The shortest distance between the diagonals of a rectangular parallelepiped whose sides are $a, b, c$ and the edges not meeting it are:

\frac{a b c}{\sqrt{b^{2} + c^{2}}}

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\frac{a b c}{\sqrt{a^{2} + b^{2}}}

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\frac{c a}{\sqrt{a^{2} + c^{2}}}

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None of these

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Solution

The correct option is D None of these

Take $O$ as the origin of vectors and let $i, j, k$ denote unit vectors along $\to O A$ , $\to O B$ and $\to O C$ respectively.

Then, $- - \to O A =$ a $i$ , $- - \to O B =$ b $j$ , $- - \to O C =$ c $k$

Also $- - \to O P =$ $a i + b j + c k$ . The edges which do not meet the diagonal $O P$ and $B L$ , $B N$ , $A N$ and this parallels $A M$ , $C M$ and $C L$ .

Suppose we are to find the distance between the diagonal $O P$ and the edge $A N$ .

Now, $O P$ is the line passing through $O$ whose position vector is $\to O$ and parallel to the vector $a i + b j + c k$ .

And $A N$ is the line through $A$ whose position vector is $a i$ and parallel to $j$ . Hence, the shortest distance between $O P$ and $A N$ is given by P $= \frac{[\to O - a i, j, a i + b j + c k]}{| j \times (a i + b j + c k) |}$

Now, $[\to O - a i, j, a i + b j + c k]$ $= ∣ ∣ ∣ ∣ \begin{matrix} - a & 0 & 0 0 & 1 & 0 a & b & c \end{matrix} ∣ ∣ ∣ ∣ = - c a$

and $j \times (a i + b j + c k) = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 0 & 1 & 0 a & b & c \end{matrix} ∣ ∣ ∣ ∣ = c i - a k$

So that $| j \times (a i + b j + c k) | = \sqrt{c^{2} + a^{2}}$

$∴ P = \frac{c a}{\sqrt{c^{2} + a^{2}}}$

Similarly, it can be shown that the shortest distance between $O P$ and $B N$ is $\frac{c a}{\sqrt{c^{2} + a^{2}}}$ and that between $O P$ and $B L$ is $\frac{a b}{\sqrt{a^{2} + b^{2}}}$

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