A triangle has vertices at the points A(5,0,0), B(0,5,0), and C(0,0,5). Identify the point in the interior of the triangle that is closest to the origin.

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- May 30th 2009, 10:47 AMdalbir4444Question on 3-d vectors
A triangle has vertices at the points A(5,0,0), B(0,5,0), and C(0,0,5). Identify the point in the interior of the triangle that is closest to the origin.

- May 30th 2009, 11:30 AMearboth
1. The points A, B, C define a plane:

$\displaystyle p:(x,y,z)=(5,0,0)+r(5,-5,0+s(5,0,-5)$

2. p in normal form:

$\displaystyle \vec n = (5,-5,0) \times (5,0,-5) = (5,5,5)$ Thus

$\displaystyle p: x+y+z=5$

3. The point F lies on the line through the origin perpendicular to p:

$\displaystyle (x,y,z)=r (1,1,1)$

4. Calculate the point of intersection:

$\displaystyle r+r+r = 5~\implies~r=\dfrac53$

Therefore $\displaystyle F\left(\dfrac53\ ,\ \dfrac53\ ,\ \dfrac53\right)$