1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Tom Dec 2 '14 at 1:44
  • $\begingroup$ 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 $t$. $\endgroup$ – eigenchris Dec 2 '14 at 1:48
3
$\begingroup$

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. :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.