### 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.

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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