Point intersecting two planes closest to origin

Solution 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 $$\left(\begin{matrix}2\\-\frac12\\-\frac12\end{matrix}\right)$$

Solution 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.