Find the point(s) at the surface such that the tangent plane goes through (3,0,0) and (0,3,0)

Question

I am having a really troubling time trying to solve this multivariable calculus problem. We are supposed to find the point(s) $P$ on the surface $x^2+y^2+z^2=1$ where the tangent plane at $P$ goes through the points $(3,0,0)$ and $(0,3,0)$. I have tried setting this up as a Lagrange multiplier problem like below, but I am having trouble finding the specific point(s) because of the lack of equations.

Let $\xi_1=x^2+y^2+z^2-1$ and $\xi_2=ax+by+cz-d$. Now define: $\nabla(\xi_1-\lambda\xi_2)=0$ subject to the constraint $x^2+y^2+z^2=1$ and solve. However, here I encounter the troubling part, namely, I have eight variables and only 5 equations. By using the given points we reduce down the variables to six since I can show $a=b$ and $x=y$.

I just need some help figuring out how I am supposed to find all variables or if there is a way to skip solving some since I am only looking for the specific point(s)?

Answer 1

Take $(x_0,y_0,z_0)$ on your surface. The tangent plane through $(x_0,y_0,z_0)$ is$$x_0(x-x_0)+y_0(y-y_0)+z_0(z-z_0)=0,$$which means$$x_0x+y_0y+z_0z=1.$$Now, you want it to contain $(3,0,0)$ and $(0,3,0)$. So, solve the system$$\left\{\begin{array}{l}3x_0=1\\3y_0=1\\{x_0}^2+{y_0}^2+{z_0}^2=1.\end{array}\right.$$

Answer 2

Given $p = (x,y,z)$ and $p_0 = (x_0,y_0,z_0)$ such that $\vert\vert p_0 \vert\vert = 1$

and $L\rightarrow p = p_1+\lambda \vec v$ with $p_1 = (3,0,0)$ and $\vec v = (3,-3,0)$

then the sough plane is

$< p-p_0, p_0 > = 0$ and then $< p_1-p_0+\lambda \vec v, p_0 > = 0$ or

$<p_1-p_0,p_0> + \lambda < \vec v, p_0> = 0$

This should be true for all $\lambda$ then follows

$$ < p_1-p_0, p_0 > = 0\\ < \vec v , p_0 > = 0\\ \vert\vert p_0\vert\vert^2 = 1 $$

Solving those equations we get

$p_0 = (1/3,1/3,\pm \sqrt7/3)$ as the tangency points.

Answer 3

The tangent plane at $(x,y,z)$ goes through $(3,0,0)$ if and only if the vector $(x-3,y,z)$ is tangent to the sphere at $(x,y,z)$, which is equivalent to $$\left<(x-3,y,z),(x,y,z)\right>=0$$ which can be rewritten $3x=1$ by using the fact that $x^2+y^2+z^2=1$.

So $x=\frac13$. Doing the same for the other vector gives you $y$, and finally using $x^2+y^2+z^2=1$ you get $z$.

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Edit: regarding Lagrange multipliers, I don't see how you could use them here.