
Analytic Geometry (Coordinate Geometry) in Threedimensional Space 

Line
and Plane

Point,
Line and Plane  orthogonal
projections, distances, perpendicularity of line and plane

Through a given point pass a line perpendicular to a given plane

Given a line and a point, through
the point lay a plane perpendicular to the line






Point,
line and plane – orthogonal
projections, distances, perpendicularity of line and plane

Through a given point pass a line perpendicular to a given plane 
In this case, the normal vector
N
of a plane is
collinear or coincide
with the direction vector 
s =
ai + bj + ck
of a line, 
that is,
s =
N =
Ai + Bj + Ck. 
The given point
A(x_{0},
y_{0},
z_{0})
plugged into rewritten
equation of the line gives 





Example:
Determine the equation of a line which passes through
the point A(3, 5,
1) perpendicular to the
plane 2x 
y + 4z  3 =
0.

Solution: By
plugging the point A(3, 5,
1)
and the components of the normal vector

N =
s
= 2i 
j + 4k
of the given plane into
the above equation of the line obtained is



Given a line and a point, through
the point lay a plane perpendicular to the line

The direction vector
s
of a line is now collinear or coincide
with
the normal vector N of a plane
so that 
N =
s =
ai + bj + ck. 
Coordinates of the given point
A(x_{0},
y_{0},
z_{0})
is plugged into the
rewritten equation
of a plane 
P
::
a · x_{0} + b · y_{0} + c · z_{0} + D = 0 
to determine the parameter
D. 




Example:
Given is a line 

and a point
A(2, 1,
4), through the point lay a plane


perpendicular to the line. 
Solution: In the above equation of the line, the zero in the denominator denotes that the direction
vector's
component c =
0, it does not mean division by zero. Consider this as symbolic
notation.

It means that the given line is parallel with the
xy coordinate plane on the distance
z = 3 that is, the coordinate
z
of each point of the line has the value 3.

Since N =
s
then, N =
i
+ 3j
or
N =
Ai + Bj + Ck.

The coordinate of the point must satisfy the equation of the plane
Ax + By + Cz +
D = 0
that is,

A(2, 1,
4) => Ax + By + Cz + D = 0
gives 1 · (2)
+ 3 · 1
+ 0 · 4 +
D = 0, D =
5.

Therefore, P
:: 2x +
3y 
5 = 0 is the plane through the given point perpendicular to the given line.









Precalculus contents
J 



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