This might seem strange, but I can't really understand how to get direction vector for a given edge.
For example: $6(x+10)=7(y+20)=7z$, and the direction vector should be $(7,6,6)$.
Another example: $7(12-x)=8(11+y)=8(13+z)$, direction vector should be $(8,-7,-7)$. How do I get normal vector, if I have direction vector, and vice versa?
What if I want to give direction or normal vector of something like this: $6x-7y+7z=52$?
Thanks for your help!
$$6(x+10)=7(y+20)=7z$$ is a line. To find a vector parallel to that line, it might be easier to put in it this form: $$\pmatrix{x \\ y \\z} = \pmatrix{x_0 \\ y_0 \\ z_0} + t\pmatrix{v_x \\ v_y \\ v_z}$$ In that form we know that $\pmatrix{x_0 \\ y_0 \\ z_0}$ is a point on the line and $\pmatrix{v_x \\ v_y \\ v_z}$ is a vector parallel to the line.
The way to do this is to set the triple equality above equal to a variable $t$. Then just solve for $x$, $y$, and $z$:
$$6(x+10)=7(y+20)=7z = t \\ \implies 6x +60= t \implies x = \frac 16 t -10 \\ \implies 7(y+20) = t \implies y = \frac 17t -20 \\ \implies 7z = t \implies z=\frac 17t$$
Therefore $$\pmatrix{x \\ y \\ z} = \pmatrix{-10 \\ -20 \\ 0} + t\pmatrix{\frac 16 \\ \frac 17 \\ \frac 17}$$
So a vector parallel to this line is $\pmatrix{\frac 16 \\ \frac 17 \\ \frac 17}$, or scaling this to get rid of the fractions, $\pmatrix{7 \\ 6 \\ 6}$.
$$6x−7y+7z=52$$ is not line. It's a plane. To tell which vector is normal to it, let's rewrite it in this form
$$(n_x, n_y, n_z) \cdot (x-x_0, y-y_0, z-z_0) = 0$$
Then by the definition of orthogonality, $(n_x, n_y, n_z)$ will be a vector orthogonal (normal) to this plane.
It's pretty easy to see how to get it in this form, just move the $52$ over and subtract it from one of the variables -- like $z$ for instance:
$$6x−7y+7z=52 \\ \implies 6x-7y+(7z-52)=0 \\ \implies 6x - 7y + 7(z-\frac {52}{7}) = 0 \\ \implies (6, -7, 7)\cdot (x, y, z-\frac {52}{7})=0$$
So clearly $(6,-7,7)$ is a vector normal to this plane.
If you have a line written in the form $\displaystyle \frac{x-h}{a_1}=\frac{y-k}{a_2}=\frac{z-l}{a_3}$,
$\;\;\;$then $\langle a_1, a_2, a_3\rangle$ is a direction vector for the line.
You can write the given lines in the form $\frac{x+10}{7}=\frac{y+20}{6}=\frac{z}{6}$ and $\frac{x-12}{-8}=\frac{y+11}{7}=\frac{z+13}{7}$.
If you have a plane written in the form $ax+by+cz=d$,
$\;\;\;$then $\langle a, b, c\rangle$ is a normal vector for the plane.
(Lines have direction vectors, and planes have normal vectors.)
If you can remember the Form of Plane : $$a(X-X_0)+b(Y-Y_0)+c(Z-Z_0)=0 $$or $$aX+bY+cZ+d=0$$ so the number with $x$,$y$,$z$ are normal vector's point! So The answer is $(6,-7,7)$ actually.
