Question

The volume of a tetrahedron whose vertices are
$4 \hat{ i }+5 \hat{ j }+\hat{ k },-\hat{ j }+\hat{ k }, 3 \hat{ i }+9 \hat{ j }+4 \hat{ k }$ and
$-2 \hat{ i }+4 \hat{ j }+4 \hat{ k }$ is (in cubic units)

• A. $\frac{14}{3}$
• B. 5
• C. 6
• D. 30

Answer 1

Answer: (B)

Given vertices of a tetrahedron are
$A(4 \hat{ i }+5 \hat{ j }+\hat{ k }), B(-\hat{ j }+\hat{ k })$
$C(3 \hat{ i }+9 j +4 k ), D(-2 \hat{ i }+4 \hat{ j }+4 \hat{ k })$
So, $A B =-4 \hat{ i }-6 \hat{ j } \quad A C =-\hat{ i }+4 \hat{ j }+3 \hat{ k }$
$A D =-6 \hat{ i }-\hat{ j }+3 \hat{ k }$
Now, volume of the given tetrahedron is
$v=\frac{1}{6}| \begin{vmatrix}
-4 & -6 & 0 \\
-1 & 4 & 3 \\
-6 & -1 & 3
\end{vmatrix} \mid$
$=\frac{1}{6} \times|[-4(12+3)+6(-3+18)+0]|$
$=\frac{1}{6}|[-60+90]|=\frac{30}{6}=5 $ cubic units