AcademyVectors In Rn
Academy
Norms And Distances
Level 1 - Linear Algebra topic page in Vectors In Rn.
Principle
The norm of a vector is its length. The distance between two vectors is the length of their difference. In \(\mathbb R^n\), both are computed from the scalar product and generalise the usual Pythagoras theorem.
Notation
\(|\mathbf v|\)
the norm or length of \mathbf v
\(\mathbf v\cdot\mathbf v\)
the scalar product of \mathbf v with itself
\(d(\mathbf u,\mathbf v)\)
the distance between vectors \mathbf u and \mathbf v
\(v_i\)
component i of \mathbf v
\(\hat{\mathbf v}\)
a unit vector in the direction of non-zero \mathbf v
The Core Method
The norm is the square root of the sum of squared components:
Norm
\[|\mathbf v|=\sqrt{v_1^2+v_2^2+\cdots+v_n^2}\]
Equivalently,
Norm from scalar product
\[|\mathbf v|^2=\mathbf v\cdot\mathbf v\]
The distance between \(\mathbf u\) and \(\mathbf v\) is
Distance
\[d(\mathbf u,\mathbf v)=|\mathbf u-\mathbf v|\]
To turn a non-zero vector into a unit vector in the same direction, divide by its norm:
Unit vector
\[\hat{\mathbf v}=\frac{1}{|\mathbf v|}\mathbf v\]
Do not divide by \(|\mathbf v|\) when \(\mathbf v=\mathbf 0\), because \(|\mathbf 0|=0\).
Worked Cases
Question
Find the norm of \(\mathbf v=(2,-3,6)\).
Answer
Use \(|\mathbf v|=\sqrt{v_1^2+v_2^2+v_3^2}\). Substitute the components: \(|\mathbf v|=\sqrt{2^2+(-3)^2+6^2}\). Square each component: \(2^2=4\), \((-3)^2=9\), and \(6^2=36\). Add them: \(4+9+36=49\). Therefore \(|\mathbf v|=\sqrt{49}=7\).
Examples
Question
Find a unit vector in the direction of \(\mathbf v=(3,4)\).
Answer
First find the norm: \(|\mathbf v|=\sqrt{3^2+4^2}=\sqrt{25}=5\). Divide by the norm: \(\hat{\mathbf v}=\frac{1}{5}(3,4)=(3/5,4/5)\). Its norm is \(\sqrt{(3/5)^2+(4/5)^2}=\sqrt{9/25+16/25}=1\).