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Wize High School Algebra II Textbook (Common Core) > Vectors (Extension Topic)

Length of a Vector (Vector Norm)

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Vector Norm

The norm of a vector v⃗\vec{v} is the length or magnitude of v⃗\vec{v}, and it is denoted: ∥v⃗∥\|\vec{v}\|

If v⃗=⟨v1, v2, v3,...,vn⟩∈Rn\vec{v}=\lang v_1,\ v_2,\ v_3, ..., v_n \rang \in \reals^n, then we calculate the norm of v⃗\vec v as:
∥v⃗∥=v12+v22+⋯+vn2\boxed{\quad \lVert \vec{v} \rVert =\sqrt{v_1^2+v_2^2+\cdots+v_n^2} \quad}

This looks like the Pythagorean theorem... because it is!

Example 1
Find the norm of the vector from the origin to the point A(3,2)∈R2A(3,2) \in \reals^2.

Then ∥A⃗∥=32+22=13\lVert \vec A \rVert = \sqrt{3^2 + 2^2} = \sqrt{13}


Properties of the Norm

  • ∥v⃗∥\|\vec{v}\| is a scalar
  • ∥v⃗∥\|\vec{v}\| is always non-negative (∥v⃗∥≥0)(\lVert \vec{v}\rVert \geq 0)
  • ∥v⃗∥=0\|\vec{v}\|=0 if and only if v⃗=0⃗\vec{v} = \vec 0

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Norm and Scalar Multiplication

For any vector v⃗\vec{v} and scalar kk:
∥kv⃗∥=∣k∣⋅∥v⃗∥\boxed{\quad \lVert k\vec{v}\rVert = \lvert k \rvert \cdot \lVert \vec{v} \rVert \quad}

Watch Out!
Since the norm is non-negative, we must take the absolute value of kk so that negative scalars become positive.


Example 2
Let u⃗=[13−2]\vec u = \begin{bmatrix} 1\\ 3\\ -2 \end{bmatrix} and let v⃗=−2u⃗\vec v = -2 \vec u . Find ∥v⃗∥\lVert \vec v \rVert.

Start by finding ∥u⃗∥=12+32+(−2)2=14\lVert \vec u \rVert = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{14}
Then
∥v⃗∥=∥−2u⃗∥=∣−2∣⋅∥u⃗∥=214\begin{aligned} \lVert \vec v \rVert &= \lVert -2 \vec u \rVert \\ &= \lvert-2 \rvert \cdot \lVert \vec u \rVert \\ &= 2\sqrt{14} \end{aligned}
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Distance Between Two Points

If PP and QQ are two points in Rn\mathbb{R}^n, then the distance between points PP and QQ is the norm of the vector PQ→\overrightarrow{PQ}:

d(P,Q)=∥PQ→∥d(P,Q)=\lVert \overrightarrow{PQ} \rVert

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Example: Vector Norm and Distance

Part A)

Find the norm of w⃗=⟨1,3,−1,1⟩∈R4\vec w=\lang 1,3,−1,1 \rang \in \reals^4.

Watch Out!
You may be asked to simplify radical expressions (square roots).

∥w⃗∥=12+32+(−1)2+12=1+9+1+1=12\lVert \vec w \rVert = \sqrt{1^2+3^2+(-1)^2+1^2} = \sqrt{1+9 +1+1} = \sqrt{12}

To simplify, think of a perfect square (e.g. 1, 4, 9, 16, ...) that can divide the number under the square root.
Since 4 is a perfect square that divides 12, we can write:
12=4×3=4×3=23\sqrt{12} = \sqrt{4\times3} = \sqrt{4} \times \sqrt{3} = 2\sqrt3
∴  ∥w⃗∥=23\therefore \ \ \boxed{\lVert \vec w \rVert = 2\sqrt3}
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Part B)

Find the distance between the vectors u⃗=[123]\vec u= \begin{bmatrix} 1\\ 2\\ 3\\ \end{bmatrix} and v⃗=[−3−20]\vec{v}= \begin{bmatrix} -3\\ -2\\ 0\\ \end{bmatrix}.

d(u⃗,v⃗)=(−3−1)2+(−2−2)2+(0−3)2=16+16+9=41\begin{aligned} d(\vec{u},\vec{v}) &=\sqrt{(-3-1)^2+(-2-2)^2+(0-3)^2}\\[0.5em] &=\sqrt{16+16+9}\\[0.5em] &=\boxed{\sqrt{41}} \end{aligned}

The answer is already in simplest form.

Practice: Vector Length

Given v⃗=[−3k]\vec{v}= \begin{bmatrix} -3\\ k\\ \end{bmatrix} and ∥v⃗∥=5\lVert \vec v \rVert = 5 , find all possible values of kk.

Practice: Properties of the Norm

Consider u⃗, v⃗, w⃗ ∈Rn\vec{u},\ \vec{v},\ \vec{w}\ \in \reals^n. Select all of the statements that are always true.

Practice: Distance Between Two Points/Vectors

Find the distance between the vectors u⃗=⟨5,0,−1⟩\vec u = \lang 5,0,−1 \rang and v⃗=⟨0,1,1⟩\vec{v}=\lang0,1,1\rang.