# Need some help with vector and parametric form

1. Aug 13, 2011

### paperdoll

1. The problem statement, all variables and given/known data
I have three questions regarding vectors in parametric/circle form. I understand that there is a starting point and a direction vector, but I just can't seem to get my head around this

2. Relevant equations
1. Rewrite y=3x-1 in vector form.
2. Rewrite r=ti+(2t-1)j in parametric form
3. Where does the circle |r-(i+2j)|=root 5 cut the coordinate axes?

3. The attempt at a solution
For 1. I got -j+t(2i+5j) but I'm dubious my answer since I tried it after watching a bunch of youtube videos regarding such questions.

For no. 2, I am really lost where to even start :S

For question 3, I know the radius is root 5, and the centre point is i+2j but not sure where to go from here...thanks if you can help ^^ I'm new here

2. Aug 13, 2011

### I like Serena

Welcome to PF, paperdoll!

For 1: no your expression is not correct.
How did you get your expression? That would help me understand what I should explain.

For 2: the parametric form is r = p + t q.
You need to find what needs to be multiplied by t, and what is not multiplied by t.

For 3: Did you try to draw it?
For starters, what is the distance of the center of the circle to the origin?

Last edited: Aug 13, 2011
3. Aug 13, 2011

### HallsofIvy

Staff Emeritus
Additional. In 1, x is NOT a constant. In 2, separate the vectors into x and y components.

4. Aug 13, 2011

### paperdoll

For 1, I substituted numbers into the expression y=3x-1. In this case, x=2, then y=5, because I read that you needed to find a vector parallel so then I got 2i+5j. and I somehow got -j in the beginning because I thought the starting point of the vector was "-1".

For 2, I have factorised the t out of it to get r=-j+t(i+2j)...not sure where to go from here.

For 3, I would say that the distance from the centre to the origin will be:
1^2+2^2=x^2 x=root 5

does this mean it touches at the origin then? since the original radius is root 5. I drew a picture of it, seems like it touches on the origin, once on the y-axis and once again on the x-axis. from that, I got it touches at x=2, y=0 and y=4, x=0...but this is from diagram, not from the equation

Thank you for replying to my post I like Serena :)

5. Aug 13, 2011

### HallsofIvy

Staff Emeritus
The vector equation for any line is of the form $\vec{r}(t)= \vec{p}+ t \vec{v}$ where $\vec{p}$ is the position vector of a point on the line and $\vec{v}$ is a vector in the direction of the line.

That is going the wrong way! The original equation said r= ti+ (2t-1)j.
So x= ? and y= ?

I would consider the "distance from the centre to the origin" as not particularly relevant.
You know that the center is at (1, 2) and the radius is $\sqrt{5}$, yes, that means that circle passes through the origin but there are two other points where the circle crosses the x and y axes. I presume you see, from the centre and origin information, that $(x- 1)^2+ (y- 2)^2= 5$
Where it crosses the x-axis y= 0 so $(0-1)^2+ (y- 2)^2= 5$. Solve for y.
Where it crosses the y-axis x= 0 so $(x- 1)^2+ (0- 2)^2= 5$. Solve for x.

Last edited: Aug 14, 2011
6. Aug 13, 2011

### I like Serena

You were right that a starting point of the line was "-1".
So your -j is right!

You also found another point on the line 2i+5j.

However, the vector you need is the vector from your starting point to this other point on the line.
So you need to subtract the one from the other.

Hmm, as far as I'm concerned that is the parametric form.
But HallsofIvy has a point.
Your teacher no doubt intended you to find an expression for x and an expression for y.
Sorry to set you on the wrong way here.

So you need to gather everything that is multiplied by i. That is your x.
And you need to gather everything that is multiplied by j. That is your y.

Yes!
In my opinion, these things are best learned from first drawing what you are doing.
And as you can see you already have the solution.
I let you calculate the distance to the origin so you'd have an instant handle on the thing. ;)

Alternatively you can calculate them from the equation.
That will be what your teacher intended.

To do that you need to fill in x=0 (or r=yj) in your equation and solve it.
That will give you all intersections with the y-axis.

After that you need to fill in y=0 (or r=xi) in your equation and solve it again.
That will give you all intersections with the x-axis.

Last edited: Aug 13, 2011
7. Aug 13, 2011

### paperdoll

Last edited: Aug 13, 2011
8. Aug 13, 2011

### I like Serena

As for number 2, you need to leave out "i" in your expression for x, but what you have is right!

Here's a picture for number 1.
(I like pictures! )

http://www.netcomuk.co.uk/~jenolive/vecline.gif

I borrowed it from: http://www.netcomuk.co.uk/~jenolive/vect3.html

In this picture "a" is the position vector on a point, and b is the "direction vector" in the direction of the line.
To find "a" you can take any point, for instance the one you already had: -j.
To find a suitable "b", you can take any 2 points on the line and subtract them from each other.

Can you think of 2 points? And subtract them?

9. Aug 13, 2011

### paperdoll

okey, so hmm two points {x=2, y=5} and {x=4, y=11} so I resulted in 2i+6j which can simplify to i+3j? I am a bit confused what do you mean by "subtracting points" because the diagram looks like the "b" value is not on the r line itself O__o

LOL at *I like pictures*, so do I.

10. Aug 13, 2011

### I like Serena

Yep! That's it!

You have just learned vector subtraction!
When you subtract one point from the other, you get the vector that connects the 2 points.
"a" and "r" are vectors from the origin to a point on the line.
"b" is a vector from a point on the line to a point on the line.

So what's the final expression?

(Btw, you could also have used {x=0, y=-1}.)

11. Aug 13, 2011

### paperdoll

final expression I got is -j+t(i+3j) :shy:

thank you so much for the help ^^ it helped so much!

12. Aug 13, 2011

### I like Serena

You're welcome!

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