Vector parametric equation of a line

[roam]

Find a vector parametric equation of the line in R^{2} with equation 2x-3y = 4



Attempt at a solution

I haven't seen this type of question before so I don't know where to start. I suppose that the equation 2x-3y = 4 is a vector equation of that line and is in the form x = x₀ + tv. I just don't know how to split this vector equation up and try to write the parametric vector equation.

Can someone please explain to me what to do. Many Thanks.

 

[danago]

2x - 3y = 4 is the cartesian equation, not the vector equation.

If a line has the vector equation r = x₀ + tv, then it passes through the point x₀ and is parallel to the vector v.

Can you find a point that the line 2x - 3y = 4 passes through? Can you find a vector to which it is parallel?

 

[tiny-tim]

Find a vector parametric equation of the line in R^{2} with equation 2x-3y = 4
…
the form x = x₀ + tv.

Hi roam!

Yes, that is the right form.

x₀ can be any point on the line.

So just choose some point on the line (at random), and chug away.

 

[HallsofIvy]

One of your difficulties is that there are many different parametric equations for the same line.

Yes, you want x= at+ b, y= ct+ d. But you can choose "t" anyway you want. You might, for example, choose t to be x itself: that is x= t. Now y= ?

 

[roam]

Hi Tim!

I think I should even write the equation 2x-3y=4 in the form:

y = \frac{2x -4}{3} for some t

And I choose some random point on the line, (2,0)

Now for x = x₀ + bt, y = y₀+ bt

If x = t, like Hall said, y = \frac{2t -4}{3}

is this the vector parametric equation of the line? Do I need to specify a "t" as well?

 

[roam]

Is it (t, \frac{2t -4}{3}) ?

 

[tiny-tim]

vector parametric equation

Hi roam!

Hi Tim!

I think I should even write the equation 2x-3y=4 in the form:

y = \frac{2x -4}{3} for some t

And I choose some random point on the line, (2,0)

So far, so good.

(though you could have got that straight from 2x-3y=4, couldn't you? )

At this point I'm going to disagree with HallsofIvy, and say that I think the question is asking for a pure vector equation, rather than a coordinate-based one.

Coordinate-based would be of the form x = f(t), y = g(t).

Pure vector would be of the form r(t) = a + bt, where a and b are both constant vectors.

So the answer would be r(t) = (2,0) + (?,?)t.

 

[roam]

So the answer would be r(t) = (2,0) + (?,?)t.

Hi Tim,

How do we determine the direction vector? Thanks!

 

[tiny-tim]

Hi Tim,

How do we determine the direction vector? Thanks!

oh come on!

just look at the original equation: 2x-3y = 4 …

if x goes up by 1, then y goes up by … ?

 

[HallsofIvy]

Very little difference between "parametric equations", x= f(t), y= g(t), as I gave, and "vector parametric equation", \vec{r}= f(t)\vec{i}+ g(t)\vec{j}.

 

[roam]

oh come on!

just look at the original equation: 2x-3y = 4 …

if x goes up by 1, then y goes up by … ?

Would it be 3/2?

 

[BoundByAxioms]

Would it be 3/2?

Hmm, try again.

You yourself said that y=\frac{2x-4}{3}. What is the slope of that line? How would you express that slope in the form of a direction vector?

 

[roam]

Hmm, try again.

You yourself said that y=\frac{2x-4}{3}. What is the slope of that line? How would you express that slope in the form of a direction vector?

m = 2/3 ?

 

[roam]

if x goes up by 1, then y goes up by … ?

y = \frac{2x-4}{3}
if x = 1 then y = -10/3


Is that it? can we plug it in the equation (2,0) + t(*,*)?

 

[tiny-tim]

m = 2/3 ?

Yes!

y = \frac{2x-4}{3}
if x = 1 then y = -10/3

Is that it? can we plug it in the equation (2,0) + t(*,*)?

uh? wots happening?

(x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2).

 

[BoundByAxioms]

Yes!


uh? wots happening?

(x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2).

You're golden.

 

[BoundByAxioms]

y = \frac{2x-4}{3}
if x = 1 then y = -10/3


Is that it? can we plug it in the equation (2,0) + t(*,*)?

If x=1, then y=-2/3. Just for clarification. Where did you get the -10/3 from?