point to a line - Punkt zur Geraden
- determine the vector between the point P and a random point A on the line.
- scalar product of the direction vector of the line to vector AP has to be equal to 0.
- find the parameter
- calculate the magnitude of vector AP
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
- find the vector between the 2 position vectors of the line
$\vec{AB}$
. - use the direction vectors to find the cross poduct
$\vec{u} \times \vec{v}$
. - find the magnitude of the cross-product
$|\vec{u} \times \vec{v}|$
. - calculate the distance as the projection of a vector from point on first line to a point on the second line.
$$\text{distance d } = \left(\frac{\left|\overrightarrow{AB} \cdot \left(\vec{u} \times \vec{v} \right)\right|}{|\vec{u} \times \vec{v}|}\right)$$
example
$$ g: \vec{x} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$
$$ h: \vec{x} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} + s \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix}$$
- Find the vector between one point on each line
$\vec{AB} = \begin{pmatrix} {3 - 0} \\ {-1 - 1} \\ {4 -2} \end{pmatrix} = \begin{pmatrix} {3} \\ {-2} \\ {2} \end{pmatrix} $
$\begin{align}\vec{u} \times \vec{v} & = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} \\ & = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \end{align}$
$|\vec{u} \times \vec{v}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
$$\begin{align}\text{distance d } & = \left(\frac{\left|\overrightarrow{AB} \cdot \left(\vec{u} \times \vec{v} \right)\right|}{|\vec{u} \times \vec{v}|}\right) \\ & = \left| \begin{pmatrix} {3} \\ {-2} \\ {2} \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\right| \cdot \frac{1}{\sqrt{2}} \\ & = |-1| \cdot \frac{1}{\sqrt{2}} \\ & = \frac{\sqrt{2}}{2} \text {units (LE)}\end{align}$$
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.