
Analytic Geometry (Coordinate Geometry) in Threedimensional Space 

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

Distance between
point and line example 
Distance between point and plane
example 





Distance between point and line

The
distance d
between a point and a line we calculate as the distance between the given point
A(x_{1},
y_{1},
z_{1})
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 AŽ(x_{p},
y_{p},
z_{p}) 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) =>
P
::
x

y + D
= 0 gives
1 · (3)

1 · 5 + D
= 0, D
= 2 so,
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 AŽ(3,
5, 0). 
The
distance between the point A
and the line equals the distance between points, A
and AŽ
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 earlier in this chapter,
they are 
 the Hessian normal form,
d = x_{0}cosa
+ y_{0}cosb
+ z_{0}cosg

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.


Example:
Given is a point A(4,
13, 11) and a plane x +
2y + 2z 
4 = 0, find the distance between the point
and the plane. 
Solution: Through the given point draw a line orthogonal to the plane, so that
s = N =
i + 2 j + 2k


The intersection of the line and the plane,


The distance between a point and the plane equals the distance between the point
A
and the intersection (or projection of the point
A onto the plane), thus


Check the result using the above formula,










Precalculus contents
J 



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