1

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 ?

user1611542
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  • possibly duplicate for: http://math.stackexchange.com/questions/13734/how-to-find-shortest-distance-between-two-skew-lines-in-3d?rq=1 – Zau Aug 15 '16 at 13:36
  • Yes it is. But applying the method to this does not give the answer! – user1611542 Aug 15 '16 at 13:41

3 Answers3

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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 $10$.

Christian Blatter
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  • I think it's easier to solve for $z$ in terms of $x$ (one still has $y=-10$). Then $x^2+y^2=x^2+100$ which is obviously minimal when $x=0$. (Though in either case the labor is minimal.) – Semiclassical Aug 15 '16 at 14:56
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$${1+2A\over 1}={1+3A\over B}={1+4A\over 0}$$ Has the solution $$A=-{1\over 4},B={1/4\over 1/2}={1\over 2}$$ As far as 3d geometry is concerned!

Qwerty
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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.

  • Hey! thanks for the answer. If possible, can you also solve by above method ? – user1611542 Aug 15 '16 at 14:40
  • When $A = -\frac{1}{2}$ we get the plane $y-10 = 0$ that contains the given line. Note that $y=0$ is also a plane containing the $z$ axis and parallel to this plane. Hence the shortest distance is 10. –  Aug 15 '16 at 15:06
  • Thanks! It works this way. – user1611542 Aug 15 '16 at 15:09
  • For A = -1/2, we get B as infinity. Which means (1) is reduced to $y=0$ by dividing by B. Thanks! This is the solution I needed. – user1611542 Aug 15 '16 at 15:14