Parametric equations and symmetric equations

[maff is tuff]

Homework Statement

Find parametric equations and symmetric equations for the line through the points (0,1/2,1) and (2,1,-3)

Homework Equations

The Attempt at a Solution

I started out graphing the points and connecting them with a straight line. I called the first point P and second Q. So the vector PQ = <2,1/2,-4>. So my vector r_0 is <0,1/2,1> so vector r = r_0 + tv

so <x,y,z>=<0,1/2,1> +t<2,1/2,-4>

<x,y,z> = <0+2t, 1/2 + t/2, 1-4t>

So my parametric equations are:

x=2t
y=1/2 + t/2
z=1-4t

And my symmetric eqs. are:

x/2 = 2y-1 = (z-1)/-4

This answer is wrong and I've done it a few times any hints? Thanks

 

[Mark44]

Homework Statement

Find parametric equations and symmetric equations for the line through the points (0,1/2,1) and (2,1,-3)

Homework Equations

The Attempt at a Solution

I started out graphing the points and connecting them with a straight line. I called the first point P and second Q. So the vector PQ = <2,1/2,-4>. So my vector r_0 is <0,1/2,1> so vector r = r_0 + tv

so <x,y,z>=<0,1/2,1> +t<2,1/2,-4>

<x,y,z> = <0+2t, 1/2 + t/2, 1-4t>

So my parametric equations are:

x=2t
y=1/2 + t/2
z=1-4t

Looks good. This is what I get, too.

And my symmetric eqs. are:

x/2 = 2y-1 = (z-1)/-4

Instead of 2y - 1, write the expression in the middle as (y - 1/2)/(1/2). This is equal to what you have, but if your work is being computer-graded, it might not be smart enough to recognize different forms of the same thing.

This answer is wrong and I've done it a few times any hints? Thanks

 

[maff is tuff]

It is not online homework. My paper actually says what you said to put but I found it easier to type so I multiplied by 2. So what do you think is wrong? Or is it right and there are multiple answers? Thanks.

 

[Mark44]

What you have is also correct, because 2y - 1 = (y - 1/2)/(1/2). If you multiply the expression on the right by 2/2, you get the expression on the left.

 

[stallionx]

Homework Statement

Find parametric equations and symmetric equations for the line through the points (0,1/2,1) and (2,1,-3)

Homework Equations

The Attempt at a Solution

I started out graphing the points and connecting them with a straight line. I called the first point P and second Q. So the vector PQ = <2,1/2,-4>. So my vector r_0 is <0,1/2,1> so vector r = r_0 + tv

so <x,y,z>=<0,1/2,1> +t<2,1/2,-4>

<x,y,z> = <0+2t, 1/2 + t/2, 1-4t>

So my parametric equations are:

x=2t
y=1/2 + t/2
z=1-4t

And my symmetric eqs. are:

x/2 = 2y-1 = (z-1)/-4

This answer is wrong and I've done it a few times any hints? Thanks

Form a vector between the points:

say : B-A= <2, 0.5, -4 >

<x,y,z> - <2,1,-3> = k < 2, 0.5,-4>

and so on...

 

[Mark44]

Form a vector between the points:

say : B-A= <2, 0.5, -4 >

<x,y,z> - <2,1,-3> = k < 2, 0.5,-4>

and so on...

stallionx, if you had read the thread before posting, you would have discovered that the OP had already arrived at the solution.