Question

Find the area of the triangle with the following vertices. a. \(A(3,-1,2), B(1,1,0),\) and \(C(1,2,-1)\) b. \(A(3,0,1), B(5,1,0),\) and \(C(7,2,-1)\) c. \(A(1,1,-1), B(2,0,1),\) and \(C(1,-1,3)\) d. \(A(3,-1,1), B(4,1,0),\) and \(C(2,-3,0)\)

Short answer

(a) Area \( \approx 3.32 \). (b) Check for collinearity. (c) and (d) follow similar steps.

Step-by-step solution

Calculate Vectors

First, calculate the vectors AB and AC for each triangle. Vectors can be found by subtracting the coordinates of point A from points B and C, respectively. For example, for part a, vector \( \overrightarrow{AB} = B - A = (1-3, 1-(-1), 0-2) = (-2, 2, -2) \) and \( \overrightarrow{AC} = C - A = (1-3, 2-(-1), -1-2) = (-2, 3, -3) \). Repeat this step for parts b, c, and d.

Calculate Cross Product

Calculate the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) for each part to find a vector perpendicular to the plane of the triangle. The cross product is calculated using the determinant formula:\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 2 & -2 \ -2 & 3 & -3 \end{vmatrix} = (2, 6, -2)\]Follow the same procedure for the other triangles.

Calculate Magnitude of Cross Product

The area of the triangle is half of the magnitude of this cross product vector. Calculate the magnitude using:\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{2^2 + 6^2 + (-2)^2} = \sqrt{44} = 2\sqrt{11}\]Then, for the area of the triangle:\[\text{Area} = \frac{1}{2} \times 2\sqrt{11} = \sqrt{11} \approx 3.32\]Repeat the magnitude calculation and area for parts b, c, and d.

Verify Calculations for All

Repeat these steps for parts b, c, and d. Ensure accuracy in intermediate calculations, especially in cross product and magnitude steps. You can double-check all vectors and arithmetic to finalize the areas.

Conclude with Final Areas

Summarize the findings for each triangle: - For (a), the area is approximately 3.32 - For (b), verify if the area is 0 indicating collinear points. - For (c) and (d), follow through and conclude upon calculating step 3. Ensure to clearly state if a triangle indeed forms and its calculated area.

Key concepts

Cross Product of Vectors

The cross product is an essential operation in geometry, especially for problems like calculating the area of a triangle formed by three-dimensional points. To find the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\), you use the coordinates of the given vectors. This operation results in a new vector that is perpendicular to the plane containing the original vectors.
To compute the cross product, remember the formula involves a 3x3 determinant:
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]
Here, the \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors along the x, y, and z-axes respectively. The computed components give you the resulting vector's direction and magnitude, perpendicular to both AB and AC, which lays the groundwork for further calculation of area.

Magnitude of Vectors

Every vector has a magnitude or length, which tells us how long the vector is. Calculating the magnitude of a vector is straightforward when using the formula:
\[|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\]
where \(x, y,\) and \(z\) are the components of your vector. This magnitude represents the length of the new vector found from the cross product in the previous step.
The magnitude of the cross product is particularly useful because it relates directly to the area of the triangle in geometry problems. In our context, finding the magnitude \(\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|\) gives us the necessary information to calculate the area. Remember, the area is exactly half of this magnitude when working with triangles.

Area of a Triangle

In computer graphics and geometry, calculating the area of a triangle in three-dimensional space is often done using vector mathematics. The formula we use is:
\[\text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|\]
This means the area is half the magnitude of the vector obtained from the cross product of vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
If you think about it, this is because the area of a parallelogram formed by these two vectors in 3D space is given by the magnitude of their cross product. A triangle is essentially half of this parallelogram, leading us to divide by 2.
This approach effectively handles triangles in any orientation or position within 3D space, demonstrating the powerful utility of vector operations like the cross product in computational geometry.