# Is this graph correct?

1. Dec 1, 2012

### f22archrer

1. The problem statement, all variables and given/known data

Write the parametric and vector equations of:

x + 2 / 3 = y − 1 / 2 = z + 5/ -3

x=-1+5t
and plot a graph of this line when t = −2, −1, 0, 1 and 2.

2. Relevant equations

vector:
r= [-2,1,-5]+t[3,2,-3]
parametric
x=3t-2
y=2t+1
z=-3t-5

using parametric equation:: i drew the graph, please check if the graph is correct at -2,-1,0,1,,2

3. The attempt at a solution

File size:
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Views:
100
2. Dec 1, 2012

### LCKurtz

Not sure where that x = -1+5t came from. But your parametric equations look correct. I didn't check all your points, but if you plotted the points to scale as apparently you did, and they came out on a straight line, aren't you pretty confident in the answers?

3. Dec 2, 2012

### f22archrer

actually i was worried about the axis ... in specific the z axis.. i just had this gut feeling like i took -z the wrong direction ...

4. Dec 2, 2012

### f22archrer

x=-1+5t is part of the question .

5. Dec 2, 2012

### Staff: Mentor

Here are the equations, using LaTeX.

$$\frac{x + 2}{3} = \frac{y - 1}{2} = \frac{z + 5}{-3}$$

When you write them in plain text, they should look like this:

(x + 2)/3 = (y - 1)/2 = (z + 5)/(-3)

What you wrote means the following:
x + $\frac{2}{3}$ = y - $\frac{1}{2}$ = z + $\frac{5}{-3}$

6. Dec 2, 2012

### Dick

Then doesn't it seem odd that you are saying x=3t-2 in the solution? I assume when you are writing "x + 2 / 3" you mean (x+2)/3, not x+(2/3). But even assuming that, your parametric form looks screwed up. Put x=(-1)+5t into (x+2)/3=(y-1)/2 and solve for y in terms of t. The 't' in your equations is the wrong parameter. Your are going to get the same line, but the points corresponding to a given value of t will be different.

Last edited: Dec 2, 2012
7. Dec 17, 2012

### f22archrer

hey , sorry for late response but if i put x=-1+5t in [x+2]/3= [y-1]/2=[z+5]/-3 what will i get after... how do i solve for y

8. Dec 17, 2012

### SammyS

Staff Emeritus
Are you asking how to solve
$\displaystyle \frac{(-1+5t)+2}{3}=\frac{y-1}{2}$​
for y ?

Give it a try.

What do you get?

9. Dec 18, 2012

### f22archrer

i didnt mean how i will solve it..rather i was concerned about what too do after i get the Y value

y =[5+10t]/3

10. Dec 18, 2012

### SammyS

Staff Emeritus
That depends.

What is it you're trying to do next?

11. Dec 18, 2012

### HallsofIvy

Staff Emeritus
Then you have two different expessions for x a function of t. What are you supposed to use?

12. Dec 18, 2012

### f22archrer

next i need to find vector and parametric equation and then plot for t ...|

13. Dec 18, 2012

### LCKurtz

This seems like a very strange way to phrase the problem, making me wonder what the exact statement of the problem is.

Taking the problem as you have quoted, don't use $t$ here. You apparently should write:
x=3s-2
y=2s+1
z=-3s-5
using s for your parameter. Now if x = -1+5t you can put that in and get s in terms of t. Then use that in these last three equations. Of course, this all assumes I understand what you are trying to do.