Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
     Point, Line and Plane - orthogonal projections, distances, perpendicularity of line and plane
      Projection of a line onto a plane, example
Projection of a line onto a plane
Orthogonal projection of a line onto a plane is a line or a point. If a given line is perpendicular to a plane, its projection is a point, that is the intersection point with the plane, and its direction vector s is coincident with the normal vector N of the plane.
If a line is parallel with a plane then it is also parallel with its projection onto the plane and orthogonal to the normal vector of the plane that is
s ^ N   =>    s  N = 0.
Projection of a line which is not parallel nor perpendicular to a plane, passes through their intersection B and through the projection A of any point A of the line onto the plane, as shows the right figure.
Example:   Determine projection of the line onto the plane
13x - 9y + 16z - 69 = 0.
Solution:  First determine coordinates of the intersection point of the line and the plane,
plug these variable coordinates of the line into the plane
x = 15t + 15,   y = -15t - 12 and  z = 11t + 17   =>    13x - 9y + 16z - 69 = 0,
that is,    13 (15t + 15) - 9 (-15t - 12) + 16 (11t + 17) - 69 = 0   =>    t = -thus,
 thus,        x = 15t + 15 = 15 (-1) + 15 = 0,     y = -15t - 12 = -15 (-1) - 12 = 3
       and    z = 11t + 17 = 11 (-1) + 17 = 6   therefore, the intersection B(0, 3, 6).
Then, find the projection A of a point A(15, -12, 17) of the given line, onto the plane, as the intersection of the normal through the point A, and the plane.
So, write the equation of the normal
Repeat the same procedure to find the projection Aas for the intersection B, that is
plug these variable coordinates of the normal into the equation of the given plane to find the projection A, so
x = 13t + 15,   y = -9t - 12 and  z = 16t + 17   =>     13x - 9y + 16z - 69 = 0,
13(13t + 15) - 9(-9t - 12) + 16(16t + 17) - 69 = 0,    t = -1.
Thus,  x = 13 (-1) + 15 = 2,    y = -9 (-1) - 12 = -3  and   z = 16 (-1) + 17 = 1,   A(2, -3, 1).
Finally, as the projection of the given line onto the given plane passes through the intersection B and the projection A then, by plugging their coordinates into the equation of the line through two points
obtained is the equation of the projection.
Example:   Projection of the line  onto the plane 13x - 9y + 16z - 69 = 0,  
the same as in the above example, can be calculated applying simpler method.
Solution:  Intersection of the given plane and the orthogonal plane through the given line, that is, the plane through three points, intersection point B, the point A of the given line and its projection A onto the plane, is at the same time projection of the given line onto the given plane, as shows the below figure.
The direction vector N1, of the plane determined by three  points A, B and A, is the result of the vector product of the  normal vector of the given plane and the direction vector s of  the given line, that is
 
By plugging the point A(15, -12, 17) into the equation of the plane,
A(15, -12, 17)  => 141x + 97y - 60z + D = 0  =>  141 15 + 97 (-12) - 60 17 + D = 0,  D = 69
obtained is the equation of the plane  P1 ::  141x + 97y - 60z + 69 = 0.
Finally, the line of the intersection l of the given plane 
P :: 13x - 9y + 16z - 69 = 0 and the plane P1 ::  141x + 97y - 60z + 69 = 0 
is at the same time the projection of the given line onto the given plane.
To check the obtained result, write the vector product of normal vectors of planes P and P1
therefore,  N N1 = ls  the vector product is collinear with the direction vector of the intersection line, what proves the result.
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