Distance

point to a line - Punkt zur Geraden

video

projection

If a and b are two vectors, the projection is the length of the component of a projected in direction of b.

$$\begin{align}\text{the projection of } \vec{a} \text{ on } \vec{b} & = |\vec{a}| \cdot \cos{\gamma} \newline & = |\vec{a}| \cdot \frac{\vec{a}\cdot\vec{b}}{|\vec{a}| \cdot | \vec{b}|}\newline & = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\end{align}$$

This is the magnitude of $\overline{OP}$ in the diagram.

projection vector

is the length of the projection multiplied by the unit vector of $\vec{b}$:

$$\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right) \cdot \frac{\vec{b}}{|\vec{b}|} $$

remember the definition of the angle from the scalar product.

$$\cos{\gamma} = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}| \cdot | \vec{b}|}$$

two skew lines - windschiefe Geraden

$$\text{distance d } = \left(\frac{\left|\overrightarrow{AB} \cdot \left(\vec{u} \times \vec{v} \right)\right|}{|\vec{u} \times \vec{v}|}\right)$$

example

$$ g: \vec{x} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$

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

explanation

the vector from A to B is projected onto the cross product. The result is the length of the component of AB in direction of the normal to both lines.

Hesse normal form - Hesse’sche Normalform (HNF)

The same method is known as the “Hesse’sche Normalform” (HNF) in the German Abitur.

Note that $$ \frac{ \vec{u} \times \vec{v} }{|\vec{u} \times \vec{v}|} $$ is a unit vector perpendicular to plane spanned by the direction vectors $\vec{u}$ and $\vec{v}$

It is also equivalent to the unit normal vector $\vec{n_0} = \frac{1}{|\vec{n}|} \cdot \vec{n}$ in the Hess’sche Normalform in the German textbooks.

definition

If

$$\vec{n} \cdot \vec{x} - \vec{n} \cdot {\overrightarrow{OP}} = 0$$

is a normal vector equation of a plane E (where P is a point in the plane), then

$$\frac{1}{|\vec{n}|} \cdot \left(\vec{n} \cdot \vec{x} - \vec{n} \cdot {\overrightarrow{OP}}\right) = 0$$

is called the Hesse normal form (German: Hesse’sche Normalform (HNF) der Ebene E)

Example - distance of a point to the plane

Calculate the distance of point $A(9,-3,2)$ to the plane $E:3x+y-2z=-12$ .

Solution

Write the cartesian equation of the plane in normal form.

$$\begin{pmatrix} {3} \\ {1} \\ {-2} \end{pmatrix} \cdot \vec{x} = -12$$

Evaluate the magnitude of the normal vector.

$$\left| \begin{pmatrix} {3} \\ {1} \\ {-2} \end{pmatrix} \right| = \sqrt{14}$$

In the Hesse normal form.

$$\frac{1}{\sqrt{14}}\left[ \begin{pmatrix} {3} \\ {1} \\ {-2} \end{pmatrix} \cdot \vec{x} + 12 \right]= 0$$

To calculate the distance of point A to the plane E, substitute the coordinates of A into the normalised vector equation (HNF).

$$\text{distance(A,E)}=\frac{1}{\sqrt{14}}\left| \begin{pmatrix} {3} \\ {1} \\ {-2} \end{pmatrix} \cdot \begin{pmatrix} {9} \\ {-3} \\ {2} \end{pmatrix} + 12 \right| = \frac{32}{\sqrt{14}} \approx 8.6$$

So, the distance of A to the plane E is approximately 8.6 units.

Alternative method

Dropping a perpendicular line from the point on to the plane in order to determine the parameter for the point where the normal line intersects the plane (denoted as: foot)

two parallel lines

method 1

method - using the properties of the cross-product

The magnitude of the cross-product of two vectors is also the area of a parallelogram spanned by the two vectors.

$$\text{Area} = |\vec{a} \times \vec{b}| = |\vec{a}|\cdot|\vec{b}|\cdot \sin{\theta}$$

proof:

This property is used to find the distance of a point in 3D space to a line.

the altitude is equal to $d = \frac{\text{area}}{\text{the base vector of the parallelogram}}$.

So,

$$d = \frac{|\vec{a} \times \vec{b}|}{|\vec{b}|}$$

method 2 - lines are parallel lines

Using the scalar product equals to zero.

Linear Algebra with Applications


(c) 2019 sebastian.williams[at]sebinberlin.de - impressum und datenschutz - Powered by MathJax & XMin & HUGO & jsxgraph & mypaint