25
 
Coordinate geometry (or Analytic geometry) in three-dimensional space
 Plane laid through a given point, such that be parallel with two skew lines

  The normal vector of planes that are parallel with two skew lines is collinear to the vector product of

direction vectors of these lines, so we can write
N = s1 ´ s2
The plane passes through a given point if the coordinates of the point A(x0, y0, z0) satisfy the equation of the plane that is
Ax0 + By0 + Cz0 + D = 0
this condition determines the parameter D of the plane to be found.
Example:  Given are skew lines, and the point
A(3, 2, 5), lay a plane through the given point parallel with the given lines.

Solution:  From the equations of the lines, s1 = 2i - j + 4k  and  s2 = -3i - 2 j + k, so the normal vector of the plane
Plug the point A into the equation of the plane to determine D,
A(3, 2, 5)   =>   7x - 14y - 7z + D = 0   =>   7 · 3 - 14 · 2 - 7 · 5 + D = 0,   D = 42

therefore, the equation of the plane  7x - 14y - 7z + 42 = 0.
 Plane laid through a given point, such that be parallel with two parallel lines

In this case, the normal vector of the planes, which are parallel with two parallel lines, is collinear with the

vector product of the direction vector of one line and the vector T1T2 that represents the difference of the radius vectors r2 - r1, of any of the points of the given lines, that is
N = s1 ´ (r2 - r1)

The given point M(x0, y0, z0) is a point of the plane, so it must satisfy the condition,
Ax0 + By0 + Cz0 + D = 0.

And this is the way to determine the parameter D of the plane to be found.
 Distance between point and line

The distance d between a point and a line we calculate as the distance between the given point

A(x1, y1, z1) and its orthogonal projection onto the given line using the formula for the distance between two points. 
The projection of the point onto the line is at the same time the intersection point A´(xp, yp, zp) of the given line and a plane which passes through the given point orthogonal to the given line, thus
where d is the distance between the given point and the given line.
Example:  Find the distance between a point A(-3, 5, -2) and a line
Solution:  Through the given point lay a plane perpendicular to the given line, then 
as     N = s = -i - j    and   A(-3, 5, -2)  substituted into   P ::  - x - y + D = 0
gives   -1 · (-3) - 1 · 5 + D = 0,    D = 2  therefore,  the plane  P ::  - x - y + 2 = 0.
Then, rewrite the given line into the parametric form to get its intersection with the plane P,
plug these variable coordinates of a point of the line into the plane to find parameter t that determine the intersection point,
x = - t - 8  and   y = - t   =>    - x - y + 2 = 0 so that  - (-t - 8) - (- t) + 2 = 0,   t = -5
therefore,  x = - (- 5) - 8 = -3  and  y = -t = - (- 5) = 5,  the intersection A´(-3, 5, 0).
The distance between the point A and the line equals the distance between points, A and A´  thus,
 Distance between point and plane
We use the formulas for the distance from a point to a plane that are given in the two forms
- the Hessian normal form,   d = x0cosa + y0cosb + z0cosg - p
- and when a plane is given in general form,
The distance between a point and a plane can also be calculated using the formula for the distance between two points, that is, the distance between the given point and its orthogonal projection onto the given plane.
 Distance between parallel lines
Distance between two parallel lines we calculate as the distance between intersections of the lines and a plane orthogonal to the given lines.
The direction vector of the plane orthogonal to the given lines is collinear or coincides with their direction vectors that is
N = s = ai + b j + ck
By plugging any of two points, which are included in the equations of the given lines, into the equation of the plane, determined is the parameter D, as well as determined is one of the intersection points.
Example:  Find the distance between given parallel lines,

Solution:  The direction vector of a plane orthogonal to the parallel lines is collinear with the direction vectors of these lines, so  N = s = 2i - 9 j - 2k.

Let the plane passes through the point A´2(-5, -3, 6)  of the second line, then
A´2(-5, -3, 6)  =>   Ax + By + Cz + D = 0  or  2 · (-5) - 9 · (-3) + (-2) · 6 + D = 0,    D = -5
thus, the equation of the plane  2x - 9y - 2z - 5 = 0.  Besides, the point  A´2(-5, -3, 6) is the intersection of the second line and the plane.
The intersection of the first line and the plane we calculate as follows,
The distance between the intersection points A´1 and A´2 is at the same time the distance between given lines, thus
 Distance between two skew lines

Through one of a given skew lines lay a plane parallel to another line and calculate the distance between any point of that line and the plane

The direction vector of planes, which are parallel to both lines, is coincident with the vector product of direction vectors of given lines, so we can write
N = s1 ´ s2 .

A plane to lie on the one of the given lines, we should plug the point, that is included into the equation of the line, into the equation of the plane and so determined is the parameter D of the plane.

Example:  Find the distance between given skew lines,

Solution:  Through the line l1 lay a plane parallel to the line l2. The direction vector of the plane,

The point included in the equation of the line l1 must satisfy the equation of the plane, so
P1(7, -10, -5)  =>    Ax + By + Cz + D = 0,   20 · 7 +  (-4) · (-10) + (-22) · (-5) + D = 0,
  D = -290  thus, the equation of the plane  20x - 4y - 22z - 290 = 0.
The distance between the point T2(-6, -1, 2), of the line l2, and the plane which is parallel to it
The sign of the square root is taken to be opposite to the sign of the parameter D. Therefore, the negative value of the result only informs us that the point is positioned at the same side as the origin of the coordinate system.
 
 
 
 
 
 
 
Contents C
 
 
 
 
Copyright © 2004 - 2020, Nabla Ltd.  All rights reserved.