# What is room X 2

## Positional relationships

Positional relationships of lines and planes are part of the overarching topic of vector calculation and you will come across it sooner or later in school. We have prepared all information on the subject of location relationships for you in easy-to-understand language and supplemented it with learning videos.

1,99€

### Basics of positional relationships

Every straight line can be described in \$ \ mathbb {R} ^ 3 \$ by an equation of the form

\ begin {align *}
g: \ \ vec {x} = \ left (\ begin {array} {c} a_1 \ a_2 \ a_3 \ end {array} \ right)
+ t \ cdot \ left (\ begin {array} {c} u_1 \ u_2 \ u_3 \ end {array} \ right), \ quad t \ in \ mathbb {R} \ notag
\ end {align *}

represent. Special positions arise when the support vector and the direction vector have zeros and ones as coordinates. For example, there is a straight line with

• \$ a_1 = a_2 = a_3 = 0 \$ a straight line through the origin
• \$ u_2 = u_3 = 0 \$ a parallel to the \$ x_1 \$ axis
• \$ u_1 = 0 \$ a parallel to the \$ x_2x_3 \$ level
• \$ u_1 = u_2 = 1, u_3 = 0 \$ a parallel to one of the bisectors between the \$ x_1 \$ axis and the \$ x_2 \$ axis
• \$ u_1 = u_2 = u_3 = 1 \$ a straight line that has an angle of \$ 45 ^ o \$ to each axis

Each level can be expressed by an equation of form

\ begin {align *}
E: \ vec {x} = \ left (\ begin {array} {c} p_1 \ p_2 \ p_3 \ end {array} \ right)
+ r \ cdot \ left (\ begin {array} {c} u_1 \ u_2 \ u_3 \ end {array} \ right)
+ s \ cdot \ left (\ begin {array} {c} v_1 \ v_2 \ v_3 \ end {array} \ right) \ notag
\ end {align *}

represent.

One level with

• \$ p_1 = p_2 = p_3 = 0 \$ goes through the origin
• \$ u_3 = v_3 = 0 \$ is parallel to the \$ x_1x_2 \$ level
• \$ u_1 = u_2 = 0 \$ is parallel to the \$ x_3 \$ axis

If the equation is given in coordinate form, one recognizes the special position of a plane immediately:
If a \$ x_i \$ is missing, the plane is parallel to its axis.

1,99€

### Positional relationships straight-straight Overview - position of two straight lines to one another

Special case: \$ g \$ and \$ h \$ intersect and are orthogonal.
Check for orthogonality: the scalar product of the direction vectors is zero.

### Examples

Investigate the position of the straight lines \$ g \$ and \$ h \$ with

\ begin {align *}
g: \ \ vec {x} = \ begin {pmatrix}
2 \\ 0 \\ 2
\ end {pmatrix} + t \ cdot \ begin {pmatrix}
1 \\ 2 \\ 1
\ end {pmatrix} \ quad \ textrm {and} \ quad h: \ \ vec {x} = \ begin {pmatrix}
4 \\ 4 \\ 4
\ end {pmatrix} + s \ cdot \ begin {pmatrix}
-1 \\ -2 \\ -1
\ end {pmatrix}.
\ end {align *}

First we check the direction vectors of the two straight lines for collinearity, i.e. whether they are multiples of each other.

We see that the direction vector of the straight line \$ g \$ results from that of \$ h \$ if this is multiplied by \$ -1 \$. If you don't have the all-seeing eye, you can choose the approach \$ \ vec {u} = r \ cdot \ vec {v} \$ and get:

\ begin {align *}
\ begin {pmatrix}
1 \\ 2 \\ 1
\ end {pmatrix} = -1 \ cdot \ begin {pmatrix}
-1 \\ -2 \\ -1
\ begin {array} {cccccc}
1 & = & r \ cdot (-1) & \ quad \ Rightarrow \ quad r = -1 \
2 & = & r \ cdot (-2) & \ quad \ Rightarrow \ quad r = -1 \
1 & = & r \ cdot (-1) & \ quad \ Rightarrow \ quad r = -1
\ end {array}
\ end {align *}

If \$ r \$ takes on the same value in all lines, the direction vectors are collinear.

