Math Problem Statement

Find the area of the triangle ABC with vertices A(-1, 0, -1), B(1, 2, 3), and C(-1, 1, 2).

Solution

To solve the problem of finding the area of the triangle △ABC\triangle ABC with vertices A(−1,0,−1)A(-1, 0, -1), B(1,2,3)B(1, 2, 3), and C(−1,1,2)C(-1, 1, 2), we use the formula for the area of a triangle in 3D space:

Area=12∥AB⃗×AC⃗∥\text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \|

Step 1: Compute vectors AB⃗\vec{AB} and AC⃗\vec{AC}

AB⃗=B−A=(1−(−1),2−0,3−(−1))=(2,2,4)\vec{AB} = B - A = (1 - (-1), 2 - 0, 3 - (-1)) = (2, 2, 4) AC⃗=C−A=(−1−(−1),1−0,2−(−1))=(0,1,3)\vec{AC} = C - A = (-1 - (-1), 1 - 0, 2 - (-1)) = (0, 1, 3)

Step 2: Compute the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}

The cross product of two vectors is given by:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 4 \\ 0 & 1 & 3 \end{vmatrix}$$ Expand the determinant: $$\vec{AB} \times \vec{AC} = \mathbf{i} \begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 4 \\ 0 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 2 \\ 0 & 1 \end{vmatrix}$$ $$= \mathbf{i} (2 \cdot 3 - 4 \cdot 1) - \mathbf{j} (2 \cdot 3 - 4 \cdot 0) + \mathbf{k} (2 \cdot 1 - 2 \cdot 0)$$ $$= \mathbf{i} (6 - 4) - \mathbf{j} (6 - 0) + \mathbf{k} (2 - 0)$$ $$= 2\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}$$ Thus, $$\vec{AB} \times \vec{AC} = (2, -6, 2)$$. ### Step 3: Compute the magnitude of $$\vec{AB} \times \vec{AC}$$ $$\| \vec{AB} \times \vec{AC} \| = \sqrt{2^2 + (-6)^2 + 2^2} = \sqrt{4 + 36 + 4} = \sqrt{44} = 2\sqrt{11}$$ ### Step 4: Compute the area of the triangle $$\text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} (2\sqrt{11}) = \sqrt{11}$$ ### Final Answer: The area of the triangle is: $$\boxed{\sqrt{11}}$$ Let me know if you'd like a further breakdown of any step! Here are some related questions you might want to explore: 1. How do we compute the cross product in 3D space? 2. What are some geometric properties of the cross product? 3. How can we generalize this formula to polygons in 3D? 4. Why do we divide by 2 in the area formula? 5. What is the interpretation of the magnitude of a cross product geometrically? **Tip:** Always verify your vector arithmetic carefully, as small errors can propagate!

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Math Problem Analysis

Mathematical Concepts

Geometry
Vectors
Cross Product
3D Geometry

Formulas

Area of triangle in 3D space: (1/2) || AB × AC ||
Cross product of two vectors: i(det(jk)) - j(det(ik)) + k(det(ij))
Magnitude of a vector: ||v|| = sqrt(v1^2 + v2^2 + v3^2)

Theorems

Properties of the cross product
Pythagorean theorem in vector space

Suitable Grade Level

Grades 10-12

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