How can I tell if a vector is normal (orthogonal) to the plane?

Question

I have to write equation of plane given 2 points $A=(-5,4,2)$ and $B(-3,2,-1)$ and vector $u=(1,2,-3)$. What I did was get a vector from points $A$ and $B$ , $AB=(2,-2,-3)$ and then I found cross product between $AB$ and $u$, $n= AB \times u = (12,3,6)$ so my first question is: can I divide the new vector $n$ by 3 and get $n=(4,1,2)$ and is this vector normal to the plane? If yes how can I tell that it is ?

Answer 1

It sounds like you are given the two points $A$ and $B$ and the vector $\vec{u}$. You want to find an equation of the plane that contains $A$ and $B$ and is parallel to the vector $\vec{u}$. Is that correct?

If so, then you are doing the right thing. The vector $\vec{AB}$ will be parallel to the plane and indeed the cross product $\vec{n} = \vec{AB}\times\vec{u}$ will be a normal vector for the plan. For any non-zero scalar $c$, the vector $c\vec{n}$ will also be a normal vector.

It looks like your calculations are correct.

And equation for the plane is $$ 4(x+5) + 1(y-4) + 2(z - 2) = 0. $$