Remember the section on linear independence! Since the values ​​of \$ r \$ are the same in this case, the straight lines are either identical or parallel. In order to be able to decide, we do a point test and insert e.g. the position vector from \$ h \$ into \$ g \$:

\ begin {align *}
\ begin {pmatrix}
4 \\ 4 \\ 4
\ end {pmatrix} = \ begin {pmatrix}
2 \\ 0 \\ 2
\ end {pmatrix} + t \ cdot \ begin {pmatrix}
1 \\ 2 \\ 1
\ end {pmatrix} \ quad \ Rightarrow \ quad \ begin {array} {cccccc}
4 & = & 2 & + & t \ cdot 1 & \ quad \ Rightarrow \ quad t = 2 \
4 & = & 0 & + & t \ cdot 2 & \ quad \ Rightarrow \ quad t = 2 \
4 & = & 2 & + & t \ cdot 1 & \ quad \ Rightarrow \ quad t = 2
\ end {array}
\ end {align *}

If \$ t \$ has the same value in all lines, then the position vector of \$ h \$ lies on the straight line \$ g \$ and in this case the straight lines are identical. Note: If the same value for \$ t \$ does not come out everywhere at this point, it is a parallel straight line!

Investigate the position of the straight lines \$ g \$ and \$ h \$ with

\ begin {align *}
g: \ \ vec {x} = \ begin {pmatrix}
-3 \\ -4 \\ -1
\ end {pmatrix} + t \ cdot \ begin {pmatrix}
2 \\ 2 \\ 1
\ end {pmatrix} \ quad \ textrm {and} \ quad h: \ \ vec {x} = \ begin {pmatrix}
4 \\ 3 \\ 1
\ end {pmatrix} + s \ cdot \ begin {pmatrix}
-1 \\ -1 \\ 1
\ end {pmatrix}.
\ end {align *}

We first check whether the direction vectors are multiples of each other:

\ begin {align *}
\ vec {u} = r \ cdot \ vec {v} \ quad \ Rightarrow \ quad
\ begin {array} {cclcccccc}
2 & = & r \ cdot (-1) & \ quad \ Rightarrow \ quad & r & = & -2 \
2 & = & r \ cdot (-1) & \ quad \ Rightarrow \ quad & r & = & -2 \
1 & = & r \ cdot 1 & \ quad \ Rightarrow \ quad & r & = & 1
\ end {array}
\ end {align *}

Since the same value for \$ r \$ does not come out in all lines, the direction vectors are not collinear. This means that it is either two intersecting or crooked straight lines. We check this by equating the two straight line equations. We receive a LGS, which we dissolve using the procedures known to us. The result is:

\ begin {align *}
\ begin {array} {ccccccc}
-3 & + & 2 t & = & 4 & - & s \
-4 & + & 2 t & = & 3 & - & s \
-1 & + & t & = & 1 & + & s
\ end {array} \ quad \ Rightarrow \ quad t = 3, \ s = 1
\ end {align *}

If we put the values ​​of \$ t \$ and \$ s \$ in the top equation, we get the true statement \$ 3 = 3 \$. Since the statement is true, there is an intersection and there are two intersecting straight lines. If a wrong statement comes out here, the straight lines are skewed. The point of intersection can be determined by substituting \$ t = 3 \$ in \$ g \$ or \$ s = 1 \$ in \$ h \$: \$ S (3 | 2 | 2) \$.

Daniel explains the location of 2 straight lines again in his tutorial video.

Position of 2 straight lines, vector geometry, compare parameter shapes, process | Math by Daniel Jung

14 tasks + solutions ✓ PDF available immediately ✓ preparing for Abiˈ21 ✓

New!

