Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
     Point, Line and Plane - orthogonal projections, distances, perpendicularity of line and plane
      Plane laid through a given point, such that be parallel with two skew lines, example
      Plane laid through a given point, such that be parallel with two parallel lines, example
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. 
 
Example:   Through the point O(0, 0, 0) lay a plane parallel to lines,
Solution:  Given lines are parallel since their direction vectors are collinear, i.e.,
s1 = -i + 2 j - 3k  and  s2 = 2i - 4 j + 6k = -2 (-i + 2 j - 3k) = -2s1.
Offered points, T1(-1, 4, -3) and T2(0, -5, 1), in the equations of lines, connected are by the vector,
T1T2 = r2 - r1 = ( -5 j + k) - (-i + 4 j - 3k) = i - 9 j + 4k.
Therefore, the normal vector of the plane
The given point (the origin O), plugged into the equation of a plane, 
O(0, 0, 0)   =>   Ax + By + Cz + D = 0  gives  -19 0 + 1 0 + 7 0 + D = 0,    D = 0
so the equation of the plane to be found is,   -19x + y + 7z = 0.
Coordinate geometry contents
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