# Lot (mathematics)

Plumb line from one point to a straight line with plumb point${\ displaystyle l}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle L}$

In geometry, a perpendicular is a segment or straight line that is perpendicular to a given straight line or plane . Depending on whether it is a straight line or a line, one speaks of a vertical line or a vertical line. The intersection of the perpendicular with the given straight line or plane is called the perpendicular base . The plumb line can be constructed geometrically in different ways with a compass and ruler . It can be calculated by determining a normal vector of the straight line or plane or by orthogonal projection of a point outside the straight line or plane. The length of the vertical line is then the distance (normal distance) of a point from the straight line or plane.

## definition

A segment or straight line is called perpendicular to a straight line or plane , if ${\ displaystyle l}$${\ displaystyle g}$ ${\ displaystyle E}$

${\ displaystyle l \ perp g}$   or.   ${\ displaystyle l \ perp E}$

applies if it is perpendicular to the straight line or plane and thus forms a right angle with it. The plumb line is then the point of intersection or the perpendicular with the straight line or plane. ${\ displaystyle l \ cap g}$${\ displaystyle l \ cap E}$

## Geometric constructions

In two dimensions, the plumb line can be easily constructed on a straight line with a compass and ruler . Depending on whether a given point lies on the straight line or outside, one speaks of erecting or felling the perpendicular. ${\ displaystyle P}$${\ displaystyle g}$

### Building the Lot

If there is a point on the straight line , then the perpendicular line through this point can be found as follows: ${\ displaystyle P}$${\ displaystyle g}$

The compass is inserted into the point and, by drawing any arc of a circle, two points are determined at the same distance from . Then you increase the angle of the compass, pierce it in each of the two points found and by drawing two arcs you find a point (of two possible) outside the straight line with the same distance from the two points. The straight line that runs through this point and the given point is then the perpendicular to through . ${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$

An alternative to erecting a plumb line on a straight line through the point with limited space is shown in the picture on the right. The simple construction can be described in the following way: You draw a circular arc with the radius around a freely selectable point until it intersects the straight line in (e.g. you can choose so that an imaginary line from to with the straight line an angle of approx. 45 °). This is followed by drawing a line through until it intersects the arc in . The final line that runs through and is then the perpendicular to through . ${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle M}$${\ displaystyle {\ overline {MP}}}$${\ displaystyle g}$${\ displaystyle A}$${\ displaystyle M}$${\ displaystyle M}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle A}$${\ displaystyle M}$${\ displaystyle P '}$${\ displaystyle P}$${\ displaystyle P '}$${\ displaystyle g}$${\ displaystyle P}$

 Establishing a perpendicular as a perpendicular to two points. Building a lot with the help of the Thaleskreis . The position of the point is freely selectable.${\ displaystyle M}$

### Felling the Lot

Felling the Lot
Alternative method of felling the lot

Is a point outside of the straight line given, then one finds by the solder on as follows: It pierces the circle in the point and is determined by drawing a circular arc with a large radius corresponding to two points equidistant from . Then you pierce one of the two points found and by drawing two arcs (with a sufficiently large radius) you find another point with the same distance from the two points. The straight line that runs through this point and the given point is then the plumb line to through and the intersection of this plumb line with is the plumb line . ${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle F}$

An alternative construction to drop the perpendicular on a straight line from a given point is to insert the compass at any two points and on the straight line and to draw the circle that runs through the given point . These two circles then intersect at another point outside of the straight line and the line that runs through and is then the perpendicular through . This construction can also be used for reflections . ${\ displaystyle M_ {1}}$${\ displaystyle M_ {2}}$${\ displaystyle P}$${\ displaystyle P '}$${\ displaystyle P}$${\ displaystyle P '}$${\ displaystyle P}$

## calculation

In analytic geometry , points in the Euclidean plane or in Euclidean space are represented by position vectors with the help of a Cartesian coordinate system

${\ displaystyle {\ vec {OP}} = (x, y)}$   or.   ${\ displaystyle {\ vec {OP}} = (x, y, z)}$

described. Lines in the plane are typically characterized by a linear equation in parametric form

