Example 28: $|\vec a+\vec b+\vec c|$ when each is perpendicular to the sum of the other two

Basics you need

Solution

1. The conditions as equations

   $$\begin{aligned} \vec a\cdot(\vec b + \vec c) &= 0, \\ \vec b\cdot(\vec c + \vec a) &= 0, \\ \vec c\cdot(\vec a + \vec b) &= 0. \end{aligned}$$
   "Each one of them being perpendicular to the sum of the other two" says three things at once, one per vector, and perpendicular vectors have scalar product 0.
   $\vec u\cdot\vec v = 0$ for perpendicular vectors (Section 10.6.1, Observation 2). Lesson: the dot product

2. Square the length

   $$|\vec a + \vec b + \vec c|^2 = (\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c).$$
   A length is hard to handle directly, but its square is a scalar product, and scalar products can be expanded. The book writes the same thing as $(\vec a + \vec b + \vec c)^2$⁠.
   $\vec r\cdot\vec r = |\vec r|^2$ (Section 10.6.1, Observation 3). Lesson: the dot product

3. Expand, keeping each vector with the other two

   $$\begin{aligned} &(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) \\ &= \vec a\cdot(\vec a + \vec b + \vec c) + \vec b\cdot(\vec a + \vec b + \vec c) \\ &\quad + \vec c\cdot(\vec a + \vec b + \vec c) \\ &= \vec a\cdot\vec a + \vec a\cdot(\vec b + \vec c) + \vec b\cdot\vec b \\ &\quad + \vec b\cdot(\vec c + \vec a) + \vec c\cdot\vec c + \vec c\cdot(\vec a + \vec b). \end{aligned}$$
   Distributing the first bracket gives one product per vector; then each of those splits as "itself" plus "the sum of the other two". The grouping is chosen on purpose, so that the three given expressions appear exactly.
   Distributivity and commutativity of the scalar product (Section 10.6.1, Property 1, Observation 7). Basics: dot products obey the same rule

4. Use the conditions, then take the square root

   The three bracketed products are 0 by Step 1, and $\vec a\cdot\vec a = |\vec a|^2$ and so on, so

   $$\begin{aligned} |\vec a + \vec b + \vec c|^2 &= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 \\ &= 9 + 16 + 25 = 50, \end{aligned}$$ $$|\vec a + \vec b + \vec c| = \sqrt{50} = 5\sqrt2.$$
   A magnitude is never negative, so only the positive square root is taken. $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2$⁠.
   $|\vec r| \ge 0$ (Section 10.2). Lesson: Examples 26–30

5. What the conditions really say

   Write $p = \vec a\cdot\vec b$⁠, $q = \vec b\cdot\vec c$⁠, $r = \vec c\cdot\vec a$⁠. The conditions are $p + r = 0$⁠, $q + p = 0$ and $r + q = 0$⁠. Adding all three gives $2(p + q + r) = 0$⁠, and subtracting each condition from $p + q + r = 0$ gives

   $$q = 0,\quad r = 0,\quad p = 0.$$

   So $\vec a$⁠, $\vec b$⁠, $\vec c$ are mutually perpendicular, like the three edges at one corner of a box with edges 3, 4 and 5, and $\vec a + \vec b + \vec c$ is the diagonal of that box.

   This explains the answer: for mutually perpendicular vectors, Pythagoras in space gives the length of the sum as $\sqrt{3^2 + 4^2 + 5^2}$⁠. Any three mutually perpendicular vectors of lengths 3, 4, 5 form the same box, just turned, so the answer does not depend on which ones we pick.
   Length of the diagonal of a box. Basics: use Pythagoras twice

Every fact used, and where it comes from