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Two non-parallel lines $L_1$ and $L_2$ in $\mathbb{R}^3$ have respective equations ${\bf{r}} \wedge {\bf{a_1}} = {\bf{b_1}}$ and ${\bf{r}} \wedge {\bf{a_2}} = {\bf{b_2}}$. For $i = 1,2$, let $\prod_i$ denote the plane of the form ${\bf{r}} \cdot ({\bf{a_1}} \wedge {\bf{a_2}}) = k_i$ which contains $L_i$. Show that $k_1 = {\bf{b_1}} \cdot {\bf{a_2}}$ and find $k_2$. Hence show that the least distance between the lines equals

$\dfrac{|{\bf{a_1}} \cdot {\bf{b_2}} + {\bf{a_2}} \cdot {\bf{b_1}}|}{|{\bf{a_1}} \wedge {\bf{a_2}}|}$

I don't really know how to begin this question. An earlier part of the question was showing that ${\bf{r}} \wedge {\bf{a}} = {\bf{b}}$ has no solutions unless ${\bf{a}} \cdot {\bf{b}} = 0$, but I can't see how to apply it to this question (I've tried taking various dot products/cross products to get the expressions for $k_i$ to fall out, but to no avail). Thanks for any help!

Noble.
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1 Answers1

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The first thing you should do is to realize that $L_i$ is a line with direction vector $\mathbf a_i$ in a plane (through the origin) orthogonal to $\mathbf b_i$. Thus, the parallel planes $\Pi_i$ containing $L_i$ have normal vector $\mathbf a_1\wedge\mathbf a_2$.

The obvious way to compute $k_i$ is to think of the distance from the origin to the plane $\Pi_i$. Indeed, $k_i = \mathbf r\cdot (\mathbf a_1\wedge\mathbf a_2)$ for any $\mathbf r$ in the plane. The vector $\mathbf x$ in the plane $\Pi_i$ closest to the origin is, indeed, a scalar multiple of $\mathbf a_1\wedge\mathbf a_2$. On the other hand, this closest vector is also a scalar multiple of $\mathbf a_i\wedge\mathbf b_i$. What scalar multiple? We must have $\|\mathbf x\|\|\mathbf a_i\| = \|\mathbf b_i\|$ (why?). My final hint is this: Do you know a formula for $(\mathbf a\wedge\mathbf b)\cdot(\mathbf a\wedge\mathbf c)$?

Ted Shifrin
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