Parametrization of lines PDF

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Parametrization of Lines and Planes As explained in class, a line in the plane is presented in a parametrized form if the coordinates of points on the line are expressed as a function of some variable, called a parameter (let’s denote it by t). That is, every point on the line is of the form ] [ x(t) for some t ∈ R. y(t) A simple way of doing this is to write [ ] [ ] [ ] v1 p1 x = + t v2 y p2 where p=

[

    [ ]  is any non-zero  ]  is a fixed  v1 p1 and v = point on vector parallel v2  p2    the line to the line

EXAMPLE [1]:

We can parametrize the line 2x + 3y = 4

as follows: 2x + 3y = 4 ⇕ x = 2 − (3/2)y ⇕ x = 2 − (3/2) y y = 0 + 1 y ⇕ [

x y

]

=

[

2 0

] 1

+ y

[

3/2 1

]

.

That is, all points which lie on the line 2x + 3y = 4 can be expressed in the form [ ] [ ] [ ] 2 3/2 x = + y y 0 1 Here

p=

[

2 0

  ]  

is a point on this line corresponding    to y = 0

   

and v =

  

[

  ]  is a vector  3/2 parallel to . 1   this line

The variable y is playing the role of the parameter t in this parametrization, it is completeley arbitrary but once it is specified a point on the line is specified.

Similarly a plane in R3 can be parametrized, however since there are two degrees of freedom on a plane we will need two arbitrary parameters. EXAMPLE [2]:

We can parametrize the plane x + 3y − 4z = 5

as follows: x + 3y − 4z = 5 ⇕ x = 5 − 3y + 4z ⇕ x = 5 − 3y + 4z y = 0 + 1y + 0z z = 0 + 0y + 1z ⇕        −3 4 5 x  y  =  0  + y  1  + z 0  0 1 0 z 

2

.

5    In this example, y and z are the arbitrary parameters, p = 0 is a point on the

plane corresponding to both parameters being zero, i.e. y = z =0 0 and the vectors     −3 4 u =  1  and v =  0  0 1

are vectors which are parallel to the plane. Note, in case you need convincing that u and v above are indeed parallel to the plane 1x + 3y−4z = 5, observe (as we have seen in class) that the vector   1 is normal to this plane and furthermore, n=  3 −4 ⟨ ⟨n, u⟩ = 

and

⟨n, v⟩ =

   1 −3 ⟩ 3  ,  1  = −3 + 3 + 0 = 0 −4 0

  ⟩ 4 1  3 ,  0  = 4 + 0 − 4 = 0. 1 −4

⟨

The Line of Intersection of two Planes in R3 : Just as we did for a line in R2 we can parametrize a line in R3 by writing:       x p1 v1  y  =  p 2  + t  v2  z p3 v3

where

     v1  is any non-zero  p1  is a fixed  point on vector parallel and v =  v2  p =  p2      the line to the line v3 p3 

We now show how to do this when the line arises as the intersection of two planes in R3 . 3

EXAMPLE [3]:

Parametrize the line of intersection of the two planes: x − 2y + 5z = 3 2x − 5y + z = −1

Solution: To explain the operations that we will perform below we note that: • If we multiply across any equation given in the box above (let’s call these equations rows) by a non-zero constant then we do not change the solution set of the system of two equations given in the box. • If we add a multiple of one row to another row then we do not change the solution set of the system of two equations given in the box. • If we interchange the two rows then we do not change the solution set of the system of two equations given in the box. Now, x + 3y + 5z = 3 2x + 5y + z = −1

Denote the first row by R1 Denote the second row by R2

⇕ R1 is left unchanged R2 is replaced by (R2 − 2R1 )

x + 3y + 5z = 3 − y − 9z = −7 ⇕

R1 is left unchanged R2 is replaced by −R2

x + 3y + 5z = 3 y + 9z = 7 ⇕ x

R1 is replaced by (R1 − 3R2 ) R2 is left unchanged

− 22z = −18 y + 9z = 7

So what we have shown is that 4

x + 3y + 5z = 3 2x + 5y + z = −1

⇐⇒

x

− 22z = −18 y + 9z = 7

⇐⇒

x = −18 + 22z y = 7 − 9z

⇐⇒

x = −18 + 22z y = 7 − 9z z = 0 + 1z

⇐⇒

     x −18 22  y = 7  + z  −9  z 0 1 

Thus, the solution set of our system of equations x + 3y + 5z = 3 2x + 5y + z = −1 is the parametrized line 

     22 x −18  y  =  7  + z  −9  1 z 0 where      22  is a vector  −18  is a fixed  . point on parallel to 7 and v =  −9  p=     this line this line 1 0 

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