Question

Volume of the tetrahedron whose vertices are represented by the position vectors, A(0, 1, 2); B(3, 0, 1); C(4, 3, 6) & D(2, 3, 2) is

• A. 3
• B. 6
• C. 36
• D. none

Answer 1

Answer: (B)

6

Let

$A=(0,1,2)$

$B=(3,0,1)$

$C=(4,3,6)$

$D=(2,3,2)$

Then

$AB=\overline{a}=3i-j-k$

$BC=\overline{b}=i+3j+5k$

$CD=\overline{c}=-2i+0j-4k$

Now

$(\overline{a}\times \overline{b})=-2i-16j+10k$

Now

$\overline{c}.(\overline{a}\times \overline{b})$

$=(-2i-16j+10k).(-2i-4k)$

$=4-40$

$=-36$

Hence the required volume of the tetrahedron will be

$=\frac{|36|}{6}$
$=6$ unit cube.