Math Problem Statement

Find the area of the triangle with vertices (-4, 2, -5), (-5, 1, -5), and (-2, 0, -4).

Solution

To find the area of the triangle with the given vertices in 3D space (−4,2,−5)(-4, 2, -5), (−5,1,−5)(-5, 1, -5), and (−2,0,−4)(-2, 0, -4), we can use the following formula:

Formula:

The area of a triangle in 3D space is given by: Area=12∥AB×AC∥\text{Area} = \frac{1}{2} \| \mathbf{AB} \times \mathbf{AC} \| where:

  • AB\mathbf{AB} and AC\mathbf{AC} are vectors formed by subtracting the coordinates of the vertices.
  • AB×AC\mathbf{AB} \times \mathbf{AC} is the cross product of these two vectors.
  • ∥AB×AC∥\|\mathbf{AB} \times \mathbf{AC}\| is the magnitude of the cross product.

Step 1: Calculate Vectors

Let the vertices be: A=(−4,2,−5),B=(−5,1,−5),C=(−2,0,−4)A = (-4, 2, -5), \quad B = (-5, 1, -5), \quad C = (-2, 0, -4)

Vectors: AB=B−A=(−5−(−4),1−2,−5−(−5))=(−1,−1,0)\mathbf{AB} = B - A = (-5 - (-4), 1 - 2, -5 - (-5)) = (-1, -1, 0) AC=C−A=(−2−(−4),0−2,−4−(−5))=(2,−2,1)\mathbf{AC} = C - A = (-2 - (-4), 0 - 2, -4 - (-5)) = (2, -2, 1)

Step 2: Compute Cross Product AB×AC\mathbf{AB} \times \mathbf{AC}

The cross product of two vectors (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is given by:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 0 \\ 2 & -2 & 1 \end{vmatrix} = \mathbf{i}((-1)(1) - (0)(-2)) - \mathbf{j}((-1)(1) - (0)(2)) + \mathbf{k}((-1)(-2) - (-1)(2))$$ $$= \mathbf{i}(-1 - 0) - \mathbf{j}(-1 - 0) + \mathbf{k}(2 - (-2))$$ $$= -\mathbf{i} - (-\mathbf{j}) + \mathbf{k}(4)$$ $$= -\mathbf{i} + \mathbf{j} + 4\mathbf{k}$$ Thus: $$\mathbf{AB} \times \mathbf{AC} = (-1, 1, 4)$$ --- ### Step 3: Compute Magnitude of Cross Product $$\|\mathbf{AB} \times \mathbf{AC}\| = \sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$$ --- ### Step 4: Compute Area of Triangle $$\text{Area} = \frac{1}{2} \|\mathbf{AB} \times \mathbf{AC}\| = \frac{1}{2} \cdot 3\sqrt{2} = \frac{3\sqrt{2}}{2}$$ --- ### Final Answer: The area of the triangle is: $$\boxed{\frac{3\sqrt{2}}{2}}$$ Let me know if you want further details or additional explanations! --- ### Related Questions: 1. What is the geometric interpretation of the cross product in this problem? 2. How can the area of a triangle in 3D space be derived using a determinant approach? 3. How do you find the normal vector of the plane containing the triangle? 4. Can this triangle's area be verified using other methods like projection? 5. What are the properties of the cross product useful in this context? --- ### Tip: When solving 3D geometry problems, always sketch the problem or verify vector calculations step by step to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
3D Geometry
Cross Product
Area of a Triangle in 3D Space

Formulas

Area = 1/2 |AB × AC|
AB = B - A
AC = C - A
|AB × AC| = sqrt((i^2) + (j^2) + (k^2))

Theorems

Properties of Cross Product

Suitable Grade Level

Grades 11-12

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