Functions - Uplift Education

Vectors (8)
•2D
Vector Equations
•point of intersect
•3D
Vector Equations
•do they intersect
In 2D : lines either …….
Are parallel
intersect
Or are the same
Intersect of 2D lines in vector form - 1
For example
[] [] []
x
y
=
1
+s
3
2
2
and
[] [] []
x
y
6
+t
-2
=
-1
4
If the lines intersect, there must be values of
s and t that give the position vector of the point of intersection.
x part: 1 + 2s = 6 - t
y part: 3 + 2s = -2 + 4t
Subtract x from y : 2 = -8 + 5t
5t = 10
t=2
Substitute: 1 + 2s = 6 - 2
2s = 3
s = 1.5
[] [] []
1
3
=
+
[] [] []
[y ] = [ ]
x
y
x
y
x
= 1 +1.5 2
3
2
3
3
4
6
position vector of the point of intersection
Intersect of 2D lines in vector form - 2
Point of intersection?
r = (i + 2j) +  (4i - 2j)
s = (2i - 6j) +  (-2i + j)
i coefficients : 1 + 4 = 2 -2
j coefficients : 2 - 2 = -6 + 
x2 : 4 - 4 = -12 + 2
add
5 = -10
… doesn’t work
Direction vectors:
(4i - 2j) and (-2i + j) are parallel
….. lines are parallel
Example
The lines r and s have the equations ...
 2  4 
   
r   3      1
5  3 
   
and
 4
 2 
 
 
s   7     2
 2
 3 
 
 
Show they intersect and
find the point of intersection
If the lines intersect, there must be values of
 and  that give the position vector of the point of intersection.
 2  4   4
 2 
     
 
 3      1   7      2 
 5  3   2
 3 
     
 
x+y : 5 + 3 = 11
Substitute
x : 2 + 4 = 4 +2
y : 3 -  = 7 - 2
z : 5 + 3 = 2 +3
3 = 6
=2
x : 2 + 4x2 = 4 +2
 =3
Check the values in the 3rd equation
z : 5 + 3x2 = 2 +3x3
11 = 11 Satisfied!
Hence a point exists
common to both lines
Example (continued)
The lines r and s have the equations ...
 2  4 
   
r   3      1
5  3 
   
and
 4
 2 
 
 
s   7     2
 2
 3 
 
 
Show they intersect and
find the point of intersection
If the lines intersect, there must be values of
 and  that give the position vector of the point of intersection.
=2
 =3
Values satisfy all equations
Hence a point exists
common to both lines
 2   4   2   8  10 
         
Substitute r   3   2  1   3     2    1 
 5   3   5   6   11
         
The 2 lines intersect at (10, 1, 11)