# Length of a Vector – Definition, Formulas, and Examples

The length of a vector allows us to understand how large the vector is in terms of dimensions. This also helps us understand vector quantities such as displacement, velocity, force, and more. Understanding the formula for calculating the length of a vector will help us in establishing the formula for the arc length of a vector function.

The length of a vector (commonly known as the magnitude) allows us to quantify the property of a given vector. To find the length of a vector, simply add the square of its components then take the square root of the result.

In this article, we’ll extend our understanding of magnitude to vectors in three dimensions. We’ll also cover the formula for the arc length of the vector function. By the end of our discussion, our goal is for you to confidently work on different problems involving vectors and vector functions’ lengths.

## What Is the Length of a Vector?

The length of the vector represents the distance of the vector in the standard position from the origin. In our previous discussion on vector properties, we’ve learned that the length of a vector is also known as the magnitude of the vector.

 Suppose that $\textbf{u} = x \textbf{i}+y \textbf{j}$, we can calculate the length of the vector using the formula for magnitudes as shown below: \begin{aligned}|\textbf{u}| = \sqrt{x^2 +y^2}\end{aligned} We can extend this formula for vectors with three components -$\textbf{u} = x \textbf{i}+ y \textbf{j} + z\textbf{k}$ : \begin{aligned}|\textbf{v}| = \sqrt{x^2 +y^2 + z^2}\end{aligned}

In fact, we can extend our understanding of three-coordinate systems and vectors to prove the formula for the vector length in space.

### Proof of Vector Length Formula in 3D

Suppose that we have a vector, $\textbf{u} = x_o \textbf{i} + y_o \textbf{j} +z_o \textbf{k}$, we can rewrite the vector as the sum of two vectors. Hence, we have the following:

\begin{aligned}\textbf{v}_1 &= \\ \textbf{v}_2 &= <0 , 0, z_o>\\\textbf{u} &= <x_o,y_o ,z_o>\\&= +<0 ,0, z_o>\\&=\textbf{v}_1+ \textbf{v}_2\end{aligned}

We can calculate the lengths of the two vectors, $\textbf{v}_1$ and $\textbf{v}_2$, by applying what we know of magnitudes.

\begin{aligned}|\textbf{v}_1| &= \sqrt{x_o^2 +y_o^2}\\ |\textbf{v}_2| &= \sqrt{z_o^2}\end{aligned}

These vectors will form a right triangle with $\textbf{u}$ as the hypotenuse, so we can use Pythagorean theorem to calculate the length of the vector, $\textbf{u}$.

\begin{aligned}|\textbf{u}| &= \sqrt{|\textbf{v}_1|^2 +|\textbf{v}_2|^2}\\&= \sqrt{(x_o^2 + y_o^2) + z_o^2}\\&= \sqrt{x_o^2 +y_o^2 +z_o^2}\end{aligned}

This means that for us to calculate the length of the vector in three dimensions, all we have to do is add the squares of its components then take the square root of the result.

### Arc Length of a Vector Function

We can extend this notion of length to vector functions – this time, we’re approximating the distance of vector function over an interval of $t$.  The length of the vector function, $\textbf{r}(t)$, within the interval of $[a, b]$ can be calculated using the formula shown below.

 \begin{aligned}\textbf{r}(t) &= \left\\\text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2]}\phantom{x}dt\\\\\textbf{r}(t) &= \left\\\text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] + [z\prime(t)]^2]}\phantom{x}dt\end{aligned}

From this, we can see that the arc length of the vector function is simply equal to the magnitude of the vector tangent to $\textbf{r}(t)$. This means that we can simplify our arc length’s formula to the equation shown below:

\begin{aligned}L &= \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x} dt\end{aligned}

We’ve now covered all the fundamental definition of vector lengths and vector function lengths, it’s time for us to apply them to calculate their values.

## How To Calculate the Length of a Vector and a Vector Function?

We can calculate the length of a vector by applying the formula for the magnitude. Here’s a breakdown of the steps to calculate the vector’s length:

• List down the components of the vector then take their squares.
• Add the squares of these components.
• Take the square root of the sum to return the length of the vector.

This means that we can calculate the length of the vector, $\textbf{u} = \left<2, 4, -1\right>$, by applying the formula, $|\textbf{u}| = \sqrt{x^2 + y^2 + z^2}$, where $\{x, y, z\}$ represents the components of the vector.

\begin{aligned}|\textbf{u}| &= \sqrt{x^2 + y^2 + z^2}\\ &= \sqrt{(2)^2 + (4)^2 + (-1)^2}\\&=\sqrt{4 + 16 + 1}\\&= \sqrt{21}\end{aligned}

Hence, the length of the vector, $\textbf{u}$, is equal to $\sqrt{21}$ units or approximately equal to $4.58$ units.

As we have shown in our earlier discussion, the arc length of the vector function depends on the tangent vector. Here’s a guideline to help you in calculating the arc length of the vector function:

• List down the components of the vector then take their squares.
• Square each of the derivatives then add the expressions.
• Write the square root of the resulting expression.
• Evaluate the integral of the expression from $t = a$ to $t = b$.

Let’s say we have the vector function, $\textbf{r}(t) = \left<(4t – 1), (2t +4)\right>$. We can calculate its arc length by from $t = 0$ to $t = 4$ using the formula, $L = \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x} dt$, where $\textbf{r}\prime(t)$ represents the tangent vector.

