Example 25: Area of the parallelogram on $3\hat i+\hat j+4\hat k$ and $\hat i-\hat j+\hat k$

Basics you need

Solution

1. Area as the length of a vector product

   Take $\vec a$ as the base. The height is the distance of the opposite side from the base, $|\vec b|\sin\theta$⁠, so

   $$\text{area} = |\vec a|\,|\vec b|\sin\theta = |\vec a\times\vec b|.$$
   By definition $\vec a\times\vec b = |\vec a||\vec b|\sin\theta\,\hat n$ with $\hat n$ a unit vector, so its length is $|\vec a||\vec b|\sin\theta$ (with $\sin\theta \ge 0$ because $0 \le \theta \le \pi$⁠). That is the base times the height.
   Area of a parallelogram $= |\vec a\times\vec b|$ (Section 10.6.3, Observation 9). Lesson: the cross product

2. The vector product

   $$\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} \\ &= \hat i(1 + 4) - \hat j(3 - 4) \\ &\quad + \hat k(-3 - 1) \\ &= 5\hat i + \hat j - 4\hat k. \end{aligned}$$
   The minors are $(1)(1) - (4)(-1) = 1 + 4$⁠, $(3)(1) - (4)(1) = 3 - 4$ and $(3)(-1) - (1)(1) = -3 - 1$⁠. The middle one, $-1$⁠, takes a minus sign and becomes $+1$⁠, so the $\hat j$ component is $+1$⁠.
   Expansion along the top row, signs $+, -, +$⁠. Basics: a determinant measures a signed area

3. Check: perpendicular to both sides

   $$\begin{aligned} (5\hat i + \hat j - 4\hat k)\cdot\vec a &= 15 + 1 - 16 = 0, \\ (5\hat i + \hat j - 4\hat k)\cdot\vec b &= 5 - 1 - 4 = 0. \end{aligned}$$
   $\vec a\times\vec b$ must be perpendicular to both $\vec a$ and $\vec b$⁠; both scalar products are 0, so the determinant work is consistent.
   $\vec u\cdot\vec v = 0$ for perpendicular vectors (Section 10.6.1, Observation 2). Lesson: the dot product

4. The length gives the area

   $$\begin{aligned} |\vec a\times\vec b| &= \sqrt{5^2 + 1^2 + (-4)^2} \\ &= \sqrt{25 + 1 + 16} = \sqrt{42}. \end{aligned}$$

   So the area is $\sqrt{42}$ square units. Since $42 = 2 \times 3 \times 7$ has no square factor, $\sqrt{42}$ is already in simplest form.

   An area is a length squared, so it is measured in square units; the vector product's direction (along $5\hat i + \hat j - 4\hat k$⁠) is not needed, only its length.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

5. A check with the scalar product only

   $|\vec a|^2 = 9 + 1 + 16 = 26$⁠, $|\vec b|^2 = 1 + 1 + 1 = 3$ and $\vec a\cdot\vec b = 3 - 1 + 4 = 6$⁠. Since $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2\sin^2\theta$ $= |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2$⁠,

   $$\begin{aligned} |\vec a\times\vec b|^2 &= (26)(3) - 6^2 \\ &= 78 - 36 = 42. \end{aligned}$$
   This uses no determinant at all, so it is an independent check of Steps 2 and 4. It comes from $\sin^2\theta = 1 - \cos^2\theta$ and $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$⁠.
   $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ (Section 10.6.1, Observation 6). Lesson: Examples 22–25

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