### Relationships between straight and plane

3 cases are possible for the position of a straight line \$ g \$ and a plane \$ E \$:

1. \$ g \$ and \$ E \$ intersect,
2. \$ g \$ and \$ E \$ are really parallel,
3. \$ g \$ is in \$ E \$.

Special case:

The straight line \$ g \$ intersects the plane \$ E \$ orthogonally. This is the case when a normal vector of \$ E \$ is a multiple of a direction vector of \$ g \$.

### example

Right now is in parameter and plane in coordinate form

Investigate the position of the line \$ g \$ and the plane \$ E \$ with

\ begin {align *}
g: \ vec {x} = \ left (\ begin {array} {c} 2 \ 1 \ 3 \ end {array} \ right) + r \ cdot \ left (\ begin {array} {c} 0 \ 2 \ 1 \ end {array} \ right) \ quad \ textrm {and} \ quad E: 2x_1 - x_3 = 4 \ notag
\ end {align *}

Action:
1. Rewrite the parametric shape of the straight line.
2. Insert \$ x_1 \$, \$ x_2 \$ and \$ x_3 \$ in the coordinate form of the plane.
3. Adjust according to the parameters of the straight line.
4. Interpret the result.

We first rewrite the parametric shape of the straight line

\ begin {align *}
\ textrm {I} \ quad x_1 & = 2 \ notag \
\ textrm {II} \ quad x_2 & = 1 + 2 r \ notag \
\ textrm {III} \ quad x_3 & = 3 + 1r \ notag
\ end {align *}

and put in \$ E \$:

\ begin {align *}
\ quad 2 \ cdot x_1 - x_3 & = 4 \
\ Leftrightarrow \ quad 2 \ cdot 2 - (3 + 1r) \ & = 4 \
\ Leftrightarrow \ quad 4-3-1r & = 4 \
\ Leftrightarrow \ quad r & = -3
\ end {align *}

We now insert the result \$ r = -3 \$ in the parametric form of the straight line \$ g \$ and we get with

\ begin {align *}
\ vec {x} = \ left (\ begin {array} {c} 2 \ 1 \ 3 \ end {array} \ right) + (-3) \ left (\ begin {array} {c} 0 \ \ 2 \ 1 \ end {array} \ right) = \ left (\ begin {array} {c} 2 \ -5 \ 0 \ end {array} \ right) \ notag
\ end {align *}

a unique solution and thus we know that the straight line intersects the plane at point \$ S (2 | -5 | 0) \$.
What other possible solutions are there?
True statements e.g.

\ begin {align *}
0 & = 0 \ notag \
4 & = 4 \ notag \
3 & = 3 \ notag
\ end {align *} Wrong statements e.g.

\ begin {align *}
0 & = 4 \ notag \
1 & = 2 \ notag \
-3 & = 1 \ notag
\ end {align *} ### example

The straight line and plane are in parametric form

Investigate the position of the line \$ g \$ and the plane \$ E \$ with

\ begin {align *}
g: \ vec {x} = \ left (\ begin {array} {c} 2 \ -3 \ 2 \ end {array} \ right) + r \ cdot \ left (\ begin {array} {c} 1 \ -1 \ 3 \ end {array} \ right) \ quad \ textrm {and} \ quad E: \ vec {x} = \ left (\ begin {array} {c} -3 \ 1 \ \ 1 \ end {array} \ right) + s \ cdot \ left (\ begin {array} {c} 1 \ -2 \ -1 \ end {array} \ right)
+ t \ cdot \ left (\ begin {array} {c} 0 \ -1 \ 2 \ end {array} \ right) \ notag
\ end {align *}

Action:

1. Equate parametric forms.
2. Set up and release the LGS. Alternatively: Write down in matrix notation and put in step form.
3. Interpret the result.

