Finding Area of Parallelogram Using Diagonals

The problem asks us to find the area of a parallelogram when the lengths and directions of its diagonals are given as vectors. We are given two vectors representing the diagonals of the parallelogram.

Let the diagonals be \(\vec{d_1}\) and \(\vec{d_2}\).

Given:

• \(\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}\)
• \(\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}\)

The area of a parallelogram can be calculated using the magnitude of the cross product of its diagonals. The formula is:

Area \(= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|\)

First, we need to calculate the cross product \(\vec{d_1} \times \vec{d_2}\). The cross product of two vectors \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\) and \(\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\) is given by the determinant of a matrix:

\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}\)

Substituting the components of \(\vec{d_1}\) and \(\vec{d_2}\):

\(\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}\)

Expanding the determinant:

\(\vec{d_1} \times \vec{d_2} = \hat{i}((1)(4) - (-2)(-3)) - \hat{j}((3)(4) - (-2)(1)) + \hat{k}((3)(-3) - (1)(1))\)

\(\vec{d_1} \times \vec{d_2} = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1)\)

\(\vec{d_1} \times \vec{d_2} = \hat{i}(-2) - \hat{j}(12 + 2) + \hat{k}(-10)\)

\(\vec{d_1} \times \vec{d_2} = -2\hat{i} - 14\hat{j} - 10\hat{k}\)

Next, we need to find the magnitude of this resulting vector. The magnitude of a vector \(\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k}\) is \(|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}\).

\(|\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2}\)

\(|\vec{d_1} \times \vec{d_2}| = \sqrt{4 + 196 + 100}\)

\(|\vec{d_1} \times \vec{d_2}| = \sqrt{300}\)

We can simplify \(\sqrt{300}\):

\(\sqrt{300} = \sqrt{100 \times 3} = \sqrt{100} \times \sqrt{3} = 10\sqrt{3}\)

So, \(|\vec{d_1} \times \vec{d_2}| = 10\sqrt{3}\).

Finally, we calculate the area of the parallelogram using the formula:

Area \(= \frac{1}{2} |\vec{d_1} \times \vec{d_2}| = \frac{1}{2} (10\sqrt{3})\)

Area \(= 5\sqrt{3}\) square units.

This result matches one of the given options.

Let's compare this with the options:

Our calculated area is \(5\sqrt{3}\) square units, which corresponds to Option 3.

Revision Table: Key Concepts for Parallelogram Area

Additional Information: Vectors and Parallelograms

Understanding vectors is crucial for solving geometry problems in three dimensions. Here are some related points:

• Vectors as Geometric Objects: Vectors represent both magnitude (length) and direction. In this problem, the diagonals are represented as vectors in a 3D coordinate system using \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) unit vectors.
• Cross Product: The cross product of two vectors results in a new vector that is perpendicular to both original vectors. Its magnitude is related to the area of the parallelogram formed by the original vectors as adjacent sides. Specifically, \(|\vec{a} \times \vec{b}|\) is the area of the parallelogram with sides \(\vec{a}\) and \(\vec{b}\).
• Diagonals of a Parallelogram: The diagonals of a parallelogram bisect each other. If the adjacent sides are \(\vec{a}\) and \(\vec{b}\), the diagonals are \(\vec{d_1} = \vec{a} + \vec{b}\) and \(\vec{d_2} = \vec{a} - \vec{b}\). The formula used in this problem, Area \(= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|\), can be derived from the side formula using this relationship between sides and diagonals.

Calculating the cross product and its magnitude are standard vector operations essential for solving problems involving areas, volumes, and perpendicular vectors in 3D space.