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?

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.

• The "plane" that you're describing in your solution is the span of the vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, which is all of $\mathbb{R}^3$. What the question is really asking for is a parametrization of the plane which contains the points with the given coordinates. – Tom Dec 2 '14 at 1:44
• I'd start by getting the equation of the plane in the form $ax + by + cz = d$. From there you can parametrize the plane by setting $x=s$, $y=t$, and then solving for $z$ in terms of $s$ and $t$. – eigenchris Dec 2 '14 at 1:48

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. :)