Vector equation of a plane through (-2,2,1) with parallel vectors (1,1,2) and (2,1,-1)

[kent davidge]

Homework Statement

What's the vector equation for a plane which contains the point $(-2,2,1)$ and whose vectors $(1,1,2)$ and $(2,1,-1)$ are parallel to it?

Homework Equations

I think the relevant here is

- The plane equation.

The Attempt at a Solution

[/B]
We can go through demanding that a vector on it can always be expressed as $(-2,2,1) + a(1,1,2) + b(2,1,-1)$ but I did in a different manner.

I first took the vector product of the two given vectors, which is $(-3,5,-1)$ and then constructed the equation of the plane, passing on the given initial point $(-2,2,1)$. It is $-3(x+2)+5(y-2)-(z-1) = 0$. Then I considered an arbitrary point in the plane having "coordinates" $(a,b,c)$ and I substituted these components into the equation above. Calling the "point" (vector) $P$ I ended up with $P(t,u) = (t,u,-3t+5u-15)$ with $\{t,u \} \in \mathbb{R}$.

It seems to satisfy the plane equation as it's constructed to satisfy it. Never the less, I would like to know whether this is a valid answer or not.

 

[Orodruin]

You can check this by expressing a and b in terms of t and u using two of the components or vice versa. If the third component is then also the same it is the same plane.