The text book question is $f(x,y)=xy$, find the gradient vector $\nabla f(3,2)$ and use it to find the tangent line to the level curve $f(x,y)=6$ at the point $(3,2)$.
My solution is $$\nabla f(x,y)=<y,x>$$ $$\nabla f(3,2)=<2,3>$$ and the equation of normal line is $$x=2t+3$$ $$y=3t+2$$
Because we are looking for tangent line to the curve, I looked for orthogonal line to the normal curve and used $m \times n=-1$ So $$ x=-\frac{1}{2}t +3$$ $$y=-\frac{1}{3}t+2$$ When I solve for t and combine the two equations it becomes $2x-3y=0$
My solution is of course wrong. The text book solution uses dot product as below $$ \nabla f(3,2) \cdot <x-3,y-2>=0$$ $$<2,3> \cdot <x-3,y-2>=0$$ $$2x+3y=12$$
I wonder why I cannot use $m \times n =-1$ and have to use dot product in this case.
Best Answer
If you have a line with parametric form $$x = at+b, y= ct+d,$$ then replacing $a$ with $-1/a$ and $c$ with $-1/c$ will not result in a line that is perpendicular to the original. This is because the original line has direction vector $\langle a, c\rangle$ while the latter has direction vector $\langle -1/a, -1/c\rangle$, and these are not perpendicular vectors (check their dot product).
The $mn=-1$ thing applies to the slope of a line, rather than the lines' parametric form (involving the $t$). That is, if you have a line $y = mx+b$, then $y = nx+b$ where $mn = -1$ is perpendicular to this line. This is the case because a line $y = mx+b$ has direction vector $\langle 1,m\rangle$, and $y = nx+b$ has direction vector $\langle 1,n\rangle $. These are perpendicular exactly when $mn=-1$ (check the dot product).