Miscellaneous · Q15: $(\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2$ if and only if $\vec a\perp\vec b$

Basics you need

Solution

1. Expand the left side

   $$\begin{aligned} &(\vec a + \vec b)\cdot(\vec a + \vec b) \\ &= \vec a\cdot\vec a + \vec a\cdot\vec b + \vec b\cdot\vec a + \vec b\cdot\vec b \\ &= |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2. \qquad (\text{i}) \end{aligned}$$
   Distributivity multiplies each term of the first bracket by each term of the second: four products. Two of them are squares of lengths, and the two mixed ones are equal because the scalar product is commutative, so they combine into $2\,\vec a\cdot\vec b$⁠. This is the vector form of $(x + y)^2 = x^2 + 2xy + y^2$⁠. Both parts of the proof use this one identity.
   Section 10.6.1, Property 1, Observations 3 and 7. Basics: dot products obey the same rule

2. The "if" part: perpendicular gives the equation

   Suppose $\vec a \perp \vec b$⁠. Then $\vec a\cdot\vec b = 0$⁠, and (i) becomes

   $$\begin{aligned} (\vec a + \vec b)\cdot(\vec a + \vec b) &= |\vec a|^2 + 0 + |\vec b|^2 \\ &= |\vec a|^2 + |\vec b|^2. \end{aligned}$$
   For perpendicular vectors the angle is $\frac{\pi}{2}$⁠, so $\vec a\cdot\vec b = |\vec a||\vec b|\cos\frac{\pi}{2} = 0$⁠, and the middle term of (i) disappears.
   Section 10.6.1, Observation 2. Lesson: the dot product

3. The "only if" part: the equation gives perpendicular

   Suppose $(\vec a + \vec b)\cdot(\vec a + \vec b) = |\vec a|^2 + |\vec b|^2$⁠. Comparing with (i),

   $$\begin{aligned} |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2 &= |\vec a|^2 + |\vec b|^2 \\ 2\,\vec a\cdot\vec b &= 0 \\ \vec a\cdot\vec b &= 0. \end{aligned}$$

   Since $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$⁠, $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = 0$ forces $\cos\theta = 0$⁠, so $\theta = \frac{\pi}{2}$⁠: $\vec a \perp \vec b$.

   Subtracting $|\vec a|^2 + |\vec b|^2$ from both sides leaves $2\,\vec a\cdot\vec b = 0$⁠. The lengths $|\vec a|$ and $|\vec b|$ are positive because the vectors are nonzero, so we may divide by them, and the only angle in $[0, \pi]$ with cosine 0 is $\frac{\pi}{2}$⁠.
   $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$⁠, $0 \le \theta \le \pi$ (Section 10.6.1, Definition 2). Basics: cos θ is a shadow

4. Why the vectors must be nonzero

   If $\vec a = \vec 0$⁠, then $\vec a\cdot\vec b = 0$ by definition and the equation holds for every $\vec b$⁠, but the angle between $\vec 0$ and $\vec b$ is not defined, so "perpendicular" has no meaning. The condition $\vec a \ne \vec 0, \vec b \ne \vec 0$ is exactly what lets "$\vec a\cdot\vec b = 0$" be read as "perpendicular" in Step 3.

   The book defines $\vec a\cdot\vec b = 0$ when either vector is zero, because no angle is available. The equivalence between "scalar product 0" and "perpendicular" is stated only for two nonzero vectors.
   Section 10.6.1, Definition 2 and Observation 2. Lesson: the dot product

5. The picture: Pythagoras in vector form

   Put $\vec b$ at the tip of $\vec a$⁠; then $\vec a + \vec b$ closes the triangle. The equation says the square on the third side equals the sum of the squares on the other two, which is Pythagoras' theorem and its converse:

   In the triangle, the angle opposite to $\vec a + \vec b$ is $\pi - \theta$⁠. Identity (i) is the cosine rule written with vectors; the correction term $2\,\vec a\cdot\vec b$ vanishes exactly for a right angle.
   Triangle law (Section 10.4). Lesson: adding arrows, head to tail

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