Question

find the volume of tetrahedron whose vertices are A(3,2,6),B(6,7,8),C(7,11,23),D(9,12,17)

Answer 1

Answer:

49/6 UNIT³

Step-by-step explanation:

Given position vectors at each vertex are

a=3i+2j+6k

b=6i+7j+8k

c=7i+11j+23k

d=9i+12j+17k

Now we need to find vectors of sides of tetrahedron

AB=b-a=(6i+7j+8k)-(3i+2j+6k)

=3i+5j+2k

AC=c-a=(7i+11j+23k)-(3i+2j+6k)

=4i+9j+17k

AD=d-a=(9i+12j+17k)-(3i+2j+6k)

=6i+10j+11k

Volume of Tetrahedron whose coterminous edges are AB,AC,AD is  $1/6\left[\begin{array}{ccc}3&5&2\\4&9&17\\6&10&11\end{array}\right]$

=1/6(49)

HENCE 49/6 UNIT³ IS THE VOLUME OF GIVEN TETRAHEDRON