A plane P passes trough the points (1,1,1) and (2,0,3) And it is perpendicular to the plane Q x+2y-3z=0 What is the equation for plane P? Please help!
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Originally Posted by techmath A plane P passes trough the points (1,1,1) and (2,0,3) And it is perpendicular to the plane Q x+2y-3z=0 $\displaystyle <1,-1,2>$ is the vector determined by the two points. So the normal to the needed plane is cross product $\displaystyle <1,-1,2>\times <1,2,-3>$.
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