Is this workaround used by my book to find the value of $1/0$ legitimate?

Question

I was doing a physics problem and found that I reached an interesting equation:

To cross the river directly from A to B (making a right angle with the velocity of the stream), how must the fisherman direct his/her boat's motion? Find the angle ($\alpha$) between the velocity vector of the stream and the velocity vector of the boat. The magnitude of the velocity of the boat is $v=2.778\mathrm{ms}^{-1}$ and that of the stream is $u=1.3889\mathrm{ms}^{-1}$.

Solution:

We know from the Parallelogram law of addition of vectors:

$$\tan\theta=\frac{Q\sin\alpha}{P+Q\cos\alpha}$$

Similarly,

$$\tan(90^{\circ})=\frac{u\sin\alpha}{v+u\cos\alpha}$$

$$\implies\frac{\sin(90^{\circ})}{\cos(90^{\circ})}=\frac{u\sin\alpha}{v+u\cos\alpha}$$

$$\implies \frac{1}{0}=\frac{u\sin\alpha}{v+u\cos\alpha}$$

$$\implies{v+u\cos\alpha}=0$$

Now, isn't $\;\dfrac10\;$ undefined? How is cross-multiplication valid here? I'm very confused.

Answer 1

It is not valid per se, because as you pointed out, cross multiplication with $\frac{1}{0}$ is to be avoided.

However, a better way to put it, is that a rational expression $$\frac{u\sin\alpha}{v+u\cos\alpha}$$ is equal to $\tan(\pi^r/2)$ which is undefined in the real domain, and the only way a rational function of the form $\frac{f(x)}{g(x)}$ can be undefined is when $g(x) = 0$ (i.e) $$ v + u\cos\alpha = 0 $$

Hope this helps.

Answer 2

Since this question originated from the Physics Stack Exchange site, let us put mathematical rigor aside to handle the equation $$\frac{1}{0}=\frac{u\sin\alpha}{v+u\cos\alpha}.$$

The left side of the equation, $\frac{1}{0}$, is $\infty$. To make the right side, $\frac{u\sin\alpha}{v+u\cos\alpha}$, also $\infty$, its denominator needs to be $0$, hence $$v+u\cos\alpha=0.$$

Answer 3

Let's redefine your equation $$\frac{\sin(90^\circ)}{\cos(90^\circ)}=\frac{u\sin\alpha}{v+u\cos\alpha}$$ as $$\frac{A}{B}=\frac{C}D$$ You rightly state that since $B=0$ (and D=0) we can't invert this relation and cross multiplication is not allowed.

If, however, you started with the relation $$\frac B A=\frac D C$$ or even $AD=BC$ then everything is fine and you get the right answer. From what I've seen the parallegram law doesn't start with an expression that involves $\tan\theta$. Can you re-do the beginning of the problem such that you get an expression like $AD=BC$ or $B/A=D/C$ without ever needing to write down $\tan\theta$?

Answer 4

$\tan \theta$ is indeed undefined at $\theta = 90^o$. However, rewrite the equation in terms of $\cot \theta$ and you have

$\theta = 90^o \\ \Rightarrow \cot \theta=0 \\ \Rightarrow v + u \cos \alpha = 0$