Reeves

2021-09-13

Find an equation of the plane. The plane through the origin and perpendicular to the vector <1, -2, 5>

hesgidiauE

Skilled2021-09-14Added 106 answers

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$\u27e81,\u20142,5\u27e9$ 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

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*(x-0) -2*(y-0)+5*(z-0) =0

Remove the brackets

x-2y+5z=0

Results:

x-2y+5z=0

nick1337

Expert2023-05-29Added 777 answers

Don Sumner

Skilled2023-05-29Added 184 answers

Vasquez

Expert2023-05-29Added 669 answers

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?

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