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## Homework Statement

Find the equation of the plane that contains the lines,

[tex]x = 2s + 2[/tex]

[tex]y = 3s - 4[/tex]

[tex]z = -5s + 2[/tex]

and

[tex]x = 4t + 3[/tex]

[tex]y = 6t + -4[/tex]

[tex]z = -10t + 5[/tex]

## 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,

[tex]\vec{PQ} = <1,0,3>[/tex]

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

[tex]\vec{n} = <x,y,z>[/tex]

That is perpendicular to,

[tex]\vec{PQ}[/tex].

Two vectors are

**perpendicular**when their dot product is 0.

So,

[tex]\vec{PQ} \cdot \vec{n} = 0[/tex]

This will give us the following equation,

[tex]1x + 0y + 3z = 0[/tex]

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

[tex]x=3[/tex]

[tex]y=0[/tex]

[tex]z=-1[/tex]

So then the vector n would be defined as follows,

[tex]\vec{n} = <3,0,-1>[/tex]

Then I could define my plane as follows,

[tex]

3(x-3) -(z-5) = 0

[/tex]

What's wrong with this approach?

**EDIT: Whoops I think I found my mistake. I want the vector,**[tex]\vec{n}[/tex]

**to be normal to,**[tex]\vec{PQ}[/tex] and [tex]\vec{v}[/tex]

**where v is the direction vector of one the lines.**

So,

[tex]\vec{n} = <-9,11,3>[/tex]

So the equation of the plane is,

[tex]-9x + 11y +3z = -56[/tex]

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