To prove that a vector $x(t)$ lies in a plane.

Question

Prove that vector $x(t)=t\,\hat{i}+\left(\dfrac{1+t}{t}\right)\hat{ j}+\left(\dfrac{1-t^2}{t})\right)\hat{k}$ lies in a curve. I am puzzled. Don't know how to approach it.

Answer 1

The vectors lie in a plane, because the functions $\{f=t,g=(1+t)/t,h=(1-t^2)/t\}$ are not linearly independent. Namely, $f-g+h=-1$.

It is obvious that they also lie in a curve: there is a single real parameter to be varied!

Answer 2

I would have used "x", "y", and "z" rather than "f", "g", and "h" but the point is valid. This curve lies in the plane x- y+ z= 1.

(I am assuming that i, j, and k are the unit vectors in the x, y, and z directions, respectively, not "quaternions".)

Answer 3

Hint: Rewrite $$ x(t) = t \hat{i} + \big{(}\frac{1+t}{t}\big{)}\hat{j} + \big{(}\frac{1-t^2}{t}\big{)}\hat{k} $$ as

$$ x(t) =  \hat{j} + t(\hat{i} - \hat{k}) + \frac{1}{t}(\hat{j} + \hat{k}). $$

Now when you revise the definition of plane and study the rewritten expression, you are able to conclude that $x(t)$ lies in a plane. Furthermore, on what plane does the curve lie on?