Given the plane $4x +3y+2z = 5$ why $\langle4,3,2\rangle$ is the vector normal to the plane?
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1Given two points on the plane, $P$ and $Q$, we have that $\vec{n}=\langle 4,3,2\rangle$ is orthogonal to $\vec{PQ}$. – Ángel Mario Gallegos May 04 '17 at 13:38
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Vector is normal to a plane if it's dot product with any vector in the plane is $0$
Take any points $P$ and $Q$ in the plane. Then form the vector $P-Q$ and find the dot product with $<4,3,2>$. It follows that it is $0$ so indeed $<4,3,2>$ is orthogonal to the plane.
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Let $(a, b, c) $ and $(d, e, f) $ be two arbitrary points on the plane. $(a-d, b-e, c-f) $ is a vector parallel to the plane. Its dot product with $(4,3,2)$ is equal to $(4a+3b+2c)-(4d+3e+2f)=5-5=0$. So it is orthogonal to $(4,3,2)$.
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