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

Concept:
In a parallelogram, diagonals bisect each other.
If position vectors of vertices satisfy \( \vec{A}+\vec{C}=\vec{B}+\vec{D} \), then the points form a parallelogram.
Area of a parallelogram formed by vectors \(\vec{u}\) and \(\vec{v}\) is \( |\vec{u}\times\vec{v}| \).
Step 1: Use the diagonal property For parallelogram \(ABCD\), \[ \vec{A}+\vec{C}=\vec{B}+\vec{D} \] \[ (a,-1,2)+(-1,1,c)=(1,b,-4)+(1,-2,8) \] Equating components: \[ a-1=2 \Rightarrow a=3 \] \[ 0=b-2 \Rightarrow b=2 \] \[ 2+c=4 \Rightarrow c=2 \]
Step 2: Find direction vectors \[ \vec{AB}=B-A=(1-3,\;2+1,\;-4-2)=(-2,3,-6) \] \[ \vec{AD}=D-A=(1-3,\;-2+1,\;8-2)=(-2,-1,6) \]
Step 3: Find the cross product \[ \vec{AB}\times\vec{AD}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 3 & -6\\ -2 & -1 & 6 \end{vmatrix} \] \[ = \mathbf{i}(18-6)-\mathbf{j}(-12-12)+\mathbf{k}(2+6) \] \[ = (12,\,24,\,8) \]
Step 4: Find the area \[ \text{Area}= \sqrt{12^2+24^2+8^2} = \sqrt{144+576+64} = \sqrt{784} = 28 \]