Shortest distance between z axis and the line $x+y+2z=3, 2x+3y+4z+4=0$

Question

z axis is : $x=0=y$

$x + B(y) = 0$ --> (1)

$x+y+2z-3+ A (2x+3y+4z+4 ) = 0$ --> (2)

These are two planes constructed through the two lines. We find the parallel planes so that it is easier to calculate the distance after that. Therefore, for planes to be parallel :

$\frac{1+2A}{1} $ = $\frac{1+3A}{B}$ = $\frac{2+4A}{0}$

But now no values of A,B will satisfy this. How to proceed further ?

Answer 1

$${1+2A\over 1}={1+3A\over B}={1+4A\over 0}$$ Has the solution $$A=-{1\over 4},B={1/4\over 1/2}={1\over 2}$$ As far as 3d geometry is concerned!

Answer 2

The intersection of the given planes is the line \begin{align*} \mathbf{r} = 13\mathbf{i} - 10\mathbf{j} + t(-2\mathbf{i}+\mathbf{k}) \end{align*} The vector perpendicular to $\mathbf{k}$ (vector along $z$ axis) and the above line is $\mathbf{k} \times (-2\mathbf{i}+\mathbf{k}) = -2\mathbf{j}$. Hence the shortest distance is given by the projection of the line joining $(0,0,0)$ and $(13, -10, 0)$ on the vector $\mathbf{j}$ and this is 10.

Answer 3

Solving the given equations for $x$ and $y$ in function of $z$ gives $x=13-2z$, $y=-10$. It follows that $$\sqrt{x^2+y^2}=\sqrt{(13-2z)^2+100}$$ is minimal when $z={13\over2}$, and the minimal value is $10$.