Find the parametric equations of Line 2

Question

I am given that the parametric equations of line 1 are $x = 3 - 3t$, $y = -2 + t$, $z = 1 + 6t$ and that line 2 passes through the point $(-6,4,-3)$ and is parallel to line one. Should I set the parametric equations to be equal to the point and solve the system ?

Answer 1

Should i set the parametric equations to be equal to the point and solve the system ?

You can, but it is better to solve the problem using geometric interpretation of the equation as below:

Write the equation of line 1 as: $$ \boldsymbol l_1(t) = (-3,1,6) t + (3,-2,1) = \boldsymbol a t + \boldsymbol b $$ Now try to interpret the meaning of the vectors $\boldsymbol a$ and $\boldsymbol b$:

1) Since $\boldsymbol l'_1(t) = \boldsymbol a$, vector $\boldsymbol a$ is tangent to, and thus, parallel to the line 1

2) Since $\boldsymbol l_1(0) = \boldsymbol b$, $\boldsymbol b$ is a point on the line.

As line 2 is parallel to line 1, thus parallel to $\boldsymbol a$, and passes $\boldsymbol c = (-6,4,-3)$, it must have the form

$$ \boldsymbol l_2(t) = \boldsymbol a t + \boldsymbol c. $$

Answer 2

The equation of the second line it's just $(-6-3s,4+s,-3+6s)$ because the vector $(-3,1,6)$ is the same for two lines.