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: If \[ \vec{AB},\quad \vec{AC},\quad \vec{AD} \] are three edges of a tetrahedron meeting at the same vertex, then its volume is \[ V=\frac{1}{6} \left| \vec{AB}\cdot(\vec{AC}\times\vec{AD}) \right|. \] That is, \[ V=\frac{1}{6} \left| \begin{vmatrix} AB_x & AB_y & AB_z AC_x & AC_y & AC_z AD_x & AD_y & AD_z \end{vmatrix} \right|. \]

Step 1: Find the vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\). \[ \vec{AB} = (-1-0)\hat{i} +(2-1)\hat{j} +(-3+2)\hat{k} = (-1,1,-1). \] \[ \vec{AC} = (2-0,\,-3-1,\,4+2) = (2,-4,6). \] \[ \vec{AD} = (3-0,\;4-1,\;5+2) = (3,3,7). \]

Step 2: Evaluate the scalar triple product. \[ \Delta = \begin{vmatrix} -1 & 1 & -1 2 & -4 & 6 3 & 3 & 7 \end{vmatrix}. \] Expanding along the first row, \[ \Delta = -1 \begin{vmatrix} -4 & 6 3 & 7 \end{vmatrix} -1 \begin{vmatrix} 2 & 6 3 & 7 \end{vmatrix} -1 \begin{vmatrix} 2 & -4 3 & 3 \end{vmatrix}. \] \[ = -1(-28-18) -(14-18) -(6+12). \] \[ = 46+4-18. \] \[ = 32. \] Thus, \[ \left| \vec{AB}\cdot(\vec{AC}\times\vec{AD}) \right| = 32. \]

Step 3: Find the volume of the tetrahedron. \[ V = \frac{1}{6}(32). \] \[ V = \frac{16}{3}. \]

Step 4: Write the final answer. \[ \boxed{\frac{16}{3}} \]