(Calculus) What would be the normal vector to 4x+6y+5z=2?

Question

jayfafr · Answer

4x+6y+5z=2 is ⟨4,6,5⟩

jayfafr · Answer

@kingpizza

wio · Answer

The normal vector to a plane is a vector that is perpendicular to the plane.

In $\mathbb R^3$ it is easy to find the normal vector to a plane from an equation, because it is just the coefficients of the variables, or in this case $\langle 4,6,5\rangle$.

But why is this the case?
To form the equation of a plane in $\mathbb R^3$ we use a normal vector $\vec n$ and some point $P_0$ which lies on the plane. We can test if any arbitrary point $P = (x,y,z) $ lies on the plane, because the vector $\vec r = P-P_0$ must be parallel to the plane and orthogonal to $\vec n$. So $\vec n \cdot \vec r = 0$.

Suppose be didn't know the equation $4x+6y+5z=2$, we just knew $\vec n = \langle 4,6,5\rangle$ and $P_0 = (-1,1,0)$. Then we can use the vector equation to find the plane equation: $$
\begin{array}{rcl}
\vec n \cdot \big( P - P_0 \big) &=&  0 \
\langle 4,6,5\rangle \cdot \big( (x,y,z) - (-1,1,0) \big) &=& 0 \
\langle 4,6,5\rangle \cdot  \langle x+1,y-1,z\rangle&=& 0 \
4(x+1) + 6(y-1) + 5z &=& 0 \
4x + 6y + 5z -2 &=& 0 \
4x + 6y + 5z &=& 2 \
\end{array}
$$
All equations of planes in $\mathbb R^3$ have the form $ax+by+cz = d$.

pup · Answer

i love math but ill try this ima say its 2