Question

I A, B, C, D are (3, 7, 4), (5, -2, 3) (-4, 5, 6) and
(1,2,3) respectively, then find the volume of the parallelopiped and tetrahedron
with AB, AC, and AD as the concurrent edges.​

Answer 1

The volume of the parallelopiped is 36 and that of tetrahedron is 6 .

• a= AB = B-A = (5i-2j+3k) - (3i+7j+4k) = 2i-9j-1k      

•  b= AC = C-A = (-4i+5j+6k) - (3i+7j+4k) = -7i-2j+2k

•  c= AD = D-A = (1i+2j+3k) - (3i+7j+4k) = -2i-5j-1k

• volume of parallelopiped = (a.b×c)

• (a.b×c) =   $\left[\begin{array}{ccc}2&-9&-1\\7&-2&2\\-2&-5&-1\end{array}\right]$  

                 = 2[(-2×-1) - (2×-5)] - (-9)[(7×-1) - (2×-2)] + (-1)[(7×-5) - (-2×-2)]

                 =36

• so volume of parallelopiped is 36
• volume of tetrahedron= 1/6 (a.b×c)
• so from the above (a.b×c)=36
• volume of tetrahedron = 6