Example 24: Area of the triangle $A(1,1,1)$⁠, $B(1,2,3)$⁠, $C(2,3,1)$

Basics you need

Solution

1. Two sides from the vertex A

   $$\begin{aligned} \overrightarrow{AB} &= (1 - 1)\hat i + (2 - 1)\hat j + (3 - 1)\hat k \\ &= \hat j + 2\hat k, \\ \overrightarrow{AC} &= (2 - 1)\hat i + (3 - 1)\hat j + (1 - 1)\hat k \\ &= \hat i + 2\hat j. \end{aligned}$$
   The formula needs two sides that start at the same vertex. Each side vector is terminal minus initial; the zero components (no $\hat i$ in $\overrightarrow{AB}$⁠, no $\hat k$ in $\overrightarrow{AC}$⁠) become zeros in the determinant.
   $\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1}$ (Section 10.5.2). Basics: three numbers fix a point

2. Why the area is $\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|$

   Take AB as the base. If $\theta$ is the angle at A, the height from C to the line AB is $|\overrightarrow{AC}|\sin\theta$⁠. So

   $$\begin{aligned} \text{area} &= \tfrac12 \times |\overrightarrow{AB}| \times |\overrightarrow{AC}|\sin\theta \\ &= \tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|. \end{aligned}$$
   The length of a vector product is $|\vec a||\vec b|\sin\theta$⁠, which is base times height of the parallelogram on $\vec a$ and $\vec b$⁠; the triangle is exactly half of that parallelogram.
   Area of a triangle $= \frac12|\vec a\times\vec b|$ (Section 10.6.3, Observation 8). Lesson: the cross product

3. The vector product

   $$\begin{aligned} \overrightarrow{AB}\times\overrightarrow{AC} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} \\ &= \hat i(0 - 4) - \hat j(0 - 2) \\ &\quad + \hat k(0 - 1) \\ &= -4\hat i + 2\hat j - \hat k. \end{aligned}$$
   The minors are $(1)(0) - (2)(2) = -4$⁠, $(0)(0) - (2)(1) = -2$ and $(0)(2) - (1)(1) = -1$⁠; the $\hat j$ term takes a minus sign, giving $+2\hat j$⁠. As a check, $(-4\hat i + 2\hat j - \hat k)\cdot(\hat j + 2\hat k) = 2 - 2 = 0$ and $(-4\hat i + 2\hat j - \hat k)\cdot(\hat i + 2\hat j) = -4 + 4 = 0$⁠.
   Expansion along the top row, signs $+, -, +$⁠. Basics: a determinant measures a signed area

4. Its length, and the area

   $$\begin{aligned} |\overrightarrow{AB}\times\overrightarrow{AC}| &= \sqrt{(-4)^2 + 2^2 + (-1)^2} \\ &= \sqrt{21}, \end{aligned}$$ $$\text{area of } \triangle ABC = \tfrac12\sqrt{21}\ \text{sq. units}.$$
   Squaring removes the signs: $16 + 4 + 1 = 21$⁠. Halving is the step that turns the parallelogram into the triangle.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

5. Check with a base and a height

   $|\overrightarrow{AB}| = \sqrt{0 + 1 + 4} = \sqrt5$ and $|\overrightarrow{AC}| = \sqrt{1 + 4 + 0} = \sqrt5$⁠, so the triangle is isosceles with apex A. The base is $\overrightarrow{BC} = \hat i + \hat j - 2\hat k$⁠, of length $\sqrt6$⁠. The midpoint of BC is M$\left(\frac32, \frac52, 2\right)$⁠, and

   $$\begin{aligned} \overrightarrow{AM} &= \tfrac12\hat i + \tfrac32\hat j + \hat k, \\ |\overrightarrow{AM}| &= \sqrt{\tfrac14 + \tfrac94 + 1} = \sqrt{\tfrac{14}{4}} = \tfrac{\sqrt{14}}{2}. \end{aligned}$$

   $\overrightarrow{AM}\cdot\overrightarrow{BC} = \frac12 + \frac32 - 2 = 0$⁠, so AM is the height, and

   $$\tfrac12 \times \sqrt6 \times \tfrac{\sqrt{14}}{2} = \tfrac{\sqrt{84}}{4} = \tfrac{2\sqrt{21}}{4} = \tfrac12\sqrt{21}.$$
   In an isosceles triangle the line from the apex to the midpoint of the base is perpendicular to the base, and the scalar product confirms it. This route uses no vector product at all, so it checks Steps 3 and 4 independently.
   Area $= \frac12 \times$ base $\times$ height; perpendicular when the scalar product is 0 (Section 10.6.1, Observation 2). Lesson: Examples 22–25

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