NIMCET 2019 Question Paper with solutions

Question : 30 of 120

Marks: +1, -0

The position vectors of the vertices A, B, C of a tetrahedron ABCD are

i↖{⋀}+j↖{⋀}+k↖{⋀}

, i and 3i respectively and the altitude from the vertex D to the opposite face ABC meets the face at E. If the length of the edge ED is 4 P JP and volume of the tetrahedron is

{2√2}/3

, then the length of DE is

Validate

Solution:

The position vector of vertices of A. B, C of tetrahedron
ABCD are

i↖{⋀}+j↖{⋀}+k↖{⋀}

i↖{⋀}

and

3i↖{⋀}

A=(i↖{⋀}+i↖{⋀}+k↖{⋀})

B=i↖{⋀}

C=3i↖{⋀}

AB=-i↖{⋀}-k↖{⋀}

AC=2i↖{⋀}-j↖{⋀}-k↖{⋀}

AB×AC=|\table i↖{⋀},j↖{⋀},k↖{⋀};0,-1,-1;2,-1,-1|=-2j↖{⋀}+2k↖{⋀}

Volume of tetrahedron

=1/6×h|AB×AC|

{2√2}/3=1/6×h√{4+4}

{2√2}/3=1/6×h×2√2

h=2

∴ Length of DE = 2

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