Exercise 10.4 · Q6: $\vec a\cdot\vec b=0$ and $\vec a\times\vec b=\vec 0$⁠: what follows?

Basics you need

Solution

1. What each condition says for nonzero vectors

   Suppose $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$⁠, at angle $\theta$⁠. Then $|\vec a||\vec b| > 0$⁠, and

   $$\begin{aligned} \vec a\cdot\vec b = 0 &\Rightarrow |\vec a||\vec b|\cos\theta = 0 \\ &\Rightarrow \cos\theta = 0, \\ \vec a\times\vec b = \vec 0 &\Rightarrow |\vec a||\vec b|\sin\theta = 0 \\ &\Rightarrow \sin\theta = 0. \end{aligned}$$
   We may divide by $|\vec a||\vec b|$ because it is positive. For the vector product, $|\vec a||\vec b|\sin\theta\,\hat n = \vec 0$ with $|\hat n| = 1$ forces the number $|\vec a||\vec b|\sin\theta$ to be 0. So the first condition says the vectors are perpendicular ($\theta = \frac{\pi}{2}$⁠) and the second says they are parallel ($\theta = 0$ or $\pi$⁠).
   Definitions of the two products (Sections 10.6.1 and 10.6.3). Lesson: the cross product

2. Both cannot hold

   $$\sin^2\theta + \cos^2\theta = 1 \ne 0 + 0.$$

   So the supposition is false: $\vec a = \vec 0$ or $\vec b = \vec 0$⁠.

   Two nonzero vectors cannot be perpendicular and parallel at the same time. Since assuming both are nonzero leads to a contradiction, at least one of them is the zero vector. "$\vec a = \vec 0$" is the same as "$|\vec a| = 0$", the form in the answer key.
   $\sin^2\theta + \cos^2\theta = 1$⁠. Basics: cos θ is a shadow, sin θ is a height

3. The conclusion is exact

   Conversely, if $\vec a = \vec 0$ (or $\vec b = \vec 0$⁠), then $\vec a\cdot\vec b = 0$ and $\vec a\times\vec b = \vec 0$ for every choice of the other vector.

   Both definitions set the product to zero when either vector is $\vec 0$ (the angle is then not defined). So the two conditions together say exactly "one of the vectors is $\vec 0$"; nothing more can be said about the other vector, which may be any vector.
   Products with the zero vector (Definitions 2 and 3, Sections 10.6.1 and 10.6.3). Lesson: the dot product

4. A second route that covers every case

   For all vectors $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ and $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$⁠, expanding both sides in components shows

   $$|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2.$$

   Here the left side is $0 + 0$⁠, so $|\vec a|^2|\vec b|^2 = 0$⁠, and $|\vec a| = 0$ or $|\vec b| = 0$⁠.

   For nonzero vectors this identity is $|\vec a|^2|\vec b|^2(\sin^2\theta + \cos^2\theta)$⁠, the same fact as Step 2; written in components it holds even when a vector is $\vec 0$⁠, so this route needs no case split. A product of two real numbers is 0 only when one of them is 0.
   Products in components (Sections 10.6.1 and 10.6.3). Lesson: the cross product, exercises

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