We equate the terms of straight line and plane and get the following LGS:

\ begin {align *}
\ begin {array} {rcccccccccc}
\ textrm {I} & 2 & + & r & = & -3 & + & s & & \
\ textrm {II} & -3 & - & r & = & 1 & - & 2 s & - & t \
\ textrm {III} & 2 & + & 3r & = & 1 & - & s & + & 2t
\ end {array}
\ Leftrightarrow
\ begin {array} {rrcrccccc}
\ textrm {I} & r & - & s & & \ & = & -5 \
\ textrm {II} & -r & + & 2 s & + & t \ & = & 4 \
\ textrm {III} & 3r & + & s & - & 2 t \ & = & -1
\ end {array}
\ end {align *}

Solve with a solution method of your choice (see chapter Solving LGS) and we get the solution \$ r = -3 \$, \$ s = 2 \$ and \$ t = -3 \$. There is an intersection of the straight line and the plane. To get this you either put \$ r \$ in the straight line equation or \$ s \$ and \$ t \$ in the plane equation. The intersection is at \$ S (-1 | \ 0 \ | -7) \$.

To deepen your knowledge, watch Daniel's video tutorial on the subject of positional relationships - straight line - plane!

Position straight line / plane, equate parametric forms, positional relationships | Math by Daniel Jung

14 tasks + solutions ✓ PDF available immediately ✓ preparing for Abiˈ21 ✓

New!

### Positional relationships level - level

3 cases are possible for the mutual position of two planes:

1. They intersect (intersection line).
2. They are really parallel.
3. They are identical.

When examining the position of two planes, the computational effort depends very much on the form in which the equations are given. If both are given in parameter form, the computational effort is usually the greatest.

Special case

The planes are orthogonal. This is the case when the scalar product of the normal vectors is zero.

Investigation of positional relationships in various forms of the plane equation

• If both equations are given in coordinate form, both are understood as an LGS with 3 variables
• If both equations are given in parameter form, the right-hand sides are set equal and an LGS with 3 equations and 4 variables is obtained.
• If there is one equation in coordinate form and one in parameter form, you insert \$ x_1, x_2 \$ and \$ x_3 \$ from the parameter equation into the coordinate equation and you get an equation with the two parameters.

### example

The planes \$ E_1 \$ in parameter form and \$ E_2 \$ in coordinate form are given

\ begin {align *}
E_1: \ vec {x} = \ left (\ begin {array} {c} 1 \ 2 \ 1 \ end {array} \ right)
+ r \ cdot \ left (\ begin {array} {c} 0 \ 1 \ 0 \ end {array} \ right)
+ s \ cdot \ left (\ begin {array} {c} 2 \ 1 \ 1 \ end {array} \ right) \ quad \ textrm {and} \ quad
E_2: x_1 - 2x_2 = 1. \ notag
\ end {align *}

Idea: Rewrite \$ E_1 \$ and insert in \$ E_2 \$:

\ begin {align *}
\ begin {array} {rccccccc}
\ textrm {I} & x_1 & = & 1 & + & 0r & + & 2s \
\ textrm {II} & x_2 & = & 2 & + & 1r & + & 1s \
\ textrm {III} & x_3 & = & 1 & + & 0r & + & 1s
(1 + 2s) - 2 \ cdot (2 + 1r + 1s) \ & = 1 \ notag
\ end {align *}

Put the result \$ r = -2 \$ in \$ E_1 \$ and we get

\ begin {align *}
g: \ vec {x} = \ left (\ begin {array} {c} 1 \ 2 \ 1 \ end {array} \ right) + (-2)
\ cdot \ left (\ begin {array} {c} 0 \ 1 \ 0 \ end {array} \ right) + s \ cdot \ left (\ begin {array} {c} 2 \ 1 \ 1 \ end {array} \ right) = \ left (\ begin {array} {c} 1 \ 0 \ 1 \ end {array} \ right) + s \ cdot \ left (\ begin {array} {c} 2 \ 1 \ 1 \ end {array} \ right) \ notag
\ end {align *}

a line of intersection \$ g \$.

