Question:
If the points \(\mathrm{A}(1,1,2), \mathrm{B}(2,1, \mathrm{p}), \mathrm{C}(1,0,3)\) and \(D(2,2,0)\) are coplanar then the value of \(p\) is
If the points \(\mathrm{A}(1,1,2), \mathrm{B}(2,1, \mathrm{p}), \mathrm{C}(1,0,3)\) and \(D(2,2,0)\) are coplanar then the value of \(p\) is
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Four points are coplanar when the scalar triple product of \(\overline{AB}, \overline{AC}, \overline{AD}\) is zero.
Updated On: Nov 13, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Points A, B, C, D are coplanar if \([\overline{AB}\ \overline{AC}\ \overline{AD}] = 0\).
Step 2: Find the vectors
\(\overline{AB} = (1, 0, p-2)\), \(\overline{AC} = (0, -1, 1)\), \(\overline{AD} = (1, 1, -2)\).
Step 3: Determinant
\(\begin{vmatrix} 1 & 0 & p-2 \\ 0 & -1 & 1 \\ 1 & 1 & -2 \end{vmatrix} = 1(2 - 1) + (p-2)(0 + 1) = p - 1\)
Step 4: Solve
\(p - 1 = 0\), so \(p = 1\).
Final Answer:
The value of p is 1. \[ \boxed{\text{(C) }1} \]
Points A, B, C, D are coplanar if \([\overline{AB}\ \overline{AC}\ \overline{AD}] = 0\).
Step 2: Find the vectors
\(\overline{AB} = (1, 0, p-2)\), \(\overline{AC} = (0, -1, 1)\), \(\overline{AD} = (1, 1, -2)\).
Step 3: Determinant
\(\begin{vmatrix} 1 & 0 & p-2 \\ 0 & -1 & 1 \\ 1 & 1 & -2 \end{vmatrix} = 1(2 - 1) + (p-2)(0 + 1) = p - 1\)
Step 4: Solve
\(p - 1 = 0\), so \(p = 1\).
Final Answer:
The value of p is 1. \[ \boxed{\text{(C) }1} \]
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