Give the parametric equations of each faces of a cube made by the vectors $\vec i, \vec j, \vec k$
The professor told me that this is how I get the under face:
$\vec x = t1 \dot\ \vec i + t2 \dot\ \vec j , 0 \le t1,t2 \le 1 $
He also told me (not sure if I understood correctly) that this is the right face :
$\vec x = [0,1,0] + t1 [1,0,0] + t2 [0,0,1] , 0 \le t1,t2 \le 1 $
So I figured, that I have to take a point + two vectors that goes in the direction to make the faces.
$\vec x = [0,1,0] + t1 [1,0,0] + t2 [0,0,1] , 0 \le t1,t2 \le 1 $
here are my other faces
$\vec x = [0,0,0] + t1 [1,0,0] + t2 [0,0,1] , 0 \le t1,t2 \le 1 $
$\vec x = [1,0,0] + t1 [0,1,0] + t2 [0,0,1] , 0 \le t1,t2 \le 1 $
$\vec x = [0,0,1] + t1 [1,0,0] + t2 [0,1,0] , 0 \le t1,t2 \le 1 $
$\vec x = [0,0,0] + t1 [0,1,0] + t2 [0,0,1] , 0 \le t1,t2 \le 1 $
Is this corrrect ?
And
Is it ok if I never use i j k explicitly in my equations ?
Best Answer
Your first two vectors are the same, and you are missing one of the faces perpendicular to the $z$-axis: $$ \vec x = [0,0,0] + t1 [1,0,0] + t2 [0,1,0] , 0 \le t1,t2 \le 1 $$ Otherwise, it looks good.
To make sure you have them all and didn't repeat, it helps to list each one in a descriptive way; e.g. $$ \begin{array}{ccc} \text{perpendicular axis}&\text{which end}&\text{parameterization}\\ \hline x&0&[0,0,0] + t1 [0,1,0] + t2 [0,0,1]\\ x&1&[1,0,0] + t1 [0,1,0] + t2 [0,0,1]\\ y&0&[0,0,0] + t1 [0,0,1] + t2 [1,0,0]\\ y&1&[0,1,0] + t1 [0,0,1] + t2 [1,0,0]\\ z&0&[0,0,0] + t1 [1,0,0] + t2 [0,1,0]\\ z&1&[0,0,1] + t1 [1,0,0] + t2 [0,1,0]\\ \end{array} $$ where $0 \le t1,t2 \le 1$.