Find the equation of the plane in xyzxyz-space through the point p=(5,5,3)p=(5,5,3) and perpendicular to the vector n=(1,2,5)

1 answer:

6 0

Given that the plane is perpendicular to the vector <1,2,5>, this means that the normal vector of the plane is <1,2,5>.

Given that the plane passes through point (5,5,3), the equation of the plane is therefore:
Π : 1(x-5)+2(y-5)+5(z-3)=0
on simplifying, x-5+2y-10+5z-15=0
Π : x+2y+5z=30

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