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### Section 12.2 : Equations of Lines

2. Give the equation of the line through the point $$\left( { - 7,2,4} \right)$$ and parallel to the line given by $$x = 5 - 8t$$, $$y = 6 + t$$, $$z = - 12t$$ in vector form, parametric form and symmetric form.

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Okay, regardless of the form of the equation we know that we need a point on the line and a vector that is parallel to the line.

We were given a point on the line so no need to worry about that for this problem.

The parallel vector is really simple to get as well since we were told that the new line must be parallel to the given line. We also know that the coefficients of the $$t$$’s in the equation of the line forms a vector parallel to the line.

So,

$\vec v = \left\langle { - 8,1, - 12} \right\rangle$

is a vector that is parallel to the given line.

Now, if $$\vec v$$ is parallel to the given line and the new line must be parallel to the given line then $$\vec v$$ must also be parallel to the new line.

Show Step 2

The vector form of the line is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle { - 7,2,4} \right\rangle + t\left\langle { - 8,1, - 12} \right\rangle = \left\langle { - 7 - 8t,2 + t,4 - 12t} \right\rangle }}$ Show Step 3

The parametric form of the line is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 7 - 8t\hspace{0.5in}y = 2 + t\hspace{0.5in}z = 4 - 12t}}$ Show Step 4

To get the symmetric form all we need to do is solve each of the parametric equations for $$t$$ and then set them all equal to each other. Doing this gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 7 - x}}{8} = y - 2 = \frac{{4 - z}}{{12}}}}$