Finding the parametric equations of the line L

[dimens]

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..

 

[tiny-tim]

hi dimens1

∏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))

(or three equations, one for each of x y and z)

 

[dimens]

I thought it was...
x=n + t
y=n+t
z=n+t

 

[tiny-tim]

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?​

 

[dimens]

I was going to substitute y into x then z=t.

 

[tiny-tim]

should work

oh, wait a mo …

∏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 !

 

[dimens]

∏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

 

[HallsofIvy]

∏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

 

[dimens]

Still that answer won't match..

 

[HallsofIvy]

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.

 

[dimens]

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.

 

[dimens]

bump?

 

[dimens]

Anyone? :(

 

[dimens]

triple bump... lol

 

[HallsofIvy]

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.

 

[tiny-tim]

That's a good way to get yourself banned!

i always thought that one bump every 24 hours was considered acceptable?

∏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

... x is meant to be..

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

weird.

ok, so put T = t/7

then

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

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

btw, can you se that if you'd eliminated x at the start (instead of y), you'd immediately have got 7y=5z ?

 

[dimens]

Cheers. Just one more question... Do I always solve for Y instead of X? Or vise versa

 

[tiny-tim]

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 )

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