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Voids

Shortest dist...

Question

Shortest distance between two tetrahedral voids is:

a \sqrt{3}

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\frac{a \sqrt{3}}{2}

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\frac{a}{2}

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\frac{a \sqrt{3}}{4}

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Solution

The correct option is D $\frac{a}{2}$
If $a$ is the edge length of the cube, then the shortest distance between two tetrahedral voids is $\frac{a}{2}$ .

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\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

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\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

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