Answer to Question #123889 in Analytic Geometry for Nikhil

Question

Show that the points (2,0,1),(0,4,-3) and
(-2,5,0) are non-collinear.Hence find the equation of plane passing through them.

Answer 1

three points in three-dimensional space. If they lie on one straight line, then the determinant of the matrix equals 0

$\begin{vmatrix} 2& 0&1 \\ 0& 4&-3\\ -2 &5&0 \end{vmatrix}$

the determinant is not equal to 0

therefore the points do not lie on one straight line

finding out the plane passing through three points

$\begin{vmatrix} x-x1 & y-y1& z-z1 \\ x2-x1& y2-y1 &z2-z1\\ x3-x1 & y3-y1 &z3-z1 \end{vmatrix}$ =$\begin{vmatrix} x-2 & y& z-1\\ 0-2 & 4-0&-3-1 \\ -2-2&5 &0-1 \end{vmatrix}$ =

= (x-2)*(4*(-1)-(-4)*5)-y*(-2*(-1)-(-4*(-4))+(z-1)*(-2*5-4*(-4))=(x-2)*16-y*(-14)+(z-1)*6=

=16x-32+18y+6z-6=0

got the plane equation

8x+7y+3z=19

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