Finding the parametric equations of the line L

In summary: Thanks :tup:In summary, the conversation discusses finding the parametric equation of the line L, which is the line of intersection of two given planes. The solution involves substituting one variable into the other to eliminate it, solving for the remaining variables, and rewriting the equations in terms of a parameter t. It is noted that there can be an infinite number of parametric equations for a single line or curve, but one possible solution is x = 2 + t, y = 5t, z = 7t. The conversation also mentions the importance of choosing a parameter that will make the equations simpler and save time during exams.
  • #1
dimens
43
0

Homework Statement


∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

Find the parametric equation of the line L...

The Attempt at a Solution



So I make them simultaneous equations...

2x + y = 4 - t (1)
3x - 2y = 6 - t (2)

multiple equation 1 by 2.

4x + 2y = 8 - 2t (1)
3x - 2y = 6 - t (2)

Solve for 'x'

7x = 14 - 3t

x = 2 - 7/3t

...

It just doesn't look right to me..
 
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  • #2
hi dimens1 :smile:
dimens said:
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

Find the parametric equation of the line L...

a parametric equation is of the form (x,y,z) = (f(t),g(t),h(t)) :wink:

(or three equations, one for each of x y and z)
 
  • #3
I thought it was...
x=n + t
y=n+t
z=n+t
 
  • #4
dimens said:
I thought it was...
x=n + t
y=n+t
z=n+t

(i assume you mean different RHSs ?)

same thing

so where's your y and z equations?​
 
  • #5
I was going to substitute y into x then z=t.
 
  • #6
should work :confused:

oh, wait a mo …
dimens said:
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

2x + y = 4 - t (1)
3x - 2y = 6 - t (2)

… should be 4 plus t ! :cry:
 
  • #7
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6


2x + y = 4 + t
3x - 2y = 6 - t

(2x + y = 4 + t)*2

4x + 2y = 8 + 2t
3x - 2y = 6 - t

7x = 14 + t


x = 2 + t/7

subt x = 2 + t/7

... 2(2+t/7) - 2y = 6 - t
4 + 2t/7 - 2y = 6 - t
-2y = 2 - 5t/7
y = - 1 + 10t/7

z=t

so...
x = 2 + t/7
y = -1 + 10t/7
z = t

.. answer says its

x = 2 + t
y = 5t
z = 7t

:s
 
  • #8
dimens said:
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6


2x + y = 4 + t
3x - 2y = 6 - t

(2x + y = 4 + t)*2

4x + 2y = 8 + 2t
3x - 2y = 6 - t

7x = 14 + t


x = 2 + t/7

subt x = 2 + t/7
Good so far!

... 2(2+t/7) - 2y = 6 - t
The equation was 3x- 2y= 6- t

4 + 2t/7 - 2y = 6 - t
-2y = 2 - 5t/7
y = - 1 + 10t/7

z=t

so...
x = 2 + t/7
y = -1 + 10t/7
z = t

.. answer says its

x = 2 + t
y = 5t
z = 7t

:s
 
  • #9
Still that answer won't match..
 
  • #10
Since you don't say what equations you got it is impossible to say whether your answer is correct or not. The fact that you get different equations than given as the answer is irrelevant. A single line can have an infinite number of different representations as parametric equations.

Let t= 0 and calculate x, y, z. Do they satisfy both equations of the original given planes? If so, that is a point on the line of intersection. Let t= 1 and calculate x, y, z. Do they satisfy both equations of the original given planes? If so that is also a point on the line of intersection? If so your equations give two points on the line and so are parametric equations for the line.
 
  • #11
HallsofIvy said:
Good so far!


The equation was 3x- 2y= 6- t

Yeah I picked up on that but still it doesn't match... x is meant to be..

x = 2 + t not x = 2 + t/7

weird.
 
  • #12
bump?
 
  • #13
Anyone? :(
 
  • #14
triple bump... lol
 
  • #15
That's a good way to get yourself banned!

By the way, your original statement of the problem was
"∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

Find the parametric equation of the line L..."
which makes no sense because you haven't said what L is! We "guessed" that you mean that L is the line of intersection of the two planes but you should have said that!

You eventually arrived at x= 2+ t/7 and then 3(2+t/7) - 2y = 6 - t, which gives 6+ 3t/7- 2y= 6- t. Then 2y= 3t/7+ t= 10t/7 so y= 5t/7.

I said that there are an infinite number of possible parametric equations for any line or curve. Here, if you choose s= t/7, you get parametric equations x= 2+ s, y= 5s.

Of course, it doesn't matter what letter you use for the parameter so those are exactly the same as x= 2+ t, y= 5t.
 
  • #16
HallsofIvy said:
That's a good way to get yourself banned!

i always thought that one bump every 24 hours was considered acceptable? :confused:
dimens said:
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

2x + y = 4 + t
3x - 2y = 6 - t

(2x + y = 4 + t)*2

4x + 2y = 8 + 2t
3x - 2y = 6 - t

7x = 14 + t


x = 2 + t/7

subt x = 2 + t/7

... 2(2+t/7) - 2y = 6 - t
4 + 2t/7 - 2y = 6 - t
-2y = 2 - 5t/7
y = - 1 + 10t/7

z=t

so...
x = 2 + t/7
y = -1 + 10t/7
z = t

.. answer says its

x = 2 + t
y = 5t
z = 7t

:s
dimens said:
... x is meant to be..

x = 2 + t not x = 2 + t/7

weird.

ok, so put T = t/7 :smile:

then

x = 2 + T
y = -1 + 10T
z = 7T

which is the given answer (apart from the y, which i assume is fixable :rolleyes:)

btw, can you se that if you'd eliminated x at the start (instead of y), you'd immediately have got 7y=5z ? :wink:
 
  • #17
Cheers. Just one more question... Do I always solve for Y instead of X? Or vise versa
 
  • #18
no, it just happens to be a lot quicker in this case

(and in the exam, saving a few minutes may get you extra marks on other questions :wink:)

(oh, and it's "vice versa" :smile: … or is it "verse vica"? :rolleyes:)​
 

1. What are parametric equations of a line?

Parametric equations of a line are a set of equations that describe the coordinates of points on a line in terms of one or more parameters. These equations can be used to graph a line and determine specific points on the line.

2. How do you find the parametric equations of a line?

To find the parametric equations of a line, you will need the slope of the line (m) and the coordinates of a point on the line (x1, y1). Then, you can use the following equations:
x = x1 + mt
y = y1 + mt

3. What is the purpose of finding parametric equations of a line?

The purpose of finding parametric equations of a line is to have a mathematical representation of the line that can be used for various calculations and applications. For example, these equations can be used to find the distance between two points on the line or to determine the intersection point of two lines.

4. Can parametric equations of a line be used for any type of line?

Yes, parametric equations of a line can be used for any type of line, including straight lines, curved lines, and even lines in three-dimensional space. As long as you have the necessary information (slope and a point), you can find the parametric equations for any line.

5. Can parametric equations of a line be converted to other forms of equations?

Yes, parametric equations of a line can be converted to other forms of equations, such as slope-intercept form or standard form. This can be done by solving for one of the variables (usually t) in terms of the other variable (x or y) and then substituting that into the other equation. However, the parametric equations are often preferred for their simplicity and ability to represent lines in multiple dimensions.

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