${\ displaystyle g: {\ vec {x}} = {\ vec {a}} + r \ cdot {\ vec {u}}}$

given, where the position vector of a straight line point (support vector), a direction vector of the straight line and a real straight line parameter . Planes in space can be given as a plane equation in parametric form. ${\ displaystyle {\ vec {p}}}$${\ displaystyle {\ vec {u}}}$${\ displaystyle r}$

${\ displaystyle E: {\ vec {x}} = {\ vec {a}} + r \ cdot {\ vec {u}} + s \ cdot {\ vec {v}}}$

given, where and are real plane parameters and and two span vectors of the plane that are not collinear . Two vectors and in the plane or in space form a right angle if their scalar product applies. ${\ displaystyle r}$${\ displaystyle s}$${\ displaystyle {\ vec {u}}}$${\ displaystyle {\ vec {v}}}$${\ displaystyle {\ vec {x}}}$${\ displaystyle {\ vec {y}}}$ ${\ displaystyle {\ vec {x}} \ cdot {\ vec {y}} = 0}$

### Building the Lot

A direction vector of the perpendicular to a given straight line or a plane is a normal vector of the straight line or plane. In the two-dimensional case, a normal vector of a straight line is obtained by interchanging the two components of the direction vector and reversing the sign of one of the two components , e.g. B. to ${\ displaystyle h}$${\ displaystyle g}$${\ displaystyle E}$ ${\ displaystyle {\ vec {n}}}$${\ displaystyle {\ vec {u}} = (u_ {1}, u_ {2})}$

${\ displaystyle {\ vec {n}} = (u_ {2}, - u_ {1})}$.

A normal vector of a plane can be found through the cross product of two non-collinear span vectors ${\ displaystyle {\ vec {n}}}$

${\ displaystyle {\ vec {n}} = {\ vec {u}} \ times {\ vec {v}}}$

to calculate. If a point with the position vector is given on the straight line, then the equation is the perpendicular${\ displaystyle P}$${\ displaystyle {\ vec {p}}}$${\ displaystyle h}$

${\ displaystyle h: {\ vec {x}} = {\ vec {p}} + r \ cdot {\ vec {n}}}$,

where the parameter traverses all real numbers. A straight line in space does not have a distinctive normal direction , instead it has a perpendicular plane at every straight line point , the normal vectors of which are collinear with the direction vectors of the straight lines. ${\ displaystyle r}$

### Felling the Lot

The perpendicular line from a point with the position vector to a straight line in the two-dimensional case or a plane in the three-dimensional case is calculated using the normal vector${\ displaystyle h}$${\ displaystyle {\ vec {p}}}$${\ displaystyle g}$${\ displaystyle {\ vec {n}}}$

${\ displaystyle h: {\ vec {x}} = {\ vec {p}} + r \ cdot {\ vec {n}}}$.

The plumb line can be calculated as the intersection of with , or . ${\ displaystyle F}$${\ displaystyle h}$${\ displaystyle g}$${\ displaystyle E}$

For a straight line in three-dimensional space, the normal vector of an auxiliary plane is used that contains the point and is perpendicular to the straight line. ${\ displaystyle H}$${\ displaystyle P}$

Alternatively, one obtains directly the nadir point to the position vector to the plane or in space with the normal vector as the orthogonal projection${\ displaystyle {\ vec {OF}}}$${\ displaystyle g}$${\ displaystyle E}$${\ displaystyle {\ vec {n}}}$

${\ displaystyle {\ vec {OF}} = {\ vec {p}} - {\ frac {({\ vec {p}} - {\ vec {a}}) \ cdot {\ vec {n}}} {{\ vec {n}} \ cdot {\ vec {n}}}} \, {\ vec {n}}}$, where is the support vector of , and .${\ displaystyle {\ vec {a}}}$${\ displaystyle g}$${\ displaystyle E}$

It is also possible to drop the perpendicular from a point in space to a straight line in space. If a direction vector is the straight line and the support vector, then the position vector of the plumb line is obtained through ${\ displaystyle {\ vec {u}}}$${\ displaystyle {\ vec {a}}}$${\ displaystyle {\ vec {OF}}}$

${\ displaystyle {\ vec {OF}} = {\ vec {a}} + {\ frac {({\ vec {p}} - {\ vec {a}}) \ cdot {\ vec {u}}} {{\ vec {u}} \ cdot {\ vec {u}}}} \, {\ vec {u}}}$.

