
Analytic Geometry (Coordinate Geometry) in Threedimensional 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 
N
= s_{1} ´
s_{2} 
The plane passes through a given point if
the coordinates of the point A(x_{0},
y_{0},
z_{0}) satisfy the equation of the
plane that is 
Ax_{0} +
By_{0} + Cz_{0} + 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, s_{1}
= 2i 
j +
4k
and s_{2}
= 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
T_{1}T_{2}
that represents the difference of the 
radius vectors r_{2}

r_{1},
of any of the points of the given lines, that is 
N
= s_{1} ´
(r_{2}

r_{1}) 
The
given point M(x_{0},
y_{0},
z_{0})
is a point of the plane, so it must satisfy the
condition, 
Ax_{0} +
By_{0} + Cz_{0} + 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.,

s_{1}
= i +
2 j 
3k and s_{2}
= 2i

4 j + 6k
= 2 · (i +
2 j 
3k)
= 2s_{1}. 
Offered points,
T_{1}(1,
4, 3)
and T_{2}(0,
5,
1), in the equations of lines, connected are by the vector, 
T_{1}T_{2}
= r_{2}

r_{1}
= ( 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.









Precalculus contents
J 



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