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

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?
