Finding the Equation of a Plane Containing Two Parallel Lines

[jegues]

Homework Statement

Find the equation of the plane that contains the lines,

x = 2s + 2

y = 3s - 4

z = -5s + 2

and

x = 4t + 3

y = 6t + -4

z = -10t + 5

Homework Equations

The Attempt at a Solution

One can quickly note that the two lines are parallel to each other, because the direction vector of the second line is simply 2 times the direction vector of the first line.

We can find a point on the first line, call it P(2,-4,2).

We can also find a point on the second line, call Q(3,-4,5)

Now if we draw a vector from P to Q then,

\vec{PQ} = <1,0,3>

Now all we need to do is find another vector call it,

\vec{n} = <x,y,z>

That is perpendicular to,

\vec{PQ}.

Two vectors are perpendicular when their dot product is 0.

So,

\vec{PQ} \cdot \vec{n} = 0

This will give us the following equation,

1x + 0y + 3z = 0

Now I could easily pick values for x,y, and z that would satisfy this. For example,

x=3

y=0

z=-1

So then the vector n would be defined as follows,

\vec{n} = <3,0,-1>

Then I could define my plane as follows,
<br /> 3(x-3) -(z-5) = 0<br />

What's wrong with this approach?

EDIT: Whoops I think I found my mistake. I want the vector, \vec{n} to be normal to, \vec{PQ} and \vec{v} where v is the direction vector of one the lines.

So,

\vec{n} = &lt;-9,11,3&gt;

So the equation of the plane is,

-9x + 11y +3z = -56

 

[ehild]

It is correct.

ehild