Miscellaneous · Q19 (MCQ): Angle at which $|\vec a\cdot\vec b|=|\vec a\times\vec b|$

Basics you need

Solution

1. Both sizes in terms of $\theta$

   $$\begin{aligned} |\vec a\cdot\vec b| &= |\vec a||\vec b|\,|\cos\theta|, \\ |\vec a\times\vec b| &= |\vec a||\vec b|\sin\theta\,|\hat n| = |\vec a||\vec b|\sin\theta. \end{aligned}$$
   The scalar product can be negative, so its absolute value needs $|\cos\theta|$⁠. The vector product is $|\vec a||\vec b|\sin\theta\,\hat n$ with $\hat n$ a unit vector and $\sin\theta \ge 0$ on $[0, \pi]$⁠, so its length is $|\vec a||\vec b|\sin\theta$ with no absolute value needed.
   Section 10.6.1, Definition 2; Section 10.6.3, Definition 3. Lesson: the cross product

2. Solve $|\cos\theta| = \sin\theta$ on $[0, \pi]$

   Divide the equation $|\vec a||\vec b|\,|\cos\theta| = |\vec a||\vec b|\sin\theta$ by $|\vec a||\vec b| > 0$⁠:

   $$|\cos\theta| = \sin\theta.$$

   $\cos\theta = 0$ is impossible here (then $\theta = \frac{\pi}{2}$ and $\sin\theta = 1 \ne 0$⁠), so we may divide by $|\cos\theta|$⁠:

   $$\begin{aligned} &\frac{\sin\theta}{|\cos\theta|} = 1 \iff \tan\theta = \pm1 \\ &\iff \theta = \frac{\pi}{4} \text{ or } \frac{3\pi}{4}. \end{aligned}$$
   For $\theta \le \frac{\pi}{2}$⁠, $|\cos\theta| = \cos\theta$ and the equation is $\tan\theta = 1$⁠; for $\theta > \frac{\pi}{2}$⁠, $|\cos\theta| = -\cos\theta$ and it is $\tan\theta = -1$⁠. Each has exactly one solution in its part of $[0, \pi]$⁠. The two angles are the ones at which a vector of $\vec b$'s length has equal shadow and height on the line of $\vec a$⁠.
   $\tan\frac{\pi}{4} = 1$⁠, $\tan\frac{3\pi}{4} = -1$ (school trigonometry). Basics: cos θ is a shadow, sin θ is a height

3. Examine every option

   (A) $\theta = 0$⁠: $|\vec a\cdot\vec b| = |\vec a||\vec b|$ but $|\vec a\times\vec b| = 0$⁠. Not equal; (A) fails.

   (B) $\theta = \frac{\pi}{4}$⁠: both equal $\frac{1}{\sqrt2}|\vec a||\vec b|$⁠. (B) is correct.

   (C) $\theta = \frac{\pi}{2}$⁠: $|\vec a\cdot\vec b| = 0$ but $|\vec a\times\vec b| = |\vec a||\vec b|$⁠. (C) fails.

   (D) $\theta = \pi$⁠: $|\vec a\cdot\vec b| = |\vec a||\vec b|$ but $|\vec a\times\vec b| = 0$⁠. (D) fails.

   At $0$⁠, $\frac{\pi}{2}$ and $\pi$ one of $\sin\theta$⁠, $\cos\theta$ is 0 and the other has size 1, so one product vanishes while the other does not. The second solution $\frac{3\pi}{4}$ is not among the options, so (B) is the only correct choice.
   Section 10.6.1, Observations 2 to 4; Section 10.6.3, Observations 2 and 3. Lesson: the multiple-choice questions

4. Check with vectors

   $\vec a = \hat i$⁠, $\vec b = \hat i + \hat j$⁠: the angle is $\frac{\pi}{4}$⁠, and

   $$\begin{aligned} \vec a\cdot\vec b &= 1, \\ \vec a\times\vec b &= \hat i\times\hat i + \hat i\times\hat j = \hat k, \end{aligned}$$

   so $|\vec a\cdot\vec b| = 1 = |\vec a\times\vec b|$⁠. With $\vec b = -\hat i + \hat j$ (angle $\frac{3\pi}{4}$⁠): $\vec a\cdot\vec b = -1$ and $\vec a\times\vec b = \hat k$⁠, again equal in size.

   Distributivity of the vector product and $\hat i\times\hat i = \vec 0$⁠, $\hat i\times\hat j = \hat k$ give the products at once. The second pair shows why the question needs $|\vec a\cdot\vec b|$ and not $\vec a\cdot\vec b$⁠.
   Section 10.6.3, Property 3 and Observation 4. Lesson: the cross product

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