
Analytic Geometry (Coordinate Geometry) in Threedimensional 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 = 1
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 A′ as
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 right figure.

The direction vector
N_{1}, 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 P_{1}
::
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 P_{1}
::
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
P_{1},


therefore,
N
´
N_{1}
= ls
the vector product is collinear with the direction vector of the intersection line,
what proves
the result. 









Precalculus contents
J 



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