Find parametric equations and symmetric equations for the line.

Question

Find parametric equations and symmetric equations for the line through $(-6,2,3)$ parallel to the line $\frac12x=\frac13y=z+1$.

Normally I'm given one equation for a line and i put the point and vector with vertices equal to the coefficient of $x,y,z$ but here I have 3 equations so I'm not sure how to proceed. If I plug in the coefficients I'm sure it will be wrong because there's a plus 1 next to $z$, I'm not sure if I'm supposed to convert this from symmetric equation to parallel equation or not. Or maybe it already is a symmetric equation. I checked the book for an example, can't find one, and the online solutions aren't explained very well. if someone could explain how to find these that would help. thank you.

Answer 1

Can you find two points belonging to line $\frac x2=\frac y3=z+1$ ?

$A=(0,0,-1)$ and $B=(2,3,0)$ for instance.

So the line passing through $C=(-6,2,3)$ has equation $M(t)=C+t\,\vec{AB}$

Answer 2

Hint:

From the symmetric equations $\frac12x=\frac13y=z+1$ we get the direction vector $\mathbf{v}=\langle 2,3,1\rangle$.

Answer 3

Instead of guessing at points on the line, you can approach this systematically by writing $$\frac12x=\frac13y=z+1=t.$$ You can now solve each of the three resulting equations for the variables $x$, $y$ and $z$: $$x=2t\\y=3t\\z=t-1$$ i.e., $(0,0,-1)+t(2,3,1)$. The line you seek is thus $(-6,2,3)+t(2,3,1)$. You can get the symmetric form of this parametric equation by reversing the above process: write $$x = 2t-6\\y=3t+2\\z=t+3$$ and solve each equation for $t$ to get ${x+6\over2}={y-2\over3}=z-3$.