What other possible solutions are there?

True statements z-B -

\ begin {align *}
0 & = 0 \ notag \
4 & = 4 \ notag \
3 & = 3 \ notag
\ end {align *} Wrong statements e.g.

\ begin {align *}
0 & = 4 \ notag \
1 & = 2 \ notag \
-3 & = 1 \ notag
\ end {align *} ### example

Both levels are in parametric form

The levels \$ E_1 \$ and \$ E_2 \$ are given in parameter form:

\ begin {align *}
E_1: \ vec {x} = \ left (\ begin {array} {c} 1 \ 0 \ 2 \ end {array} \ right)
+ r \ left (\ begin {array} {c} -1 \ 1 \ 0 \ end {array} \ right)
+ s \ left (\ begin {array} {c} 0 \ 2 \ -1 \ end {array} \ right), \
E_2: \ vec {x} = \ left (\ begin {array} {c} 1 \ 2 \ 1 \ end {array} \ right)
+ t \ left (\ begin {array} {c} -1 \ 0 \ 2 \ end {array} \ right)
+ u \ left (\ begin {array} {c} 0 \ 1 \ 0 \ end {array} \ right) \ notag
\ end {align *}

As with the position of a straight line - plane in parametric form, we first equate the terms of the plane equations and create a system of equations with 3 equations and 4 unknowns.

For our example, the LGS follows

\ begin {align *}
\ begin {array} {rccccccc}
\ textrm {I} & 1 & - & r & = & 1 & - & t \
\ textrm {II} & r & + & 2s & = & 2 & + & u \
\ textrm {III} & 2 & - & s & = & 1 & + & 2t \
\ begin {array} {rccccccc}
\ textrm {I} & -r & + & t & & & = & 0 \
\ textrm {II} & r & + & 2s & - & u & = & 2 \
\ textrm {III} & & - & s & - & 2t & = & -1 \
\ end {array} \ notag
\ end {align *}

with the solution \$ u = -3t \$. This means that the planes intersect in a straight line. To determine the line of intersection, we insert the solution in one of the two planes (here in \$ E_2 \$).

\ begin {align *}
g: \ vec {x} = \ left (\ begin {array} {c} 1 \ 2 \ 1 \ end {array} \ right)
+ t \ cdot \ left (\ begin {array} {c} -1 \ 0 \ 2 \ end {array} \ right)
-3t \ cdot \ left (\ begin {array} {c} 0 \ 1 \ 0 \ end {array} \ right) = \ left (\ begin {array} {c} 1 \ 2 \ 1 \ end {array} \ right) + t \ cdot \ left (\ begin {array} {c} -1 \ -3 \ 2 \ end {array} \ right) \ notag
\ end {align *}

### example

Planes are in the form of coordinates

If the two planes are in coordinate form, there are several options. You can convert a level equation into parameter form and work through the appropriate procedure, which promises a safe process. Alternatively, you can also get the result without converting the equations. The aim is to eliminate a coordinate (\$ x_1 \$, \$ x_2 \$ or \$ x_3 \$).

Investigate the positional relationships of the levels

\ begin {align *}
\ textrm {I} & \ quad & E_1 &: \ 2x_1-4x_2 + 6x_3 && = 8 \
\ textrm {II} & \ quad & E_2 &: \ x_1 + 4x_2-3x_3 && = - 5
\ end {align *}

We look at the two equations and see that the coordinate \$ x_2 \$ disappears when the equations are added. Then we switch to a remaining coordinate, here \$ x_1 \$.

