Understanding the Problem: Finding Triangle Area Using Vectors

The question asks us to find the area of a triangle ABC. We are given the vectors representing two adjacent sides, AB and AC. Specifically, the vector \(\vec{AB}\) is given as \(-2\vec{i} + 3\vec{j} + 2\vec{k}\) and the vector \(\vec{AC}\) is given as \(-4\vec{i} + 5\vec{j} + 2\vec{k}\). We need to use vector methods to calculate the area of this triangle.

Area of a Triangle Using Side Vectors

The area of a triangle can be calculated using the vectors representing two adjacent sides. If \(\vec{a}\) and \(\vec{b}\) are two vectors representing the adjacent sides of a triangle, the area of the triangle is half the magnitude of their cross product. The formula is:

Area \(= \frac{1}{2} |\vec{a} \times \vec{b}|\)

In this problem, our adjacent sides are represented by the vectors \(\vec{AB}\) and \(\vec{AC}\). So, the area of triangle ABC is:

Area of \(\Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}|\)

Step-by-Step Calculation of Triangle Area

Let's perform the calculation in steps:

Step 1: Find the cross product of the vectors \(\vec{AB}\) and \(\vec{AC}\).

The vectors are \(\vec{AB} = -2\vec{i} + 3\vec{j} + 2\vec{k}\) and \(\vec{AC} = -4\vec{i} + 5\vec{j} + 2\vec{k}\).

The cross product \(\vec{AB} \times \vec{AC}\) is calculated using a determinant:

\( \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 3 & 2 \\ -4 & 5 & 2 \end{vmatrix} \)

Expand the determinant:

\( \vec{AB} \times \vec{AC} = \vec{i}((3)(2) - (5)(2)) - \vec{j}((-2)(2) - (-4)(2)) + \vec{k}((-2)(5) - (-4)(3)) \)

\( \vec{AB} \times \vec{AC} = \vec{i}(6 - 10) - \vec{j}(-4 - (-8)) + \vec{k}(-10 - (-12)) \)

\( \vec{AB} \times \vec{AC} = \vec{i}(-4) - \vec{j}(-4 + 8) + \vec{k}(-10 + 12) \)

\( \vec{AB} \times \vec{AC} = -4\vec{i} - 4\vec{j} + 2\vec{k} \)

Step 2: Find the magnitude of the cross product.

The magnitude of the vector \(-4\vec{i} - 4\vec{j} + 2\vec{k}\) is given by:

\( |\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + (-4)^2 + (2)^2} \)

\( |\vec{AB} \times \vec{AC}| = \sqrt{16 + 16 + 4} \)

\( |\vec{AB} \times \vec{AC}| = \sqrt{36} \)

\( |\vec{AB} \times \vec{AC}| = 6 \)

Step 3: Calculate the area of the triangle.

Using the formula Area \(= \frac{1}{2} |\vec{AB} \times \vec{AC}|\), we get:

\( \text{Area} = \frac{1}{2} \times 6 \)

\( \text{Area} = 3 \)

The area of the triangle ABC is 3 square units.

Conclusion: Triangle Area Calculation

By calculating the cross product of the vectors representing the adjacent sides AB and AC, and then finding half of the magnitude of this cross product, we determined the area of triangle ABC. The calculation shows the area is 3 square units.

Revision Table: Key Vector Concepts for Area

Additional Information: Parallelogram Area and Vector Applications

The magnitude of the cross product of two vectors \(\vec{a}\) and \(\vec{b}\), i.e., \(|\vec{a} \times \vec{b}|\), represents the area of the parallelogram formed by these two vectors when they are placed tail-to-tail. Since a triangle formed by two adjacent sides is exactly half of the parallelogram formed by those sides, the triangle's area is indeed \(\frac{1}{2} |\vec{a} \times \vec{b}|\).

Vectors are powerful tools in geometry and physics. They can be used to represent forces, velocities, displacements, and also geometric entities like areas and volumes. The cross product, specifically, is useful in problems involving areas, torque, and angular momentum, as it naturally produces a vector perpendicular to the plane defined by the two original vectors.

Understanding how to represent geometric shapes using vectors and applying vector operations like the cross product allows us to solve complex geometric problems analytically.