Distances, angles, and dot products

Vector calculus in plane and space: Distances, Angles and Inner Product

Distances, angles, and dot products

The concepts of distance, length, and angle can be expressed in simple formulas with the aid of the concept of dot product. We work with a fixed origin in the plane or space. The length of a vector $\vec{x}$ has already been entered; it is the distance from the origin to the endpoint of the representative of $\vec{x}$ that is placed at the origin.

Distance in terms of vector length

The length or norm of a vector $\vec{v}$ is indicated by $\parallel \vec{v}\parallel$ .

The distance between two vectors $\vec{u}$ and $\vec{v}$ , is the length of the difference vector $\vec{u}-\vec{v}$ , which means that $\parallel \vec{u}-\vec{v}\parallel$ .

The distance between two points $A$ and $B$ in the space, is the length of the vector $\vec{AB}$ and corresponds to the distance between the vectors $\vec{OA}$ and $\vec{OB}$ , wherein $O$ is the origin.

The perpendicular projection of a point on a line, gives a point on the line at the shortest distance from the given point. This fact has been previously discussed in two dimensions, but is also true in the space. Something similar applies to a plane instead of a line.

Perpendicular projection of a point on a line or plane

Let $U$ be a line or a plane in the space, and let $P$ be a point. There is a unique point $Q$ on $U$ that has the shortest distance to $P$ out of all points on $U$ . This point is characterized by the property that the vector $\vec{PQ}$ is perpendicular to the directional vector(s) of $U$ .

The point $Q$ is called the perpendicular projection of $P$ on $U$ .

We will now prove that $U$ is a line. Let $V$ be the plane through $P$ perpendicular to (a directional vector of) $U$ .
$V$ then intersects the line $U$ in a unique point $Q$ . The vector $\vec{PQ}$ is located in $V$ and is, therefore, perpendicular to $U$ . If $A$ is a point of $U$ that differs from $Q$ , then $\triangle APQ$ forms a triangle with the right angle $Q$ . We can use the Pythagorean theorem: $\left|AQ\right|=\sqrt{\left|AQ\right|^2+\left|PQ\right|^2}\geq\sqrt{\left|PQ\right|^2}=\left|PQ\right|\tiny.$ The inequality shows that the distance $\left|AQ\right|$ is at least $\left|PQ\right|$ , and that equality only applies if $\left|AQ\right|=0$ , i.e., if $A=Q$ . We have now proven the theory that $U$ is a line.

Proving the case wherein $U$ is a plane, is almost the same, so it will not be described in detail. The biggest difference is that $V$ is the line through $P$ , which is perpendicular to $U$ ; the proof obtained above can then be applied almost literally.

What is the distance in $\mathbb{R}^3$ between the points $P=\rv{3,4,1}$ and $Q=\rv{8,5,0}$ ?

$3\sqrt{3}$

This can be calculated as follows:
$\begin{array}{rcl}\parallel P-Q\parallel &=& \parallel\,\rv{3-8,4-5,1-0}\,\parallel\\&=& \parallel\,\rv{-5,-1,1}\,\parallel\\ &=&\sqrt{\left(-5\right)^2+\left(-1\right)^2+\left(1\right)^2}\\ &=& \sqrt{27}\\ &=&3\sqrt{3}\end{array}$

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