1. ## first octant question

Can someone help me out on if the bottom left corner of a box is at (0,0,0 ) and the top right corner is in the first octant on the plane 3x + 2y + z = 1, how do I determine the coordinates for the top right corner??? I know x,y and z must be positive but???? It also states that this box faces parallel to the coordinate planes.

2. Originally Posted by Frostking
Can someone help me out on if the bottom left corner of a box is at (0,0,0 ) and the top right corner is in the first octant on the plane 3x + 2y + z = 1, how do I determine the coordinates for the top right corner??? I know x,y and z must be positive but???? It also states that this box faces parallel to the coordinate planes.
Since we know it is a box (all sides have the same length) the opposite corner must lie on the line $r(t)=t\vec i +t \vec j +t\vec k$

so this vector and the plane intersect at

$3t+2t+t=1 \iff t=\frac{1}{6}$ so the ordered triple is

$\left( \frac{1}{6},\frac{1}{6},\frac{1}{6}\right)$