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Let L be a line in the plane and let M be any point not on the line. Then, we define distance d from M to L as the length of line segment M P , where P is a point on L such that M P is perpendicular to L ( [link] ).

This figure has two line segments. The first line is labeled “L” and has point P on the segment. The second line segment is drawn from point P to point M and is perpendicular to line L. The second line segment is labeled “d.”
The distance from point M to line L is the length of M P .

When we’re looking for the distance between a line and a point in space, [link] still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let P be an arbitrary point on line L and let v be a direction vector for L ( [link] ).

This figure has a line segment labeled “L.” On the line segment L there is point P. There is a vector drawn from point P to another point M. Also, from M there is a line segment drawn to line L. This segment is perpendicular to line L. There is also a vector labeled “v” on line segment L. A parallelogram has been formed with vector v, line segment P M, and two other segments back to line L.
Vectors P M and v form two sides of a parallelogram with base v and height d , which is the distance between a line and a point in space.

By [link] , vectors P M and v form two sides of a parallelogram with area P M × v . Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

P M × v = v d .

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Distance from a point to a line

Let L be a line in space passing through point P with direction vector v . If M is any point not on L , then the distance from M to L is

d = P M × v v .

Calculating the distance from a point to a line

Find the distance between t point M = ( 1 , 1 , 3 ) and line x 3 4 = y + 1 2 = z 3 .

From the symmetric equations of the line, we know that vector v = 4 , 2 , 1 is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point P ( 3 , −1 , 3 ) lies on the line. Then,

P M = 1 3 , 1 ( −1 ) , 3 3 = −2 , 2 , 0 .

To calculate the distance, we need to find P M × v :

P M × v = | i j k −2 2 0 4 2 1 | = ( 2 0 ) i ( −2 0 ) j + ( −4 8 ) k = 2 i + 2 j 12 k .

Therefore, the distance between the point and the line is ( [link] )

d = P M × v v = 2 2 + 2 2 + 12 2 4 2 + 2 2 + 1 2 = 2 38 21 .
This figure is the first octant of the 3-dimensional coordinate system. There is a 3-dimensional box drawn in the octant. There is a point labeled at (1, 1, 3). There is a line segment labeled “L” inside of the box. Also, there is a perpendicular line segment from the point to line L.
Point ( 1 , 1 , 3 ) is approximately 2.7 units from the line with symmetric equations x 3 4 = y + 1 2 = z 3 .
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Find the distance between point ( 0 , 3 , 6 ) and the line with parametric equations x = 1 t , y = 1 + 2 t , z = 5 + 3 t .

10 7

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Relationships between lines

Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines    ( [link] ).

This figure has two line segments. They are 3-dimensional, are not parallel, and do not intersect. The directions are different and one is above the other.
In three dimensions, it is possible that two lines do not cross, even when they have different directions.

To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point ( [link] ).

This figure is a table with two rows and two columns. Above the columns is the question “Lines share a common point?” The first column is labeled “yes,” and the second column is labeled “no.” To the left of the rows is the question “Direction vectors are parallel?” The first row is labeled “yes,” and the second row is labeled “no.” The entries of the first row are “equal” and “parallel but not equal.” The entries in the second row are “intersecting” and “skew.”
Determine the relationship between two lines based on whether their direction vectors are parallel and whether they share a point.

Classifying lines in space

For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.

  1. L 1 : x = 2 s 1 , y = s 1 , z = s 4
    L 2 : x = t 3 , y = 3 t + 8 , z = 5 2 t
  2. L 1 : x = y = z
    L 2 : x 3 2 = y = z 2
  3. L 1 : x = 6 s 1 , y = −2 s , z = 3 s + 1
    L 2 : x 4 6 = y + 3 −2 = z 1 3
  1. Line L 1 has direction vector v 1 = 2 , 1 , 1 ; line L 2 has direction vector v 2 = 1 , 3 , −2 . Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, ( x , y , z ) , that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
    2 s 1 = t 3 ; s 1 = 3 t + 8 ; s 4 = 5 2 t .

    By the first equation, t = 2 s + 2 . Substituting into the second equation yields
    s 1 = 3 ( 2 s + 2 ) + 8 s 1 = 6 s + 6 + 8 5 s = −15 s = −3.

    Substitution into the third equation, however, yields a contradiction:
    s 4 = 5 2 ( 2 s + 2 ) s 4 = 5 4 s 4 5 s = 5 s = 1.

    There is no single point that satisfies the parametric equations for L 1 and L 2 simultaneously. These lines do not intersect, so they are skew (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
  2. Line L 1 has direction vector v 1 = 1 , −1 , 1 and passes through the origin, ( 0 , 0 , 0 ) . Line L 2 has a different direction vector, v 2 = 2 , 1 , 1 , so these lines are not parallel or equal. Let r represent the parameter for line L 1 and let s represent the parameter for L 2 :
    x = r y = r z = r x = 2 s + 3 y = s z = s + 2.

    Solve the system of equations to find r = 1 and s = 1 . If we need to find the point of intersection, we can substitute these parameters into the original equations to get ( 1 , −1 , 1 ) (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
  3. Lines L 1 and L 2 have equivalent direction vectors: v = 6 , −2 , 3 . These two lines are parallel (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
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Questions & Answers

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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