Show in parametric form plane in $R^3$ I have this question in my book.
Show in parametric form the plane of $R^3$ that determined by these points :
$$(1,0,0)$$
$$(0,1,0)$$
$$(0,0,1)$$
Does $(0,0,0)$ found on this plane?
My answer
The parametric form of these points is ($\lambda_1,\lambda_2,\lambda_3 \in R$):
$$\lambda_1(1,0,0)+\lambda_2(0,1,0)+\lambda_3(0,0,1)$$
Therefore for $\lambda_1=\lambda_2=\lambda_3=0$
$$0(1,0,0)+0(0,1,0)+0(0,0,1)=(0,0,0)$$
Therefore $(0,0,0)$ found on this plane.
But for some reason the book says it false, and I don't understand why.
Any idea?
Any help will be appreciated.
 A: I don't know what "parametric form" means, but here's one way of representing your plane.  If you need it in some other form, you can find it from this one:
One way of specifying a plane in $\Bbb R^3$ is with the vector equation: $\mathbf r(s,t) = s\mathbf v_1 + t\mathbf v_2 + \mathbf a$, where $s,t \in \Bbb R$, and $\mathbf v_1, \mathbf v_2$ are two non-collinear vectors parallel to your plane, and $\mathbf a$ is a vector pointing to $1$ particular point in your plane.
If you can find ANY $\mathbf v_1, \mathbf v_2$ parallel to your plane (but not to each other) and ANY $\mathbf a$ pointing to a point in your plane, then you've got your equation.

So let's find an equation for your plane.  First off let's label $A=(1,0,0), B=(0,1,0), C=(0,0,1)$.
You need two non-collinear vectors parallel to your plane.  Two such vectors are $\vec {AB} = (0,1,0) - (1,0,0) = (-1,1,0)$ and $\vec {AC}= (0,0,1)-(1,0,0)=(-1,0,1)$.  Do you see why?
Now you need a vector pointing to your plane.  Because points and vectors can really be thought of as the same thing, you can just choose a point in your plane -- luckily you're provided with three!  So how about we use $A$?
Putting these together we have that one equation (out of an infinite number of possible equations) which specifies your plane is $\mathbf r(s,t) = s(-1,1,0) + t(-1,0,1) + (1,0,0)$.  Now check if $(0,0,0)$ is a solution to this equation.  :)
