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

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.

• Forget about "the parallellogram law" and just find an expression for the component of the velocity at right angles to the stream velocity (which the question says is zero). Then you get one equation, with no "infinities" involved. Jun 17, 2021 at 19:29

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.

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 $$v+u\cos\alpha=0.$$

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 $$\tan\theta$$?

• "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 tanθ?" No, I do not know of any way to find the relation between the variables without needing to write tanθ. Jun 17, 2021 at 16:02
• @AbuSafwan Ok one way would be to see that the velocity in the x-direction must cancel out so $u_x+v_x=u+v\cos\alpha=0$. This formula is slightly different from yours but I tried yours and it has no solutions while mine has $\alpha\approx 120^\circ$ so maybe mine is correct. Jun 17, 2021 at 17:53
• @AbuSafwan You don't show all your steps (I don't know where the formula for $\tan\theta$ comes from) and I don't fully understand how you would come to your solution. So I can't help you right now but maybe if you start from a formula from the Wikipedia page for the parallelogram law you could find a solution. Jun 17, 2021 at 17:55

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

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