# Find an equation of the plane. The plane through the origin and perpendicular to

Find an equation of the plane. The plane through the origin and perpendicular to the vector <1, -2, 5>
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hesgidiauE
Perpendicular to plane and normal to the plane mean the same thing.
Theorem 7 states that:
The scalar equation of a plane that passes through the point (a,b,c) and has (I,m,n) as the normal vector, is given by
l(x-a)+m(y-b)+n(z-c) =0
Therefore, equation of the plane that has $⟨1,—2,5⟩$ as the normal vector and passes through the point (0,0,0) is given by
1*(x-0) -2*(y-0)+5*(z-0) =0
Remove the brackets
x-2y+5z=0
Results:
x-2y+5z=0