Vector and Parametric equations, i need help!

1. Jul 15, 2011

kathialopez

1. Determine vector and parametric equations for the line through the point A(2, 5) with direction vector = (1, −3).

2.Determine parametric equations for the line through (-2, 3) and parallel to the line with vector equation = (−2, 1) + t(6, 4).

3.Find vector and parametric equations for the line with equation 2x + y + 3 = 0.

4.Determine parametric equations for the line with scalar equation 4x – y + 5 = 0.

5. A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.

I need extreme help wiht all of these questions. i understand nothing in this unit and im teaching myself :|

2. Jul 15, 2011

timthereaper

So what don't you understand about vector and parametric equations? Maybe we can help you understand them.

3. Jul 15, 2011

jambaugh

As far as understanding what's going on is concerned, I teach my students that vectors in the plane represent actions on points... the vector V = <a,b> moves the point (x,y) to the point (x',y') = (x+a,y+b).

Thinking of them this way we then understand
a.) that vectors and points are distinct types of mathematical objects,
b.) how to add vectors (compose actions),
c.) why vectors can be placed anywhere on the plane (choosing which point to move).

Note the use of angle brackets to distinguish points and vectors.

Then we define specifically a position vector as the vector associated with a point which moves the origin to that point (and so will have the same numerical components).

Getting the above clear first one can then get a little sloppy and "identify" position vectors with the points they indicate (while in the back of one's mind keeping clear that this is dependent on a choice of origin point.) You can then drop the angle bracket notation for vectors if you like, but it is good practice to keep it up.

OK, now a line is a set of points and you can indicate it by giving a relation for the coordinates e.g. the points (x,y) such that ax+by = c.

Given a point (e.g. p=(2,5) and a direction vector v=<1,-3> ) you can indicate all other points by the fact that you can reach them by starting at p and moving in the direction of v by some amount. (say t times v's length).

Just so we work only with vectors, lets instead start at the origin (0,0) then move to p using its position vector (e.g. <2,5> ) and then move off in the v direction by t times its length. The composite action is then: <2,5> + t <1,-3> = <2,5> + <t,-3t> = <2+t,5-3t>.

Acting on the origin gives you the point (2+t, 5-3t). And that is your parametric form:
(x,y) such that x=2+t, y=5-3t.

That will help you with 1.) and you should be able to handle 2.) using the same techniques.

Now here's a handy bit of info when you're given a linear equation: $ax + by = c$.
Letting $\vec{r}=\langle x,y \rangle$ be the position vector for the point $(x,y)$, and then considering the vector $\langle a, b\rangle$ with components corresponding to coefficients of the linear equation you can rewrite the equations in vector form as:
$\langle a,b\rangle\bullet \langle x,y\rangle = c$ where $\bullet$ is the dot product.

Now consider some direction vector for this line, say $\vec{v}$. Remember how I said to think of vectors as actions moving points. The direction vector moves points on the line to other points on the line. Let $\vec{v} = \langle h,k\rangle$, where h is a change in x, and k is a change in y coordinates. Since this point is still on the line it still satisfies the equation for the line and so:
$\langle a,b \rangle \bullet( \langle x,y \rangle + \langle h,k \rangle) = c$
but since the original equation is also true this means:
$\langle a,b \rangle \bullet \langle h , k \rangle = 0$
In short the vector $\langle a,b\rangle$ must be perpendicular to the direction of the line.

This fact is a tool for solving 3.) through 5.).