Question

If \[ A(0,1,-2),\quad B(-1,2,-3),\quad C(2,-3,4) \] and \[ D(3,4,5) \] are the vertices of a tetrahedron \(ABCD\), then the volume of the tetrahedron is

• A. \(\dfrac{16}{3}\)
• B. \(32\)
• C. \(\dfrac{8}{3}\)
• D. \(16\) \bigskip

Hint:

For a tetrahedron with one vertex taken as the origin point of three edges, \[ V=\frac{1}{6}\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|. \] So first form the three edge vectors from the same vertex, compute the determinant, take its absolute value, and divide by \(6\).

Answer 1

The Correct Option is A

Concept: Volume of a tetrahedron with adjacent edges \(\vec{a},\vec{b},\vec{c}\) is \(V=\frac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]|\). Compute the scalar triple product using the determinant method with cofactor expansion.

Step 1: Determine vectors \(\vec{AB}, \vec{AC}, \vec{AD}\). \(\vec{AB}=(-1,1,-1)\), \(\vec{AC}=(2,-4,6)\), \(\vec{AD}=(3,3,7)\).

Step 2: Evaluate the determinant \(\begin{vmatrix}-1&1&-1\\2&-4&6\\3&3&7\end{vmatrix}\). Expand along row 1: \(-1\begin{vmatrix}-4&6\\3&7\end{vmatrix} - 1\begin{vmatrix}2&6\\3&7\end{vmatrix} - 1\begin{vmatrix}2&-4\\3&3\end{vmatrix} = -1(-28-18) - (14-18) - (6+12) = 46 + 4 - 18 = 32\). Absolute value is 32.

Step 3: Compute volume. \(V = \frac{1}{6} \times 32 = \frac{16}{3}\).

Step 4: Write the final answer. \(\boxed{\frac{16}{3}}\)