Point intersecting two planes closest to origin

Question

Find the point $(x_0,y_0,z_0)$ closest to the origin in $\mathbb{R}^3$ that is on both $3x+y+z=5$ and $x+y+x=1$.

My current thoughts are to solve a Lagrange where, $x+y+z-1=0, 3x+y+z-5=0$ and some equation for the intercept is satisfied.

However, I am having difficult setting up the equation for the intercept.

edit:

Let $f(x,y,z)=z$

Then, let $g_1(x,y,z) = x+y+z-1, g_2(x,y,z) = 3x+y+z-5$ So, $g(x,y,z)=(g_1(x,y,z),g_2(x,y,z)$

Then we have lagrange $L=f(x,y,z)-(\lambda,\mu) * g(x,y,z)$

With, $\frac{\partial L}{\partial x}= -\lambda-3\mu=0$

$\frac{\partial L}{\partial x}= -\lambda-\mu=0$

$\frac{\partial L}{\partial z}= 1-\lambda-\mu=0$

However, I then get $\mu = \lambda$ and $\mu = \lambda -1$, so have run into a problem.

Answer 1

An alternative method, in case you're interested, is to use vector methods. First find the line of intersection of the planes by solving them simultaneously, which gives $$\underline{r}=\left(\begin{matrix}2\\0\\-1\end{matrix}\right)+\lambda\left(\begin{matrix}0\\1\\-1\end{matrix}\right)$$ Then you can find the foot of the perpendicular from the origin by solving for $\lambda$ $$\left[\left(\begin{matrix}2\\0\\-1\end{matrix}\right)+\lambda\left(\begin{matrix}0\\1\\-1\end{matrix}\right)\right]\cdot\left(\begin{matrix}0\\1\\-1\end{matrix}\right)=0$$ From this we obtain $\lambda=-\frac12$, so the closest point required is $$\left(\begin{matrix}2\\-\frac12\\-\frac12\end{matrix}\right)$$

Answer 2

Intersect that line with the plane perpendicular to it, through the origin. This plane has the equation

$$ax+by+cz=0$$ where $(a,b,c)$ is the cross product of the normals to the two given planes.

Now you have a system of three linear equations in three unknowns.