The nadir point is the one of straight lines or layer point of the least distance to has. The length of the plumb line, which is calculated with the amount , is called the distance from to the straight line or plane . ${\ displaystyle F}$${\ displaystyle P}$${\ displaystyle | {\ vec {PF}} |}$${\ displaystyle P}$${\ displaystyle g}$${\ displaystyle E}$

### example

The plane through the support vector as well as the tension vectors and is given${\ displaystyle {\ vec {a}} = {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}}}$${\ displaystyle {\ vec {u}} = {\ begin {pmatrix} 2 \\ 1 \\ 2 \ end {pmatrix}}}$${\ displaystyle {\ vec {v}} = {\ begin {pmatrix} 2 \\ - 2 \\ - 1 \ end {pmatrix}}}$

${\ displaystyle E: {\ vec {x}} = {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}} + r {\ begin {pmatrix} 2 \\ 1 \\ 2 \ end { pmatrix}} + s {\ begin {pmatrix} 2 \\ - 2 \\ - 1 \ end {pmatrix}}}$

A normal vector of the plane is then

${\ displaystyle {\ vec {n}} = {\ begin {pmatrix} 2 \\ 1 \\ 2 \ end {pmatrix}} \ times {\ begin {pmatrix} 2 \\ - 2 \\ - 1 \ end { pmatrix}} = {\ begin {pmatrix} 3 \\ 6 \\ - 6 \ end {pmatrix}}}$

or easier . The perpendicular line through the support point with on thus has the straight line equation, for example ${\ displaystyle {\ vec {n}} = {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}}}$${\ displaystyle h}$${\ displaystyle {\ vec {a}} = {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}}}$${\ displaystyle E}$

${\ displaystyle h: {\ vec {x}} = {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}} + r \, {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}} \ ,, \; r \ in \ mathbb {R}}$.

Is the position vector of a point outside the plane given, then one obtains for the nadir point of the solder of on the support vector with ${\ displaystyle {\ vec {p}} = {\ begin {pmatrix} 8 \\ 8 \\ - 1 \ end {pmatrix}}}$${\ displaystyle F}$${\ displaystyle P}$${\ displaystyle E}$${\ displaystyle {\ vec {a}} = {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}}}$

${\ displaystyle {\ vec {OF}} = {\ begin {pmatrix} 8 \\ 8 \\ - 1 \ end {pmatrix}} - {\ frac {\ left [{\ begin {pmatrix} 8 \\ 8 \ \ -1 \ end {pmatrix}} - {\ begin {pmatrix} 1 \\ 3 \\ 4 \ end {pmatrix}} \ right] \ cdot {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}}} {{\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}} \ cdot {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}}}} \ cdot {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}} = {\ begin {pmatrix} 8 \\ 8 \\ - 1 \ end {pmatrix}} - {\ frac {27} {9}} \ cdot {\ begin {pmatrix} 1 \\ 2 \\ - 2 \ end {pmatrix}} = {\ begin {pmatrix} 5 \\ 2 \\ 5 \ end {pmatrix}}}$.

The distance of the point from the plane thus satisfies ${\ displaystyle d (P; E)}$${\ displaystyle P}$${\ displaystyle E}$

${\ displaystyle d (P; E) = d (P; F) \; {\ widehat {=}} \; | {\ vec {PF}} | = \ left | {\ begin {pmatrix} 5 \\ 2 \\ 5 \ end {pmatrix}} - {\ begin {pmatrix} 8 \\ 8 \\ - 1 \ end {pmatrix}} \ right | = \ left | {\ begin {pmatrix} -3 \\ - 6 \ \ 6 \ end {pmatrix}} \ right | = {\ sqrt {81}} = 9}$.