Question

show that the area of the triangle formed by the points (x1,y1,z1),(x2,y2,z2) and the origin is
1/2sqrt((y1z2-y2z1)2+(z1x2-z2x1)2+(x1y2-x2y1)2)

Answer 1

Conditions

show that the area of the triangle formed by the points $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and the origin is $1/2 \sqrt{2x_1 + (z_1 z_2 - z_2 x_1)^2 + (x_1 y_2 - x_2 y_1)^2}$

Solution

We must find the area of the triangle, which has 3 top points:

$$A = (x_1, y_1, z_1), B = (x_2, y_2, z_2), C = (0, 0, 0)$$

As it is known, there is a formula to find the area of triangle using coordinates of the tops:

$$S = \sqrt{S_x^2 + S_y^2 + S_z^2}$$ $$S_x = \frac{1}{2} \begin{vmatrix} y_B - y_A & z_B - z_A \\ y_C - y_A & z_C - z_A \end{vmatrix} = \frac{1}{2} \begin{vmatrix} 1 & y_A & z_A \\ 1 & y_B & z_B \\ 1 & y_C & z_C \end{vmatrix}$$ $$S_y = \frac{1}{2} \begin{vmatrix} x_A & 1 & z_A \\ x_B & 1 & z_B \\ x_C & 1 & z_C \end{vmatrix}, \qquad S_z = \frac{1}{2} \begin{vmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{vmatrix}$$

Where $\mathbf{r}_B(x_B, y_B, z_B), \mathbf{r}_C(x_C, y_C, z_C), \mathbf{r}_A(x_A, y_A, z_A)$ - top points.

So,

$$S_A = \frac{1}{2} \begin{vmatrix} 1 & y_1 & z_1 \\ 1 & y_2 & z_2 \\ 1 & 0 & 0 \end{vmatrix} = \frac{1}{2} (0 + y_1 z_2 + 0 - y_2 z_1 - 0 - 0) = \frac{1}{2} (y_1 z_2 - y_2 z_1)$$ $$S_B = \frac{1}{2} \begin{vmatrix} x_1 & 1 & z_1 \\ x_2 & 1 & z_2 \\ 0 & 1 & 0 \end{vmatrix} = \frac{1}{2} (0 + z_1 x_2 + 0 - z_2 x_1 - 0 - 0) = \frac{1}{2} (z_1 x_2 - z_2 x_1)$$ $$S_B = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ 0 & 0 & 1 \end{vmatrix} = \frac{1}{2} (0 + x_1 y_2 + 0 - x_2 y_1 - 0 - 0) = \frac{1}{2} (x_1 y_2 - x_2 y_1)$$ $$\begin{array}{l} S = \sqrt{ \frac{1}{4} (z_1 x_2 - z_2 x_1)^2 + \frac{1}{4} (z_1 x_2 - z_2 x_1)^2 + \frac{1}{4} (x_1 y_2 - x_2 y_1)^2 } \\ = \frac{1}{2} \sqrt{ (z_1 x_2 - z_2 x_1)^2 + (z_1 x_2 - z_2 x_1)^2 + (x_1 y_2 - x_2 y_1)^2 } \\ \end{array}$$

Q.E.D.