# first octant question

• February 21st 2009, 08:28 PM
Frostking
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.
• February 21st 2009, 11:12 PM
TheEmptySet
Quote:

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

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