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Q: A particle moves on a given straight line with a constant speed v. At a certain time it is at a point $P$ on its straight line path. $O$ is a fixed point. Show that (OP×v)is independent of the position P.

My solution:

I considered the axis to be X-axis I.e as 1D motion. Points P & O on it. O to be the starting point from where the particle started.

$OP = x$ $\hat{\mathbf{i}}$

$OP$ X $v$ = vi X x$\hat{i}$(OP) = 0 since $ixi$ =$0$.

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Solution from my textbook:(Different way than mine)

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My questions regarding this solution :

Q1: Why did they consider point O to be at a distance y axis ? Since about point O , no description related to its position is given. Also , Can we consider point O to be at any point ? Like somewhere in Z axis.

Q2: The Q says that we have to find $OPxV$ independent of the position P. Then , In this solution = $-y*v*k$. We have $(OPxV) $dependant upon the magnitude of y & v & direction k. The value will change w/ time.

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    $\begingroup$ Constant speed $\vec v = v\,\vec i$ means $v$ is constant and $\vec i$ doesn't change. Fixed point $O$ means $y$ doesn't change, so $-yv\vec k$ doesn't change either. A point and a line define a unique plane $(\vec i, \vec j)$ in which $OP$ always lies. $\endgroup$
    – dxiv
    Commented Jan 19, 2022 at 8:15
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    $\begingroup$ Nowhere in the question does it say that O lies somewhere on the line through which the particle moves. You have assumed this, so you have only proved the question for that special case; not for the general case that the question is asking about. $\endgroup$
    – Jaap Scherphuis
    Commented Jan 19, 2022 at 12:41
  • $\begingroup$ @JaapScherphuis Yes & nether does it say it lies somewhere outside the line. Now , when I solved it first. I assumed it to be on the line. The solution of textbook assume it to be outside the line. We both had same answers. I have explained this already in my Q with the related difficulties I have regarding it. $\endgroup$
    – S.M.T
    Commented Jan 19, 2022 at 12:42
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    $\begingroup$ So if I asked you to prove for example that all rectangles have diagonals that are of equal length, and you proved it only for squares, would you consider that a complete proof? $\endgroup$
    – Jaap Scherphuis
    Commented Jan 19, 2022 at 12:46
  • $\begingroup$ @JaapScherphuis No , I wouldn’t. $\endgroup$
    – S.M.T
    Commented Jan 19, 2022 at 12:47

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The equation of the line is $ P(t) = P_0 + t v $

$OP(t) \times v = ((P_0 - O) + t v ) \times v = (P_0 - O) \times v + t (v \times v) = (P_0 - O) \times v $

Thus the cross product is independent of $t$.

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  • $\begingroup$ What do you think about the answer I wrote & the one textbook gave. $\endgroup$
    – S.M.T
    Commented Jan 19, 2022 at 14:34
  • $\begingroup$ All answers are pretty much the same. $\endgroup$
    – Quadrics
    Commented Jan 19, 2022 at 15:54
  • $\begingroup$ If u see the comments & I also asked on chat group , they say we cannot place point O on the path where particle is moving I.e my solution. But they’re also unable to tell why is the solution of textbook the general case. Could you please answer upon that as well. It will be a huge help in understanding. $\endgroup$
    – S.M.T
    Commented Jan 19, 2022 at 15:55
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    $\begingroup$ Point $O$ can be any where in space, and that's why the textbook solution is more general than yours. In your solution the cross product will be zero, while in the textbook answer (and also my solution) it is in general a non-zero constant vector. $\endgroup$
    – Quadrics
    Commented Jan 19, 2022 at 18:05

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