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O A B C is regular tetrahedron in whi...

`O A B C` is regular tetrahedron in which `D` is the circumcentre of ` O A B` and E is the midpoint of edge `A Cdot` Prove that `D E` is equal to half the edge of tetrahedron.

Text Solution

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Let `vec(OA) = veca and vec(OB) =vecb`
since `triangleABC` is equilateral, the circumecenters, triangle coincide with the centroid, therefore,
`vec(OD) = (veca +vecb)/3`
`Also vec(OE) = (veca +vecc)/2`
`vec(DE) = (veca = 3vecc - 2vecb)/6`
`|vec(DE)|^(2)= |(veca + 3vecc - 2vecb)/6|^(2)`
Since, `|veca|= |vecb|= |vecc| and theta = 60^(@)` we have,
`|vec(DE)|^(2)= 1/36 (a^(2)+ 9a^(2)+4a^(2) +(6a^(2))/2-(12a^(2))/2-(4a^(2))/2)`
`a^(2)/4`
or `|vec(DE)|= a/2`
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