\ begin {align *}
\ text {I + II} \ quad & 3x_1 + 3x_3 \ && = 3 && \ quad | \ -3x_3 \
\ Leftrightarrow \ quad & 3x_1 \ && = 3-3x_3 && \ quad | \: 3 \
\ Leftrightarrow \ quad & x_1 \ && = 1-x_3 &&
\ end {align *}

We see that the levels are not identical (otherwise a true statement such as \$ 0 = 0 \$ would have to come out) and not parallel (otherwise a false statement such as \$ 0 = 8 \$ would have to come out). The planes intersect and have a straight line of intersection. To determine this, we insert \$ x_1 = 1-x_3 \$ into one of the two equations, here II, and convert to \$ x_2 \$. Thereby we expressed \$ x_1 \$ and \$ x_2 \$ in relation to \$ x_3 \$. It follows:

\ begin {align *}
\ Rightarrow \ quad & 1-x_3 + 4x_2-3x_3 \ && = -5 && \ quad | \ \ text {to summarize} \
\ Leftrightarrow \ quad & 1 + 4x_2 - 4x_3 \ && = -5 && \ quad | \ -1 \; + 4x_3 \
\ Leftrightarrow \ quad & 4x_2 \ && = -6 + 4x_3 && \ quad | \: 4 \
\ Leftrightarrow \ quad & x_2 \ && = -1.5 + x_3 && \ quad | \: 4 \
\ end {align *}

We can not Do more. We write the result a little differently and recognize the structure of a straight line.

\ begin {align *}
\ left (\ begin {array} {c} x_1 \ x_2 \ x_3 \ end {array} \ right) = \ left (\ begin {array} {c} 1-x_3 \ -1.5 + x_3 \ \ x_3 \ end {array} \ right) = \ left (\ begin {array} {c} 1 \ -1.5 \ 0 \ end {array} \ right) + \ left (\ begin {array} { c} -x_3 \ x_3 \ x_3 \ end {array} \ right)
\ end {align *}

Now comes the final step, which is often difficult for many to understand. We claim that \$ x_3 = t \$ (or \$ r \$ or \$ s \$ etc.) is a parameter and get the line of intersection we are looking for in the parameter equation

\ begin {align *}
g: \ vec {x} = \ left (\ begin {array} {c} 1 \ -1.5 \ 0 \ end {array} \ right) + t \ cdot \ left (\ begin {array} { c} -1 \ 1 \ 1 \ end {array} \ right).
\ end {align *}

1,99€

### Overview of cutting angles \ begin {align *}
\ cos (\ alpha) = \ frac {\ vec {a} \ bullet \ vec {b}} {| \ vec {a} | \ cdot | \ vec {b} |}
\ end {align *} \ begin {align *}
\ cos (\ alpha) = \ left | \ frac {\ overrightarrow {RV} _g \ bullet \ overrightarrow {RV} _h} {| \ overrightarrow {RV} _g | \ cdot | \ overrightarrow {RV} _h |} \ right |
\ end {align *}

with \$ \ overrightarrow {RV} _g \$ or \$ \ overrightarrow {RV} _h \$ as direction vectors of two intersecting straight lines \$ g \$ and \$ h \$. \ begin {align *}
\ sin (\ alpha) = \ left | \ frac {\ vec {n} _E \ bullet \ overrightarrow {RV} _g} {| \ vec {n} _E | \ cdot | \ overrightarrow {RV} _g |} \ right |
\ end {align *}

with \$ \ vec {n} _E \$ as the normal vector of a plane \$ E \$ and \$ \ overrightarrow {RV} _g \$ as the direction vector of a straight line \$ g \$. \ begin {align *}
\ cos (\ alpha) = \ left | \ frac {\ vec {n} _1 \ bullet \ vec {n} _2} {| \ vec {n} _1 | \ cdot | \ vec {n} _2 |} \ right |
\ end {align *}

with \$ \ vec {n} _1 \$ or \$ \ vec {n} _2 \$ as the normal vector of two levels \$ E_1 \$ and \$ E_2 \$.

Daniel's positional relationships instructional video for an overview of cutting angles.

Angle, overview, vector geometry, formulas, math help online, explanatory video | Math by Daniel Jung

14 tasks + solutions ✓ PDF available immediately ✓ preparing for Abiˈ21 ✓

New!