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
$\displaystyle \bold{u} = < 1, 4, -2>$
is on the plane. We can also find another vector on the plane, a point on the line to the point given
$\displaystyle \bold{v} = <5-(-3),0-9 , 3 -10> = <8,-9,-7>$
The normal to the plane is then obatin from
$\displaystyle \bold{n} = \bold{u} \times \bold{v}$ (*)
If $\displaystyle \bold{n} = <n_1,n_2,n_3>$
then the plane is given by
$\displaystyle n_1(x+3) + n_2(x-9) + n_3(x-10) = 0$
So all you need to do is find n *
Well, when you substitute the point $\displaystyle (-3,9,10)$ into this plane we get 0, right? And when we substitute $\displaystyle x = 5+t,\;\;y = 4t,\;\;z = 3-2t$ we get
$\displaystyle -46(5+t+3)-9(4t-9)-41(3-2t-10)=$
$\displaystyle -46(8+t)-9(4t-9)-41(-7-2t)= $
$\displaystyle -368 - 46 t +81-36t + 287 + 82t = 0$
So both the point and line lie on the plane. Did the question say that the line is perpendicular to the plane?