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 resultant velocity vector of the boat. The magnitude of the velocity of the boat is $v=2.778ms^{-1}$ and that of the stream is $u=1.3889ms^{-1}$.


We know from the Parallelogram law of addition of vectors:





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


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

  • 2
    $\begingroup$ it is not valid however as any fraction with finite numerator and denominator approaches infinity, the denominator must approach $0$ $\endgroup$ Jun 17, 2021 at 14:44
  • 6
    $\begingroup$ That representation of $\tan 90$ is not valid (mathematically, I won't speak for physicists) and the cross-multiplication is also invalid. What you could rigorously say is that since $\tan \theta$ grows without bound as $\theta \rightarrow 90^\circ$ and $\sin \alpha$ is bounded, either $v+v \cos \alpha$ must shrink to $0$ or $u$ grows without bound. Assuming you have constraints on $u$ you're then (mostly) in the position of your last line $\endgroup$
    – postmortes
    Jun 17, 2021 at 14:44
  • 5
    $\begingroup$ You're not confused; you're correct. Cross multiplication is not legitimate here. And writing $1/0$ is meaningless. What we do know is that as the argument of the tangent function appoaches $90$ degrees from the left, the term on the right-hand side tends to $\infty$. The only way for this to happen is that the denominator tends to $0$ since the speed $u$ is bounded by the speed of light. $\endgroup$
    – Mark Viola
    Jun 17, 2021 at 14:45
  • $\begingroup$ One line answer: $\tan 90$ is not $\frac{1}{0}$ I am confused whether to write this as a answer as it is or rather leave it a comment $\endgroup$
    – user876009
    Jun 17, 2021 at 14:47
  • $\begingroup$ @postmortes ya I also thought same so rather left it as a comment $\endgroup$
    – user876009
    Jun 17, 2021 at 14:48


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