# Relationship between point and line by lindash

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```									                                     Relationship between point and line

Position Vectors
Position vector is the term given to the vector which locates a point with respect to some
datum or axis system. The components of a position vector are the coordinates of the point
(see Figure 1).

The vector giving the displacement between two points is given by the difference between the
two position vectors of the points (see Figure 2). That is, the vector from P to Q is (q - p)

Note that the sum of the vectors around a closed figure must be a zero vector 0 . In Figure 2
the total displacement achieved by going around the figure is: p + (q - p) + (-q) = 0

N                                                  N              P
P                                                         q-p
p                                                  p                         Q
Figure 1
np
q
Figure 2
ep
E                                            E

Relationship between point and line - using the dot product – 2D example.

P
N                               ( u1 • v2 ) u1
u1                              S
v1
Q
v2

R            Figure 3

E
P
( u1 • v2 ) u1
N
S
v1
pp                                                      Q

ps
Figure 4
O
E
The unit vector of the vector v1 between PQ is u1 where

v1
u1 =
| v1 |

The distance PS is (u1 • v2) - from the definition of the dot product.

Therefore the vector between P and S is (u1 • v2) u1

And the position vector of S is:           ps = pp + (u1 • v2) u1

The vector at right angles to a 2-D vector

N                                   v2
δe

δn               Figure 5
v1
δn
δe

E

In Figure 5, the vectors v1 and v2 are at right angles.

 δe 
If v1 has components:                                     v1 =   ,
δn 

− δn 
Then, as shown in the figure, v2 has components:          v2 =      
 δe 

 sin θ 
Thus the unit vector along a line with bearing θ , is:              cos θ
       

− cos θ
The vector at bearing (θ - 90o) is                                   sin θ 
       
Relationship between point and line - using the cross product – 2D example.
In Figure 6, the length of the line RS is |u1*v2| u3 – from the definition of a cross product.

P u1
N
S
v1
Q
v2                          |u1*v2| u3
u3

R                Figure 6
E

u 
If the unit vector u1 has components:                 u1 =  1 
u 2 

− u 
The vector at right angles has components:            u3 =  2 
 u1 

Thus the position vector of S is:                     ps = pr + |u1*v2| u3

Intersection of vector through point P and line.
The final case to be considered is shown in Figure 7. Here a line through P along unit vector
u3 intersects the line at S . In vector terms the point S is at the intersection of the two
vectors αu1 and βu3 where α and β are unknown scalar multipliers of the unit vectors u1
and u3 .

P u1
P u1

S                                            αu1
v1                                          S
Q
v2                                                                           β u3
v2
u3
R               Figure 7                                     u3
R

In the closed figure PRS the vector equation can be written:

v2 = αu1 + βu3          which leads to the simultaneous linear equations

 v12   u 11    u 13  α 
 v  = u        u 23   β 
which can be solved for α and β
 22    21             

v                     u                         u 
Note;    v2 =  12  ,           u1 =  11  ,               u3 =  13  ,
 v 22                 u 21                      u 23 
Relationship between point and line - using the dot product – 3D example.

H
R

v1
v2                                      Q

S
P             u1                                             N

E

R

v1
v2                                     Q

S

( u1 • v2 ) u1                    Figure 8
P

As before - the position vector of S is:             ps = pp + (u1 • v2) u1

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