The Norm of a Vector

# The Norm of a Vector

 Definition: If $\vec{u} \in \mathbb{R}^n$, then the Norm or Magnitude of $\vec{u}$ denoted $\| \vec{u} \|$ is defined as the length or magnitude of the vector and can be calculated using the formula: $\| \vec{u} \| = \sqrt{u_1^2 + u_2^2 + ... + u_n^2}$.

We will note that the norm of a vector is sometimes denoted with single bars, that is $\mid \vec{u} \mid$ is a notation commonly used to denote what we have defined. We will not use this notation to prevent confusion with mistaking the norm of a vector and the absolute value of a scalar.

## Example 1

Calculate the norm of the vector $\vec{u} = (3, 4)$.

We first note that $\vec{u} \in \mathbb{R}^2$, and we will thus use the formula $\| \vec{u} \| = \sqrt{u_1^2 + u_2^2}$. In fact, this formula should make sense geometrically as it analogous to the Pythagorean theorem as illustrated: When we substitute our values in, we obtain that $\| \vec{u} \| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Thus our vector $\vec{u}$ has length 5.

## Example 2

Find the norm of the vector $\vec{u} = (2, -2, 3, -4)$.

Since $\vec{u} \in \mathbb{R}^4$, we will use the formula $\| \vec{u} \| = \sqrt{u_1^2 + u_2^2 + u_3^2 + u_4^2} = \sqrt{4 + 4 + 9 + 16} = \sqrt{33}$. So the norm of our vector $\vec{u}$ is the square root of 33.

# The Distance Between Two Points

Remember, we can write a vector that starts at some initial point $P$, and some terminal point $Q$. Recall that we can calculate this vector in 3-space with the formula $\vec{PQ} = (x_{Q} - x_{P}, y_{Q} - y_{P}, z_{Q} - z_{P})$. We can thus apply this formula to obtain the norm of $\vec{PQ}$:

(1)
\begin{align} \| \vec{PQ} \| = \sqrt{ (x_{Q} - x_{P})^2 + (y_{Q} - y_{P})^2 + (z_{Q} - z_{P})^2 } \end{align}

We note that the norm of $PQ$ is also equal to the distance between $P$ and $Q$, so an alternative way to write this formula is:

(2)
\begin{align} d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \end{align}

Where $d$ denotes the distance between the initial point $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$.

## Example 3

Find the distance between the point $P(2, 3, 4)$, and the point $Q(5, 4, 3)$:

Applying the formula we just established we obtain that:

(3)
\begin{align} d = \sqrt{ (5 - 2)^2 + (4 - 3)^2 + (3 - 2)^2 } \\ d = \sqrt{ 3^2 + 1^2 + 1^2 }\\ d = \sqrt{ 11 } \end{align}

Therefore, the distance between point $P$ and $Q$ is $\sqrt{11}$.

# The Norm of a Scalar Multiple of a Vector

Suppose that we have a vector $\vec{u}$. If we multiply this vector by a scalar $k$, then the norm of the vector $k\vec{u}$ will be k-times larger than $k$. There is a problem though. We define the norm to be the magnitude or length of the vector so the norm must be positive. Hence we draw the following relation:

(4)
\begin{align} \| k \vec{u} \| = \mid k \mid \| \vec{u} \| \end{align}

We note that $\mid k \mid$ represents the absolute value of k, that is:

(5)
\begin{align} \mid k \mid = \left\{\begin{matrix} k & \mathrm{if \: k >0}\\ -k & \mathrm{if \: k < 0} \end{matrix}\right. \end{align}

Thus if $k$ is negative, then we take the positive of $k$, otherwise $k$ stays the same. For example, $\mid -3 \mid = 3$. Note that sometimes the notation $\mathrm{abs} (k)$ will be used to denote the absolute value.

From this we can draw a relationship between a scalar multiple of a vector and its norm, that is:

(6)
\begin{align} \| k \vec{u} \| = \mid k \mid \| \vec{u} \| \end{align}

## Example 4

Given that $\| \vec{u} \| = 4$, find $\| -3\vec{u} \|$.

From the formula given, we note that $\| -3 \vec{u} \| = \mid -3 \mid \| \vec{u} \| = (3)(4) = 12$.