Example 22: $|\vec a\times\vec b|$ for $2\hat i+\hat j+3\hat k$⁠, $3\hat i+5\hat j-2\hat k$

Basics you need

Solution

1. Set up the determinant

   $$\vec a\times\vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}.$$
   The second row holds the components of $\vec a$ (the first vector) and the third row those of $\vec b$⁠. The order matters: swapping the two rows gives $\vec b\times\vec a = -\vec a\times\vec b$⁠, the opposite vector (though with the same length).
   The determinant form of $\vec a\times\vec b$ (Section 10.6.3). Lesson: the cross product

2. Expand along the top row

   $$\begin{aligned} \vec a\times\vec b &= \hat i\begin{vmatrix} 1 & 3 \\ 5 & -2 \end{vmatrix} - \hat j\begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} \\ &\quad + \hat k\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix} \\ &= \hat i(-2 - 15) - (-4 - 9)\hat j \\ &\quad + (10 - 3)\hat k \\ &= -17\hat i + 13\hat j + 7\hat k. \end{aligned}$$
   Cover the row and column of each top entry and take the 2 by 2 determinant that is left: for $\hat i$⁠, $(1)(-2) - (3)(5) = -2 - 15$⁠; for $\hat j$⁠, $(2)(-2) - (3)(3) = -4 - 9$⁠; for $\hat k$⁠, $(2)(5) - (1)(3) = 10 - 3$⁠. The middle term carries a minus sign, so $-(-13) = +13$⁠.
   Expansion along the first row, signs $+, -, +$⁠. Basics: a determinant measures a signed area

3. Check: the answer is perpendicular to both vectors

   $$\begin{aligned} (\vec a\times\vec b)\cdot\vec a &= -34 + 13 + 21 = 0, \\ (\vec a\times\vec b)\cdot\vec b &= -51 + 65 - 14 = 0. \end{aligned}$$
   The products are $(-17)(2) = -34$⁠, $(13)(1) = 13$⁠, $(7)(3) = 21$ for $\vec a$⁠, and $(-17)(3) = -51$⁠, $(13)(5) = 65$⁠, $(7)(-2) = -14$ for $\vec b$⁠. By definition $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$⁠, so both scalar products must be 0. A slip in any 2 by 2 minor almost always breaks one of them, so this is a quick and reliable test.
   $\vec u\cdot\vec v = 0$ for perpendicular vectors (Section 10.6.1, Observation 2). Lesson: the dot product

4. The length

   $$\begin{aligned} |\vec a\times\vec b| &= \sqrt{(-17)^2 + (13)^2 + (7)^2} \\ &= \sqrt{289 + 169 + 49} = \sqrt{507}. \end{aligned}$$

   Since $507 = 169 \times 3 = 13^2 \times 3$⁠, $\sqrt{507} = 13\sqrt3$⁠.

   The signs disappear when the components are squared. The book stops at $\sqrt{507}$⁠; taking out the square factor 169 gives the simpler equal form $13\sqrt3$⁠.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

5. What $\sqrt{507}$ measures, and a second check

   $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ is the area of the parallelogram with sides $\vec a$ and $\vec b$⁠. Squaring and using $\sin^2\theta = 1 - \cos^2\theta$⁠:

   $$\begin{aligned} |\vec a\times\vec b|^2 &= |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 \\ &= (14)(38) - 5^2 \\ &= 532 - 25 = 507, \end{aligned}$$

   because $|\vec a|^2 = 4 + 1 + 9 = 14$⁠, $|\vec b|^2 = 9 + 25 + 4 = 38$ and $\vec a\cdot\vec b = 6 + 5 - 6 = 5$⁠.

   This second route uses only the scalar product, so it checks the determinant work independently. The identity comes from $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2\sin^2\theta$ and $(\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2\cos^2\theta$⁠.
   Area of a parallelogram $= |\vec a\times\vec b|$ (Section 10.6.3, Observation 9). Lesson: Examples 22–25

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