Question

The area of the quadrilateral ABCD, when A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal to

• A. 9 sq. units
• B. 18 sq. units
• C. 27 sq. units
• D. 81 sq. units

Answer 1

The Correct Option is A

We can find the area of the quadrilateral ABCD by dividing it into two triangles, for example, triangle ABC and triangle ADC.
The total area will be the sum of the areas of these two triangles.
The area of a triangle with vertices P, Q, R in 3D space is given by half the magnitude of the cross product of two vectors forming adjacent sides, e.g., Area = \( \frac{1}{2} |\vec{PQ} \times \vec{PR}| \).

First, calculate the area of triangle ABC. The vertices are A(0, 4, 1), B(2, 3, -1), C(4, 5, 0). Find the vectors \(\vec{AB}\) and \(\vec{AC}\): \[ \vec{AB} = B - A = (2 - 0, 3 - 4, -1 - 1) = (2, -1, -2) \] \[ \vec{AC} = C - A = (4 - 0, 5 - 4, 0 - 1) = (4, 1, -1) \]

Calculate the cross product \(\vec{AB} \times \vec{AC}\): \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -2 \\ 4 & 1 & -1 \end{vmatrix} \] \[ = \mathbf{i}((-1)(-1) - (-2)(1)) - \mathbf{j}((2)(-1) - (-2)(4)) + \mathbf{k}((2)(1) - (-1)(4)) \] \[ = \mathbf{i}(1 + 2) - \mathbf{j}(-2 + 8) + \mathbf{k}(2 + 4) \] \[ = 3\mathbf{i} - 6\mathbf{j} + 6\mathbf{k} = (3, -6, 6) \]

Calculate the magnitude of the cross product: \[ |\vec{AB} \times \vec{AC}| = \sqrt{3^2 + (-6)^2 + 6^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9 \] The area of triangle ABC is: \[ \text{Area}(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \times 9 = 4.5 \text{ sq. units} \]

Next, calculate the area of triangle ADC. The vertices are A(0, 4, 1), D(2, 6, 2), C(4, 5, 0). Find the vectors \(\vec{AD}\) and \(\vec{AC}\): \[ \vec{AD} = D - A = (2 - 0, 6 - 4, 2 - 1) = (2, 2, 1) \] \[ \vec{AC} = (4, 1, -1) \quad \text{(calculated earlier)} \]

Calculate the cross product \(\vec{AD} \times \vec{AC}\): \[ \vec{AD} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 4 & 1 & -1 \end{vmatrix} \] \[ = \mathbf{i}((2)(-1) - (1)(1)) - \mathbf{j}((2)(-1) - (1)(4)) + \mathbf{k}((2)(1) - (2)(4)) \] \[ = \mathbf{i}(-2 - 1) - \mathbf{j}(-2 - 4) + \mathbf{k}(2 - 8) \] \[ = -3\mathbf{i} + 6\mathbf{j} - 6\mathbf{k} = (-3, 6, -6) \]

Calculate the magnitude of the cross product: \[ |\vec{AD} \times \vec{AC}| = \sqrt{(-3)^2 + 6^2 + (-6)^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9 \] The area of triangle ADC is: \[ \text{Area}(\triangle ADC) = \frac{1}{2} |\vec{AD} \times \vec{AC}| = \frac{1}{2} \times 9 = 4.5 \text{ sq. units} \]

The total area of the quadrilateral ABCD is the sum of the areas of the two triangles: \[ \text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) \] \[ \text{Area}(ABCD) = 4.5 + 4.5 = 9 \text{ sq. units} \]

Alternatively, if the quadrilateral is planar (which can be confirmed because the normal vectors \( (3, -6, 6) \) and \( (-3, 6, -6) \) are parallel), the area can be calculated as half the magnitude of the cross product of its diagonals. Diagonal \(\vec{AC}\) = (4, 1, -1) Diagonal \(\vec{BD}\) = D - B = (2 - 2, 6 - 3, 2 - (-1)) = (0, 3, 3) \[ \vec{AC} \times \vec{BD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 1 & -1 \\ 0 & 3 & 3 \end{vmatrix} \] \[ = \mathbf{i}((1)(3) - (-1)(3)) - \mathbf{j}((4)(3) - (-1)(0)) + \mathbf{k}((4)(3) - (1)(0)) \] \[ = \mathbf{i}(3 + 3) - \mathbf{j}(12 - 0) + \mathbf{k}(12 - 0) = 6\mathbf{i} - 12\mathbf{j} + 12\mathbf{k} = (6, -12, 12) \] \[ |\vec{AC} \times \vec{BD}| = \sqrt{6^2 + (-12)^2 + 12^2} = \sqrt{36 + 144 + 144} = \sqrt{324} = 18 \] \[ \text{Area}(ABCD) = \frac{1}{2} |\vec{AC} \times \vec{BD}| = \frac{1}{2} \times 18 = 9 \text{ sq. units} \] Both methods yield the same result.

The area of the quadrilateral ABCD is (A): 9 sq. units.