# Thread: scalar equation of a line -> parametric

1. ## scalar equation of a line -> parametric

The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

3x + 5y + 2 = 0

How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?

2. Originally Posted by theowne
The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

3x + 5y + 2 = 0

How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?
Draw the line. A vector in the direction of the line should now be obvious ......

3. Originally Posted by theowne
The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

3x + 5y + 2 = 0

How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?
Rewrite the equation:

$3x+5y +2 = 0 ~\implies~\langle 3,5 \rangle \cdot \langle x,y \rangle + 2 = 0$

$\vec n = \langle 3,5 \rangle$ is the normal vector of the line.

You get the direction vector of the line if you change the order of the components and change the sign at one component:

The direction vector is now: $\vec d = \langle 5,-3 \rangle$ or $\vec d = \langle -5,3 \rangle$

Remark: This procedure works only in 2D.

4. Originally Posted by theowne
...

3x + 5y + 2 = 0

... how can I get this into vector form?
Change this equation into parametric form:

$\left| \begin{array}{l}x =0+ t \\y = -2 - \frac35 t\end{array}\right.$

Now you can rewrite this form into an equation in vector form:

$\langle x,y \rangle = \langle 0,-2 \rangle + t \cdot \langle 1~,~-\frac35 \rangle$