Vectors: Distance, angle, dot product and cross product
Length and distance
Length in 2D and 3D
We denote the length of a vector \(\vec{v}=\cv{v_1\\ v_2}\) in the coordinate plane \(\mathbb{R}^2\), more precisely the Euclidean length, by \( \lVert\vec{v}\rVert\). It can be calculated via the Pythagorean theorem: \[ \lVert\vec{v}\rVert=\sqrt{v_1^2+v_2^2}\]
When we deal with a vector \(\vec{v}=\cv{v_1\\ v_2\\ v_3}\) in \(\mathbb{R}^3\), the notion of length becomes \[ \lVert\vec{v}\rVert=\sqrt{v_1^2+v_2^2+v_3^3}\]
Examples
\[\begin{aligned}\left\lVert\cv{2\\ -3}\right\rVert &=\sqrt{2^2+(-3)^2} \\ &= \sqrt{4+9}=\sqrt{13} \\ \\ \left\lVert\cv{2\\ 3\\ -z}\right\rVert &=\sqrt{2^2+3^2+(-z)^2} \\ &= \sqrt{4+9+z^2}=\sqrt{13+z^2}\end{aligned}\]
This can be generalized to the \(n\)-dimensional coordinate space \(\mathbb{R}^n\).
Length The Euclidean length of a vector \(\vec{v}=\cv{v_1\\ \vdots \\ v_n}\) in the \(n\)-dimensional space \(\mathbb{R}^n\), also called the (Euclidean) norm, we denote by \(\lVert\vec{v}\rVert\). It can be calculated as follows: \[ \lVert\vec{v}\rVert=\sqrt{v_1^2+\cdots + v_n^2}=\sqrt{\sum_{i=1}^n v_i^2}\]
Distance The distance between two vectors \(\vec{u}\) and \(\vec{v}\) is noted as \(d(\vec{u}, \vec{v})\). This is by definition the length of the difference vector \(\vec{u}-\vec{v}\), so \(d(\vec{u}, \vec{v})=\lVert\vec{u}-\vec{v}\rVert\).
Unit vector A vector \(\vec{e}\) is called a unit vector if its norm (i.e., its length) is equal to \(1\): \(\lVert\vec{e}\rVert=1\). Division of a nonzero vector \(\vec{v}\) by its norm, that is, multiplication by the reciprocal value of the norm, leads to a unit vector \(\vec{e}=\frac{1}{\|\vec{v}\|}\cdot \vec{v}\) with the same direction as \(\vec{v}\), but with length equal to \(1\).
Properties of the Euclidean norm, the Euclidean norm \(\mathbb{R}^n\) has the following properties: \[\begin{aligned} \lVert\vec{v}\rVert > 0 & \quad\text{if } \vec{v}\neq\vec{0}\\ \lVert\vec{v}\rVert = 0 & \quad\text{if } \vec{v}=\vec{0}\\ \lVert c\cdot\vec{v}\rVert =|c|\cdot \lVert \vec{v}\rVert & \quad\text{voor alle }c \in \mathbb{R}\\ \lVert\vec{u}+\vec{v}\rVert \leq \lVert\vec{u}\rVert+\lVert\vec{v}\rVert & \quad\text{for all } \vec{u}, \vec{v}\in \mathbb{R}^n\end{aligned}\text{.}\]
The properties are exactly what is needed to turn a vector space \(V\) via norm \(\lVert\;\rVert : V\longrightarrow \mathbb{R}\;\) into a so-called normed vector space.
The last inequality is called the triangle inequality or Minkowski inequality.
Two vectors \(\vec{u}\) and \(\vec{v}\) in \(\mathbb{R}^2\) or \(\mathbb{R}^3\) are perpendicular to one another if and only if \[\lVert\vec{u}\rVert^2 + \lVert\vec{v}\rVert^2= \lVert\vec{u}-\vec{v}\rVert^2\tiny.\]
Distance from a point to a line or plane Let \(U\) be a line or a plane in space and \(P\) a point. There is a unique point \(Q\) on \(U\) that has the shortest distance from all points on \(U\) to \(P\). This point is characterized by the property that the vector \(\vec{PQ}\) is perpendicular to the direction vector(s) of \(U\) state. The point \(Q\) is called the perpendicular projection or orthogonal projection of \(P\) on \(U\).
Projection on a line In the visualisation below, \(U\) is the span of the vector \(\vec{u}\) and we have drawn a perpendicular projection of a vector \(\vec{v}\) on the line \(U\). The vector \(\vec{w}\) is a scalar multiple of the vector \(\vec{u}\). The length of the vector \(\vec{v}-\vec{w}\) is minimal when this vector is perpendicular to the line \(U\). By dragging the terminal point of the vector \(\vec{w}\) along the line you can explore this.
Projection on a plane In the visualisation below, \(U\) is a plane through the origin and we have drawn a perpendicular projection of a vector \(\vec{P}\) on the plane\(U\). The length of the vector \(\vec{QP}\) is minimal under the lengths of \(\vec{RP}\) for all choices of the vectors \(\vec{R}\) in the plane.
The length of a vector \( \cv{x \\ y \\ z} \) is given by \( \sqrt{x^2+y^2+z^2} \).
In the case of \(\vec{v} = \cv{1 \\ 3 \\ -7} \), and thus:
\[\begin{aligned}
\|\vec{v}\| &= \sqrt{1^2 + 3^2 + (-7)^2} \\ \\
&= \sqrt{1 + 9 + 49} \\ \\
&=\sqrt{59}
\end{aligned}\]
