Exercise 10.3 · Q11: $|\vec a|\vec b+|\vec b|\vec a$ is perpendicular to $|\vec a|\vec b-|\vec b|\vec a$

Basics you need

Solution

1. Two vectors of equal length

   Let $\vec u = |\vec a|\vec b$ and $\vec v = |\vec b|\vec a$⁠. Then

   $$|\vec u| = |\vec a|\,|\vec b|, \qquad |\vec v| = |\vec b|\,|\vec a|,$$

   so $|\vec u| = |\vec v|$⁠.

   $|\vec a|$ is just a positive number, so $\vec u$ is $\vec b$ stretched by the factor $|\vec a|$⁠, and its length is $|\vec a|$ times $|\vec b|$⁠. In the same way $\vec v$ is $\vec a$ stretched by $|\vec b|$⁠. The two products of lengths are the same number. The question asks about $\vec u + \vec v$ and $\vec u - \vec v$⁠.
   $|\lambda\vec x| = |\lambda||\vec x|$ (Section 10.5). Lesson: components and scalar multiples

2. Expand $(\vec u + \vec v)\cdot(\vec u - \vec v)$

   $$\begin{aligned} &(\vec u + \vec v)\cdot(\vec u - \vec v) \\ &= \vec u\cdot\vec u - \vec u\cdot\vec v + \vec v\cdot\vec u - \vec v\cdot\vec v \\ &= |\vec u|^2 - |\vec v|^2 \\ &= 0. \end{aligned}$$
   Distributivity multiplies out the brackets like $(x + y)(x - y)$⁠. The middle terms cancel because $\vec v\cdot\vec u = \vec u\cdot\vec v$⁠, and a vector dotted with itself gives the square of its length. The last line uses $|\vec u| = |\vec v|$ from Step 1.
   Section 10.6.1, Properties 1 and 2, Observations 3 and 7. Basics: dot products obey the same rule

3. The same working in $\vec a$ and $\vec b$

   $$\begin{aligned} &(|\vec a|\vec b + |\vec b|\vec a)\cdot(|\vec a|\vec b - |\vec b|\vec a) \\ &= |\vec a|^2(\vec b\cdot\vec b) - |\vec a||\vec b|(\vec b\cdot\vec a) \\ &\quad + |\vec b||\vec a|(\vec a\cdot\vec b) - |\vec b|^2(\vec a\cdot\vec a) \\ &= |\vec a|^2|\vec b|^2 - |\vec b|^2|\vec a|^2 = 0. \end{aligned}$$
   This is Step 2 with $\vec u$ and $\vec v$ written out, which is how the answer is usually written in an exam. The scalars $|\vec a|$ and $|\vec b|$ come out of each scalar product, for example $(|\vec a|\vec b)\cdot(|\vec b|\vec a) = |\vec a||\vec b|(\vec b\cdot\vec a)$⁠.
   $(\lambda\vec x)\cdot(\mu\vec y) = \lambda\mu(\vec x\cdot\vec y)$ (Section 10.6.1, Property 2). Basics: dot products obey the same rule

4. Conclude: the vectors are perpendicular

   If $\vec a$ and $\vec b$ are not parallel, then $\vec u + \vec v$ and $\vec u - \vec v$ are both nonzero, and a zero scalar product means they are perpendicular.

   If $\vec a$ and $\vec b$ are parallel, one of the two vectors is $\vec 0$⁠: for the same direction $\hat b = \hat a$⁠, so $\vec u = |\vec a||\vec b|\hat b = |\vec a||\vec b|\hat a = \vec v$ and $\vec u - \vec v = \vec 0$⁠; for opposite directions $\vec u + \vec v = \vec 0$⁠. The zero vector may be regarded as having any direction, so the statement still holds.

   "Scalar product 0 means perpendicular" is stated for nonzero vectors, so the case where one vector vanishes needs a word. Write $\vec u = |\vec a||\vec b|\hat b$ and $\vec v = |\vec a||\vec b|\hat a$⁠: $\vec u - \vec v = \vec 0$ exactly when $\hat b = \hat a$⁠, and $\vec u + \vec v = \vec 0$ exactly when $\hat b = -\hat a$⁠. These are the two parallel cases.
   Section 10.6.1, Observation 2; the zero vector "may be regarded as having any direction" (Section 10.3). Lesson: six kinds of vectors

5. The picture: a rhombus

   Place $\vec v$ and $\vec u$ tail to tail. The parallelogram they span has diagonals $\vec u + \vec v$ (from the common tail to the far corner) and $\vec u - \vec v$ (from the tip of $\vec v$ to the tip of $\vec u$⁠). Equal sides make it a rhombus, and a rhombus has perpendicular diagonals: the scalar-product proof above is that fact written in vectors.
   Parallelogram law of addition (Section 10.4). Lesson: adding arrows, head to tail

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