Question:
The volume of a tetrahedron with vertices \(5\hat{i}-\hat{j}+\hat{k},7\hat{i}-4\hat{j}+p\hat{k},\hat{i}-6\hat{j}+10\hat{k}\) and \(-\hat{i}-3\hat{j}+7\hat{k}\) is 11 cubic units, then one of the values of p is
The volume of a tetrahedron with vertices \(5\hat{i}-\hat{j}+\hat{k},7\hat{i}-4\hat{j}+p\hat{k},\hat{i}-6\hat{j}+10\hat{k}\) and \(-\hat{i}-3\hat{j}+7\hat{k}\) is 11 cubic units, then one of the values of p is
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The volume is one sixth of the absolute value of the scalar triple product of the three edge vectors from one vertex.
Updated On: Oct 1, 2026
- \(1\)
- \(2\)
- \(0\)
- \(3\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a tetrahedron with vertices \(A, B, C, D\), the volume is \(\dfrac{1}{6}\left|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\right|\).
Step 2: Edge vectors.
Take \(A(5, -1, 1)\). Then \(\overrightarrow{AB} = (2, -3, p - 1)\), \(\overrightarrow{AC} = (-4, -5, 9)\), \(\overrightarrow{AD} = (-6, -2, 6)\).
Step 3: Compute the determinant.
\[ \begin{vmatrix} 2 & -3 & p-1 \\ -4 & -5 & 9 \\ -6 & -2 & 6 \end{vmatrix} = 2(-30 + 18) + 3(-24 + 54) + (p-1)(8 - 30) \]
\[ = -24 + 90 - 22(p - 1) = 66 - 22(p - 1) \]
Step 4: Use the volume.
\(\dfrac{1}{6}|66 - 22(p-1)| = 11 \Rightarrow |66 - 22(p-1)| = 66\).
Case 1: \(66 - 22(p-1) = 66\), so \(p = 1\).
Case 2: \(66 - 22(p-1) = -66\), so \(p - 1 = 6\) and \(p = 7\).
Step 5: Check the options.
One of the values of \(p\) is 1, which is option (A). The value 7 is not listed.
Final Answer:
One value is \(p = 1\), option (A). \[ \boxed{p = 1} \]
For a tetrahedron with vertices \(A, B, C, D\), the volume is \(\dfrac{1}{6}\left|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\right|\).
Step 2: Edge vectors.
Take \(A(5, -1, 1)\). Then \(\overrightarrow{AB} = (2, -3, p - 1)\), \(\overrightarrow{AC} = (-4, -5, 9)\), \(\overrightarrow{AD} = (-6, -2, 6)\).
Step 3: Compute the determinant.
\[ \begin{vmatrix} 2 & -3 & p-1 \\ -4 & -5 & 9 \\ -6 & -2 & 6 \end{vmatrix} = 2(-30 + 18) + 3(-24 + 54) + (p-1)(8 - 30) \]
\[ = -24 + 90 - 22(p - 1) = 66 - 22(p - 1) \]
Step 4: Use the volume.
\(\dfrac{1}{6}|66 - 22(p-1)| = 11 \Rightarrow |66 - 22(p-1)| = 66\).
Case 1: \(66 - 22(p-1) = 66\), so \(p = 1\).
Case 2: \(66 - 22(p-1) = -66\), so \(p - 1 = 6\) and \(p = 7\).
Step 5: Check the options.
One of the values of \(p\) is 1, which is option (A). The value 7 is not listed.
Final Answer:
One value is \(p = 1\), option (A). \[ \boxed{p = 1} \]
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