Find the equation of the plane which passes through the points A(-1,-3,1), B(1,-2,0) and C(2,1,-1).

Calculate the shortest distance between the plane and the origin of the coordinate system.

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- Aug 20th 2008, 12:09 PMBogStandardEquation of plane.
Find the equation of the plane which passes through the points A(-1,-3,1), B(1,-2,0) and C(2,1,-1).

Calculate the shortest distance between the plane and the origin of the coordinate system. - Aug 20th 2008, 12:21 PMgalactus
Since A, B, and C lie in the plane, the vectors

$\displaystyle \overline{AB}=[2,1,-1]$

and $\displaystyle \overline{AC}=[3,4,-1]$ are parallel to the plane.

Therefore, $\displaystyle \overline{AB}\times \overline{AC}=\begin{vmatrix}i&j&k\\2&1&-1\\3&4&-1\end{vmatrix}$

$\displaystyle =2i+j+5k$

is normal to the plane, since it is perpendicular to $\displaystyle \overline{AB}, \;\ and \;\ \overline{AC}$.

By using the normal and the point (-1,-3,1) in the plane, we get the

point-normal form:

$\displaystyle 2(x+1)+(y+3)+5(z-1)$

$\displaystyle 2x+y+5z=0$

To find the distance between a point and a plane, use the formula:

$\displaystyle D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2 }+c^{2}}}$