Question

Let the points \(A(a,-1,2)\), \(B(1,b,-4)\), \(C(-1,1,c)\) and \(D(1,-2,8)\) be the vertices of a parallelogram \(ABCD\). Then its area is equal to:

• A. \(2\sqrt{73}\)
• B. \(2\sqrt{51}\)
• C. \(28\)
• D. \(14\)

Hint:

For 3D parallelogram problems, always use the diagonal condition first to determine unknowns, then apply the cross product for area.

Answer 1

The Correct Option is C

We are given the vertices of parallelogram \(ABCD\) as:

• \(A(a,-1,2)\)
• \(B(1,b,-4)\)
• \(C(-1,1,c)\)
• \(D(1,-2,8)\)

The area of a parallelogram formed by two adjacent sides is equal to the magnitude of the cross product of the corresponding vectors.

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Step 1: Find vectors \(\vec{AB}\) and \(\vec{AD}\)

\[ \vec{AB} = B - A = (1-a,\; b+1,\; -6) \]

\[ \vec{AD} = D - A = (1-a,\; -1,\; 6) \]

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Step 2: Use parallelogram condition

For a parallelogram, diagonals bisect each other:

\[ \vec{AC} = \vec{BD} \]

Compute vectors:

\[ \vec{AC} = (-1-a,\; 2,\; c-2) \]

\[ \vec{BD} = (0,\; -2-b,\; 12) \]

Equating components:

• \(-1-a = 0 \Rightarrow a = -1\)
• \(2 = -2-b \Rightarrow b = -4\)
• \(c-2 = 12 \Rightarrow c = 14\)

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Step 3: Substitute values

\[ \vec{AB} = (2,\; -3,\; -6) \]

\[ \vec{AD} = (2,\; -1,\; 6) \]

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Step 4: Compute cross product

\[ \vec{AB} \times \vec{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & -6 \\ 2 & -1 & 6 \end{vmatrix} \]

\[ = \mathbf{i}((-3)(6)-(-6)(-1)) - \mathbf{j}((2)(6)-(-6)(2)) + \mathbf{k}((2)(-1)-(-3)(2)) \]

\[ = \mathbf{i}(-18-6) - \mathbf{j}(12+12) + \mathbf{k}(-2+6) \]

\[ = -24\mathbf{i} -24\mathbf{j} + 4\mathbf{k} \]

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Step 5: Find magnitude (area)

\[ \text{Area} = \left|\vec{AB} \times \vec{AD}\right| = \sqrt{(-24)^2 + (-24)^2 + 4^2} \]

\[ = \sqrt{576 + 576 + 16} = \sqrt{1168} = 4\sqrt{73} \]

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Final Answer

\[ \boxed{4\sqrt{73}} \]