Question

The foot of the perpendicular drawn from the origin to the plane is (2,5,7). FInd the equation of the plane.

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Answer 1

Answer:

now point ( 2,5,7) lie on plane as it is foot of line on plane .We know that equation of plane passing through one point is a(x-x1)+b(y-y1)+c(z-z1)=0.So equation of line passing through ( 2,5,7) is a(x-2) +b(y-5)+c(z-7)=0. But we have to find a,b,c which is dr of normal ,Now normal is passing through two points origin to foot (2,5,7)so dr of normal will be 2-0 ,5-0 ,7-0.thus putting values of a,b,c in above equation we will get 2(x-2) +5(y-5) +7(z-7)=0 and further solving this the required equation is 2x+5y+7z=78.