# Homework Help: Finding the parametric equations of the line L

1. Apr 18, 2012

### dimens

1. The problem statement, all variables and given/known data
∏1: 2x + y - z = 4
∏2: 3x - 2y +z = 6

Find the parametric equation of the line L...

3. 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..

2. Apr 18, 2012

### tiny-tim

hi dimens1
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)

3. Apr 18, 2012

### dimens

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

4. Apr 18, 2012

### tiny-tim

(i assume you mean different RHSs ?)

same thing

so where's your y and z equations?​

5. Apr 18, 2012

### dimens

I was gonna substitute y into x then z=t.

6. Apr 18, 2012

### tiny-tim

should work

oh, wait a mo …
… should be 4 plus t !

7. Apr 18, 2012

### 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

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

:s

8. Apr 18, 2012

### HallsofIvy

Good so far!

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

9. Apr 18, 2012

### dimens

10. Apr 18, 2012

### 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.

11. Apr 18, 2012

### dimens

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. Apr 20, 2012

### dimens

bump?

13. Apr 20, 2012

### dimens

Anyone? :(

14. Apr 22, 2012

### dimens

triple bump... lol

15. Apr 22, 2012

### 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.

16. Apr 22, 2012

### tiny-tim

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

17. Apr 22, 2012

### dimens

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

18. Apr 23, 2012

### 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"? )