Math Problem Statement

use the cross product to find the area of the triangle with vertices (a,0,0) (0,b,0) and (0,0,c)

Solution

The area of a triangle with vertices (a,0,0)(a, 0, 0), (0,b,0)(0, b, 0), and (0,0,c)(0, 0, c) can be found using the cross product of two vectors representing two sides of the triangle.

Step-by-Step Process:

  1. Define the vectors:

    • Let v1\mathbf{v}_1 be the vector from (a,0,0)(a, 0, 0) to (0,b,0)(0, b, 0): v1=(0a,b0,00)=(a,b,0)\mathbf{v}_1 = (0 - a, b - 0, 0 - 0) = (-a, b, 0)
    • Let v2\mathbf{v}_2 be the vector from (a,0,0)(a, 0, 0) to (0,0,c)(0, 0, c): v2=(0a,00,c0)=(a,0,c)\mathbf{v}_2 = (0 - a, 0 - 0, c - 0) = (-a, 0, c)
  2. Compute the cross product v1×v2\mathbf{v}_1 \times \mathbf{v}_2: v1×v2=ijkab0a0c\mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix} Expanding the determinant: v1×v2=i(bc0)j(ac0)+k(0(a)(b))\mathbf{v}_1 \times \mathbf{v}_2 = \mathbf{i} \left(bc - 0\right) - \mathbf{j} \left(ac - 0\right) + \mathbf{k} \left(0 - (-a)(b)\right) Simplifying: v1×v2=(bc,ac,ab)\mathbf{v}_1 \times \mathbf{v}_2 = (bc, -ac, ab)

  3. Find the magnitude of the cross product: The magnitude of v1×v2\mathbf{v}_1 \times \mathbf{v}_2 is: v1×v2=(bc)2+(ac)2+(ab)2|\mathbf{v}_1 \times \mathbf{v}_2| = \sqrt{(bc)^2 + (-ac)^2 + (ab)^2} Simplifying: v1×v2=b2c2+a2c2+a2b2|\mathbf{v}_1 \times \mathbf{v}_2| = \sqrt{b^2c^2 + a^2c^2 + a^2b^2} Factor out the common terms: v1×v2=a2b2+a2c2+b2c2|\mathbf{v}_1 \times \mathbf{v}_2| = \sqrt{a^2b^2 + a^2c^2 + b^2c^2}

  4. Area of the triangle: The area of the triangle is half the magnitude of the cross product: Area=12v1×v2=12a2b2+a2c2+b2c2\text{Area} = \frac{1}{2} |\mathbf{v}_1 \times \mathbf{v}_2| = \frac{1}{2} \sqrt{a^2b^2 + a^2c^2 + b^2c^2}

Thus, the area of the triangle with vertices (a,0,0)(a, 0, 0), (0,b,0)(0, b, 0), and (0,0,c)(0, 0, c) is:

12a2b2+a2c2+b2c2\boxed{\frac{1}{2} \sqrt{a^2b^2 + a^2c^2 + b^2c^2}}

Would you like further details or have any questions?

Here are 5 related questions for practice:

  1. What is the cross product of v1=(1,2,0)\mathbf{v}_1 = (1, 2, 0) and v2=(0,1,3)\mathbf{v}_2 = (0, -1, 3)?
  2. How can the cross product be used to find the area of a parallelogram?
  3. What is the geometrical interpretation of the magnitude of the cross product?
  4. How would you find the area of a triangle in 3D if one of the vertices is at the origin?
  5. What happens to the area if two of the vertices coincide?

Tip: The cross product is always perpendicular to the plane containing the two vectors involved, making it a useful tool for 3D geometry.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Geometry of Triangles in 3D

Formulas

Cross product of two vectors: v1 x v2 = (v1_y * v2_z - v1_z * v2_y, v1_z * v2_x - v1_x * v2_z, v1_x * v2_y - v1_y * v2_x)
Magnitude of a cross product: |v1 x v2| = sqrt((v1 x v2)_x^2 + (v1 x v2)_y^2 + (v1 x v2)_z^2)
Area of a triangle: Area = 1/2 * |v1 x v2|

Theorems

Properties of the Cross Product
Pythagorean Theorem in 3D Geometry

Suitable Grade Level

Grades 10-12