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Math Help - first octant question

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Frostking View Post
    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)
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