# Plane containing 2 lines

## Homework Statement

Find the equation of the plane that contains the lines,

$$x = 2s + 2$$

$$y = 3s - 4$$

$$z = -5s + 2$$

and

$$x = 4t + 3$$

$$y = 6t + -4$$

$$z = -10t + 5$$

## 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,
$$3(x-3) -(z-5) = 0$$

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,

$$\vec{n} = <-9,11,3>$$

So the equation of the plane is,

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

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