Calc 3: Equation of a line normal to the contour

[mtruong1999]

Homework Statement

Given g(x,y)=x² - xy + 2y², find the equation of the line normal to the contour that passes through the point (1,2).

Homework Equations

Not 100% positive, but the equation to a plane tangent to a function of 3 variables g(x,y,z) is (partial of x)(x - x₀) + (partial of y)(y - y₀) + (partial of z)(z - z₀)= 0
So, I assume that for a function of 2 variables, using this same equation (without the z component, of course) would result in a line tangent to the contour?

The Attempt at a Solution

I am having trouble picturing this. So far I took the partial derivatives with respect to x (which is 2x - y) and with respect to y (which is -x+4y).
What I think I should do is to plug in (1,2) into the partial derivatives, which
would give me 0 for the partial with respect to x, and 7 for the partial with respect to y. Then I plug into the equation 0(x-1)+7(y-2)=0 which I believe is the equation to the line tangent at (1,2)... then I don't know where to go from here...

--or--

Would I just take the partials with respect to x and y at point (1,2) which gives me 0 and 7 respectively then plug them into the symmetric equations for lines? Hmm but the symmetric for the x component would give me and undefined number, so if I wrote it in vector form instead I would get x=1 and y=2+7t which is nicer. Is this the right answer/ did I do this right.
I'm very lost on the appropriate approach to this problem.

 

[Orodruin]

Then I plug into the equation 0(x-1)+7(y-2)=0 which I believe is the equation to the line tangent at (1,2)... then I don't know where to go from here...

Apart from simplifying it to y = 2, you are done here. However, as you noted before, this is the equation of the tangent plane - not the line normal to the contour. It is the tangent because $\nabla g$ is normal to the contour and the equation $(\vec x - \vec x_0) \cdot \vec n$ describes the plane that passes through $\vec x_0$ and has $\vec n$ as its surface normal.

Would I just take the partials with respect to x and y at point (1,2) which gives me 0 and 7 respectively then plug them into the symmetric equations for lines? Hmm but the symmetric for the x component would give me and undefined number, so if I wrote it in vector form instead I would get x=1 and y=2+7t which is nicer. Is this the right answer/ did I do this right.
I'm very lost on the appropriate approach to this problem.

On vector form, the equation for a line in the direction $\vec n$ passing through $\vec x_0$ is indeed $\vec x = \vec x_0 + \vec n t$, where $t$ is a curve parameter. Now, since your normal has an $x$-component 0, this can just be summarised as the line $x = 1$, the parametrisation of $y$ really does not matter.