Question

The volume of the tetrahedron whose vertices are A\((-1,2,3)\), B\((3,-2,1)\), C\((p,1,3)\), D\((-1,-2,4)\) is \(\frac{16}{3}\) cubic units then the value of p is

• A. \(\frac{-10}{3}\)
• B. \(5\)
• C. \(8\)
• D. \(10\)

Hint:

Volume = one sixth of the absolute scalar triple product of AB, AC, AD.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The volume of a tetrahedron with vertices \(A, B, C, D\) is \(V = \frac16\left|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\right|\).

Step 2: Find the edge vectors:
\(\overrightarrow{AB} = (4, -4, -2)\), \(\overrightarrow{AC} = (p+1, -1, 0)\), \(\overrightarrow{AD} = (0, -4, 1)\).

Step 3: Triple product:
\[ \begin{vmatrix} 4 & -4 & -2 \\ p+1 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix} = 4(-1) + 4(p+1) - 2\left(-4(p+1)\right) = 12(p+1) - 4 = 12p + 8 \]

Step 4: Use the volume:
\(\frac16|12p + 8| = \frac{16}{3}\), so \(|12p + 8| = 32\).
Either \(12p + 8 = 32\), giving \(p = 2\), or \(12p + 8 = -32\), giving \(p = -\frac{10}{3}\).
Only \(p = -\frac{10}{3}\) appears among the options.

Step 5: Why the other options are wrong.
Values 5, 8 and 10 give \(12p + 8 = 68, 104, 128\), none equal to \(\pm32\).

Final Answer:
\(p = -\frac{10}{3}\), option (A). \[ \boxed{-\frac{10}{3}} \]