Finding parametric equations of a line

calculus

In my textbook there are two questions:

  1. Find parametric equations for the line through $P(-2,0,3)$ and $Q(3,5,-2)$

  2. The line through $(2,4,5)$ perpendicular to the plane $3x+7y- 5z=21$

I know that to get the parametric equations of a line, you need a vector parallel to that line and a point on the line. So question 1) seems pretty straightforward.

The vector $\vec{PQ}=<2,-1,3>$ is obviously parallel to the line since it includes the line. So the answer is

$x=1+2t$

$y=2-t$

$z=3t$

However I'm confused for how to get 2). I need a vector that is parallel to the line. But the solution says

$x=2+3t$

$y=4+7t$

$z=5-5t$

I'm just confused because I thought I needed a vector that is parallel to the line, but in this question, I have one that is perpendicular to the line.

Best Answer

You have some mistakes in your work for $(1)$.

$\vec {PQ} = (3, 5, -2) - (-2, 0, 3) = (5, 5, -5)$

So direction vector of the line is $(1, 1, -1)$ and the equation of the line is,

$(-2, 0, 3) + (1, 1, -1) t$

For $(2)$, note that if equation of a plane is $a x + by + cz = d$, vector $(a, b, c)$ is normal to the plane. Here equation of the plane is $3x+7y- 5z=21$ and hence vector $(3, 7, -5)$ is normal to the plane. As the line that we are seeking is also perpendicular to the plane, we take its direction vector as $(3, 7, -5)$.

That leads to equation of the line as $(2, 4, 5) + (3, 7, -5) t$

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