1. ## Vector Parametric Equation

How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t

2. Originally Posted by acg716
How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t
I'm guessing that the plane contains the line you mention. If so, then the vector

$\bold{u} = < 1, 4, -2>$

is on the plane. We can also find another vector on the plane, a point on the line to the point given

$\bold{v} = <5-(-3),0-9 , 3 -10> = <8,-9,-7>$

The normal to the plane is then obatin from

$\bold{n} = \bold{u} \times \bold{v}$ (*)

If $\bold{n} = $

then the plane is given by

$n_1(x+3) + n_2(x-9) + n_3(x-10) = 0$

So all you need to do is find n *

3. When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...

4. Originally Posted by acg716
When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...
Well, when you substitute the point $(-3,9,10)$ into this plane we get 0, right? And when we substitute $x = 5+t,\;\;y = 4t,\;\;z = 3-2t$ we get

$-46(5+t+3)-9(4t-9)-41(3-2t-10)=$
$-46(8+t)-9(4t-9)-41(-7-2t)=$
$-368 - 46 t +81-36t + 287 + 82t = 0$

So both the point and line lie on the plane. Did the question say that the line is perpendicular to the plane?

5. yes it is perpendicular, so sorry that i forgot to include that!

6. Originally Posted by acg716
yes it is perpendicular, so sorry that i forgot to include that!
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so

$\bold{n} = <1,4,-2>$

so the line is

$(x+3) + 4(x-9) - 2(x-10)=0$

See, way easier.

7. ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:

Find an equation of the plane through the point and perpendicular to the given line.

(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t

8. Originally Posted by acg716
ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:

Find an equation of the plane through the point and perpendicular to the given line.

(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t
What answer are you given? Are you given $x + 4y - 2z = 13$?

9. I'm not given an answer, its on my webassign homework so when I submit it it tells me if I'm right or not.

10. Originally Posted by danny arrigo
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so

$\bold{n} = <1,4,-2>$

so the line is

$(x+3) + 4(x-9) - 2(x-10)=0$

See, way easier.
Originally Posted by danny arrigo
What answer are you given? Are you given $x + 4y - 2z = 13$?
Did you try both answers? If so, I don't know what to suggest. Can some of the other regulars chime in here.

11. The second one was correct, thank you so much!!!