# Math Help - Finding an equation of the plane.

1. ## Finding an equation of the plane.

I seem to be converging on an answer for this one, and if somebody could show me the method then I'll have learned a lesson and will be able to move forward. Here's the question:

The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1 + t, z = 2 - t.

I'm still going through my notes and realize the answer might be simple.

2. Originally Posted by Undefdisfigure
I seem to be converging on an answer for this one, and if somebody could show me the method then I'll have learned a lesson and will be able to move forward. Here's the question:

The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1 + t, z = 2 - t.

I'm still going through my notes and realize the answer might be simple.
To determine a plane you need 1 fixed point which belongs to the plane and 2 vectors which span (?) the plane.

I f $\vec x = \left(\begin{array}{c}x\\y\\z\end{array}\right)$ then the equation of the line can be written like:

$\vec x = \left(\begin{array}{c}0\\3\\2\end{array}\right) + t \cdot \left(\begin{array}{c}3\\1\\-1\end{array}\right)$ . I take (0, 3, 2) as the fixed point of the plane.

Then the 2nd vector is $\vec u = \left(\begin{array}{c}1\\2\\3\end{array}\right) - \left(\begin{array}{c}0\\3\\2\end{array}\right) = \left(\begin{array}{c}1\\-1\\1\end{array}\right)$

Therefore the equation of the plane is:

$\vec x = \left(\begin{array}{c}0\\3\\2\end{array}\right) + t \cdot \left(\begin{array}{c}3\\1\\-1\end{array}\right) + s \cdot \left(\begin{array}{c}1\\-1\\1\end{array}\right)$

3. I think you've made a small mistake. When you have the parametric equation
x = 3t, y = 1 + t, z = 2 -t you get the fixed point (0, *1*, 2) not (0, *3*, 2).
Unless there is some convention in linear algebra that I don't remember now. I'm assuming you took the constants of the parametric equation to be the fixed point.

4. Originally Posted by Undefdisfigure
I think you've made a small mistake. When you have the parametric equation
x = 3t, y = 1 + t, z = 2 -t you get the fixed point (0, *1*, 2) not (0, *3*, 2).
Unless there is some convention in linear algebra that I don't remember now. I'm assuming you took the constants of the parametric equation to be the fixed point.
Ooops, you are right. Sorry for that mistake.

You have to change the 2nd direction vector of the plane too!

$
\vec u = \left(\begin{array}{c}1\\2\\3\end{array}\right) - \left(\begin{array}{c}0\\1\\2\end{array}\right) = \left(\begin{array}{c}1\\1\\1\end{array}\right)
$

The equation of the plane becomes now:

$
\vec x = \left(\begin{array}{c}0\\1\\2\end{array}\right) + t \cdot \left(\begin{array}{c}3\\1\\-1\end{array}\right) + s \cdot \left(\begin{array}{c}1\\1\\1\end{array}\right)
$