This means that we’ll need to find $\textbf{r}\prime(t)$ by differentiating each of the vector function’s component.

 \begin{aligned}x \prime(t)\end{aligned} \begin{aligned}x\prime(t) &= \dfrac{d}{dt} (4t –1)\\&= 4(1) – 0\\&= 4\end{aligned} \begin{aligned}y \prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt} (2t +4)\\&= 2(1) – 0\\&= 2\end{aligned} \begin{aligned}\textbf{r}\prime(t) &= \left\\&= \left<4, 2\right>\end{aligned}

Take the magnitude of the tangent vector by squaring the components of the tangent vector then writing down the square root of the sum.

\begin{aligned}|\textbf{r}\prime(t)| &= \sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] }\\&= \sqrt{4^2 + 2^2} \\&= \sqrt{20}\end{aligned}

Now, evaluate the integral of the resulting expression from $t = 0$ to $t = 4$.

\begin{aligned}\int_{0}^{4} \sqrt{20} \phantom{x}dt &=\int_{0}^{4} 2\sqrt{5} \phantom{x}dt\\&= 2\sqrt{5}\int_{0}^{4}  \phantom{x}dt\\&= 2\sqrt{5} [t]_0^4\\&= 2\sqrt{5}( 4 -0)\\&= 8\sqrt{5}\end{aligned}

This means that the arc length of $\textbf{r}(t)$ from $t=0$ to $t=4$ is equal to $8\sqrt{5}$ units or approximately $17.89$ units.

These are two great examples of how we can apply the formulas for vector and vector function lengths. We’ve prepared some more problems for you to try, so head over to the next section when you’re ready!

Example 1

The vector $\textbf{u}$ has an initial point at $P(-2, 0, 1 )$ and an endpoint at $Q(4, -2, 3)$. What is the vector’s length?

Solution

We can find the position vector by subtracting the components of $P$ from the components of $Q$ as shown below.

\begin{aligned}\textbf{u} &= \overrightarrow{PQ}\\&= \left<(4 – -2), (-2 – 0), (3 -1)\right>\\&= \left<6, -2, 2\right>\end{aligned}

Use the formula for the vector’s magnitude to calculate the length of $\textbf{u}$.

\begin{aligned}|\textbf{u}| &= \sqrt{(6)^2 + (-2)^2 + (2)^2}\\&= \sqrt{36+ 4+ 4}\\&= \sqrt{44}\\&= 2\sqrt{11}\\&\approx 6.63 \end{aligned}

This means that the vector, $\textbf{u}$, has a length of $2\sqrt{11}$ units or approximately $6.33$ units.

Example 2

Calculate the arc length of the vector-valued function, $\textbf{r}(t) = \left<2\cos t, 2\sin t, 4t\right>$, if $t$ is within the interval, $t \in [0, 2\pi]$.

Solution

We’re now looking for the arc length of the vector function, so we’ll use the formula shown below.

\begin{aligned} \text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] + [z\prime(t)]^2]}\phantom{x}dt\\&= \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x}dt\end{aligned}

First, let’s take the derivative of each components to find $\textbf{r}\prime(t)$.

 \begin{aligned}x\prime(t)\end{aligned} \begin{aligned}x\prime(t) &= \dfrac{d}{dt}(2 \cos t)\\&= 2(-\sin t)\\&= -2\sin t \end{aligned} \begin{aligned}y \prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt}(2 \sin t)\\&= 2(\cos t)\\&= 2\cos t\end{aligned} \begin{aligned}z\prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt}(2 4t)\\&= 4(1)\\&= 4\end{aligned} \begin{aligned}\textbf{r}\prime(t) &= \left\\&= \left<-2\sin t, 2\cos t, 4\right>\end{aligned}

Now, take the magnitude of $\textbf{r}\prime(t)$ by adding the squares of the tangent vector’s components. Write the square root of the sum to express the magnitude in terms of $t$.

\begin{aligned}|\textbf{r}\prime(t)| &= \sqrt{(-2 \cos t)^2  + (4\sin t)^2 + 4^2}\\&= \sqrt{4 \cos^2 t + 4\sin^2 t + 16}\\&= \sqrt{4(\cos^2 t + \sin^2 t) + 16}\\&= \sqrt{4(1) + 16}\\&= \sqrt{20}\\&= 2\sqrt{5}\end{aligned}

Integrate $|\textbf{r}\prime(t)|$ from $t = 0$ to $t = 2\pi$ to find the arc length of the vector.

\begin{aligned} \text{Arc Length} &= \int_{a}^{b}|\textbf{r}\prime(t)| \phantom{x}dt\\&= \int_{0}^{2\pi} 2\sqrt{5} \phantom{x}dt\\&= 2\sqrt{5}\int_{0}^{2\pi} \phantom{x}dt\\&= 2\sqrt{5}(2\pi – 0)\\&= 4\sqrt{5}\pi\\&\approx 28.10\end{aligned}

This means that the arc length of the vector function is $4\sqrt{5}\pi$ or approximately $28.10$ units.

### Practice Questions

1. The vector $\textbf{u}$ has an initial point at $P(-4, 2, -2 )$ and an endpoint at $Q(-1, 3, 1)$. What is the vector’s length?

2. Calculate the arc length of the vector-valued function, $\textbf{r}(t) = \left<t\cos t, t\sin t, 2t\right>$, if $t$ is within the interval, $t \in [0, 2\pi]$.

1. The vector has a length of $\sqrt{19}$ units or approximately $4.36$ units.
2. The arc length is approximately equal to $25.343$ units.