Question

Find the area of the triangle with the given vertices. (Hint: \(\frac{1}{2}\|\mathbf{u} \times \mathbf{v}\|\) is the area of the triangle having \(u\) and \(v\) as adjacent sides.) $$ (2,-3,4),(0,1,2),(-1,2,0) $$

Short answer

The area of the triangle is \(\sqrt{11}\).

Step-by-step solution

Find the vectors

Based on the given vertices, we can create two vectors. Let's define vector u as the difference between the second point and the first point, and vector v as the difference between the third point and the first point. So we get: \n \( \mathbf{u} = \begin{bmatrix} 0 - 2 \ 1 - (-3) \ 2 - 4 \end{bmatrix} = \begin{bmatrix} -2 \ 4 \ -2 \end{bmatrix}\) \n \( \mathbf{v} = \begin{bmatrix} -1 - 2 \ 2 - (-3) \ 0 - 4 \end{bmatrix} = \begin{bmatrix} -3 \ 5 \ -4 \end{bmatrix}\)

Cross product of vectors

Next, calculate the cross product of vector u and vector v: \n \( \mathbf{u} \times \mathbf{v} = \begin{bmatrix} (4*-4)-(-2*5) \ (-2*-4)-(-2*-3) \ -2*5-4*-3 \end{bmatrix} = \begin{bmatrix} -6 \ -2 \ 2 \end{bmatrix}\)

Magnitude of cross product

Now, find the magnitude (or length) of this cross product vector: \n \( \|\mathbf{u} \times \mathbf{v}\| = \sqrt{(-6)^2 + (-2)^2 + 2^2} = \sqrt{44}\) \n

Calculate the area

Finally, apply the formula for area of a triangle to get \n \(Area=\frac{1}{2}\|\mathbf{u} \times \mathbf{v}\|=\frac{1}{2}*\sqrt{44}= \sqrt{11}\).

Key concepts

Cross Product of Vectors

The cross product, also known as the vector product, is an operation that takes two vectors in three-dimensional space and returns a new vector that is orthogonal (at right angles) to both of the original vectors. Imagine you have two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), which can physically represent two sides of a parallelogram in space. Their cross product, \(\mathbf{u} \times \mathbf{v}\), is a vector with a direction given by the right-hand rule and a magnitude equal to the area of that parallelogram.

For computing the cross product, each vector must be in the form \(\mathbf{u} = \begin{bmatrix} u_1 \ u_2 \ u_3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix}\). The result is given by \(\mathbf{u} \times \mathbf{v} = \begin{bmatrix} u_2v_3 - u_3v_2 \ u_3v_1 - u_1v_3 \ u_1v_2 - u_2v_1 \end{bmatrix}\). The coordinates of this new vector represent the areas of the two-dimensional projections of the parallelogram on the three coordinate planes (XY, YZ, and ZX planes).

Magnitude of a Vector

The magnitude of a vector represents its length, and it is computed using the Pythagorean theorem in three-dimensional space. When you have a vector \(\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix}\), the magnitude, denoted as \(\|\mathbf{a}\|\), is calculated using the formula \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\). Essentially, if the vector represents a displacement in space, the magnitude gives us the direct straight-line distance from the starting point to the end point of that displacement.

In the context of finding the area of a triangle, the magnitude of the cross product vector gives us twice the area of the triangle. The reason is that the cross product gives the area of a parallelogram, and a triangle is precisely half of a parallelogram when you split it diagonally.

Coordinates of Vertices

Coordinates of vertices are essentially the numerical addresses that pinpoint the location of each corner, or vertex, of a shape in a space. In a three-dimensional system, coordinates are expressed as a triplet \( (x, y, z) \) where \(x\) represents the position on the horizontal axis, \(y\) on the vertical axis, and \(z\) on the depth axis, creating a three-dimensional grid system.

To find the vectors needed for calculating the area of a triangle in three dimensions, you subtract the coordinates of one vertex from another. This gives you a vector pointing from one vertex to another, capturing the distance and direction between these two points. For instance, given the coordinates of two vertices, A and B, as \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), the vector from A to B is found by \(\mathbf{AB} = B - A = \begin{bmatrix} x_2 - x_1 \ y_2 - y_1 \ z_2 - z_1 \end{bmatrix}\). In our problem, using the coordinates of vertices, we determine vectors \(\mathbf{u}\) and \(\mathbf{v}\) that represent two sides of the triangle, which are then used to find the triangle's area using the cross product.