$\vec {v}=\begin {bmatrix}2\\-3\\6\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {2/7}\\\answer {-3/7}\\\answer {6/7}\end {bmatrix}$

$\vec {v}=\begin {bmatrix}1\\3\\-2\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {1/\sqrt {14}}\\\answer {3/\sqrt {14}}\\\answer {-2/\sqrt {14}}\end {bmatrix}$

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.

Answer: $\vec {w}=\begin {bmatrix} \answer {-5/3}\\\answer {5/3}\\\answer {5\sqrt {7}/3} \end {bmatrix}$