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?
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:
$\displaystyle 3x+5y +2 = 0 ~\implies~\langle 3,5 \rangle \cdot \langle x,y \rangle + 2 = 0$
$\displaystyle \vec n = \langle 3,5 \rangle$ 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: $\displaystyle \vec d = \langle 5,-3 \rangle$ or $\displaystyle \vec d = \langle -5,3 \rangle$
Remark: This procedure works only in 2D.
Change this equation into parametric form:
$\displaystyle \left| \begin{array}{l}x =0+ t \\y = -2 - \frac35 t\end{array}\right.$
Now you can rewrite this form into an equation in vector form:
$\displaystyle \langle x,y \rangle = \langle 0,-2 \rangle + t \cdot \langle 1~,~-\frac35 \rangle$