In here the author discusses finding the extrema of a function of three variables $F(x,y,z)$ subject to a condition of the form $G(x,y,z)=0$. He then makes the statement
We can observe that this implies that the cross product $\nabla F \times \nabla G$ must be $0$, and this vector equation gives us two independent component equations that we can solve along with $G = 0$ to find the extrema.
I've worked on the first problem on the same page, in which $F=xyz-x$ and $G=2x^2+4y^2+3z^2-6$, and found that $$\nabla F \times \nabla G=(6xz^2-8xy^2) \hat{\imath}-(6yz^2-6z-4x^2y) \hat{\jmath}+(8y^2z-8y-4x^2z) \hat{k} $$ in which case it seems that indeed only two of these equations are independent. However, I can't see why this statement is true in general, and would appreciate some insights regarding this. Thank you!
Assume you have two vector fields ${\bf v}=(v_1,v_2,v_3)$ and ${\bf w}=(w_1,w_2,w_3)\ne{\bf 0}$ in $3$-space, and you want to find the points ${\bf x}$ where ${\bf v}({\bf x})=\lambda {\bf w}({\bf x})$ for some $\lambda\in{\mathbb R}$. If ${\bf p}$ is such a point, and $w_i({\bf p})\ne0$ for all $i$ (for simplicity) then necessarily $$\lambda={v_1({\bf p})\over w_1({\bf p})}={v_2({\bf p})\over w_2({\bf p})}={v_3({\bf p})\over w_3({\bf p})}\ .\tag{1}$$ Since you are not interested in the value of $\lambda$ the line $(1)$ constitutes just two equations for the point ${\bf p}$. Written denominator-free this says that ${\bf v}\times{\bf w}={\bf 0}$ is equivalent to just two equations.