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I'm sorry for the convoluted title, I couldn't figure out what to call it and am open to suggestions for edits.

I have a linear algebra textbook problem:

Show that the curve $$x=\frac{t^2+2t+2}{t^2+3t-1}, y=\frac{3t^2-6t+4}{t^2+3t-1}, z=\frac{2t^2-3t+1}{t^2+3t-1}$$

lies in a plane and determine the equation for that plane.

My solution:

If the curve is in a plane, then it should have the equation $$Ax+By+Cz=1$$

Inserting x,y,x:

$$A(t^2+2t+2)+B(3t^2-6t+4)+C(2t^2-3t+1)=t^2+3t-1$$

I made a linear equation system and solved it $$\begin{cases}(A+3B+2C)t^2=t^2\\(2A-6B-3C)t=3t\\(2A+4B+C)=-1\end{cases}$$

$A=\frac{9}{11},B=\frac{-12}{11},C=\frac{19}{11}$

The thing that bothers me about my solution is the step where I assume that all terms before $t^2, t$ and constants must be equal on both sides of the equation. I have a general sense that this is right, but how can you prove that this is the only solution to the equation? $$A(t^2+2t+2)+B(3t^2-6t+4)+C(2t^2-3t+1)=t^2+3t-1$$

Erik Eriksson
  • 863

2 Answers2

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Just to get a more algebraic perspective. Suppose you have two polynomials $p(t)$ and $q(t)$ and $p(t)=q(t)$. Then

$$r(t):=p(t)-q(t)$$

is a polynomial which is $0$ for every $t\in\mathbb{R}$. But it is not too difficult to check that a polynomial of degree $n\geq 1$ has at most $n$ roots. So it must be that $r$ is identically $0$ (i.e. all of its coefficients are $0$) and hence the coefficients of $p$ and $q$ are the same.

podiki
  • 2,273
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Take the derivatives of the left and right and evaluate at $t=0$. $L(0) = R(0)$, $\frac{dL(0)}{dt} = \frac{dR(0)}{dt}$ and $\frac{d^2L(0)}{dt^2} = \frac{d^2R(0)}{dt^2}$

You might want to study Taylor series.