I read this question in Sears and Zemansky's University Physics book:

"Two displacement vectors, S and T, have magnitudes S = 3 m and T = 4 m. Which of the following could be the magnitude of the difference vector S - T ?(There may be more than one answer.) i. 9 m; ii. 7 m; iii. 5 m; iv. 1 m; v. 0m; vi. -1m "

My question is: shouldn't the direction of both vectors be specified in order for me to solve it? Or do I just assume that S and T are positive and negative respectively as per the equation S - T

  • $\begingroup$ The question is asking you what the possible values of $\mathbf S - \mathbf T$ could be for all possible directions. So for example if one of the options was $666$m you need consider if there is any possible arrangement of the two vectors that could give $|\mathbf S - \mathbf T| = 666$. This freedom to orient the vectors in any direction means more than one of the answers can be correct. $\endgroup$ – John Rennie Jan 12 '17 at 16:20
  • $\begingroup$ That's what I thought initially, but I had doubts. Thanks for confirming! $\endgroup$ – DigiNin Gravy Jan 12 '17 at 16:23
  • $\begingroup$ Re Answer (vi): can the magnitude of a vector be negative? $\endgroup$ – DJohnM Jan 12 '17 at 17:18
  • $\begingroup$ @DJohnM I know that the magnitude can never be negative, but the vector can. $\endgroup$ – DigiNin Gravy Jan 12 '17 at 18:20
  • $\begingroup$ @DigiNinGravy A vector which is anti-parallel to, say, the $x$-axis of a coordinate system is sometimes called "a vector in the negative-$x$ direction." But it is an incorrect oversimplification to call such an object "a negative vector." $\endgroup$ – rob Jan 13 '17 at 7:26

The magnitude of the difference vectors depends on the orientation of $ \bf\vec{S} $ and $\bf \vec{T} $. If they are parallel then $|\bf \vec{S}-\bf \vec{T}|=|\,|\bf \vec{S}|-|\bf \vec{T}|\,|$ and if they are anti-parallel then $ |\bf \vec{S}-\bf \vec{T}|=|\bf \vec{S}|+|\bf \vec{T}|$.

Thus the possible values of $|\bf \vec{S}-\bf \vec{T}\| $ lie in the range:

$$ |\,|\bf \vec{S}|-|\bf \vec{T}|\,| \leq |\bf \vec{S}-\bf \vec{T}| \leq |\bf \vec{S}|+|\bf \vec{T}| $$

I'll let you work out which answers comply with this.

  • $\begingroup$ |S - T| = 5 or | |S| - |T| | = -1 or |S| + |T| = 7. I just don't understand how |S - T| = |S| + |T|. I mean to have |S| + |T| both vectors must be parallel - having the same direction - so | |S| - |T| | seems off, unless we consider T as -T? $\endgroup$ – DigiNin Gravy Jan 12 '17 at 17:09
  • $\begingroup$ @DigiNinGravy I think you gave the aswer yourself both $\bf T and $-\bf T$ have se same magnitude, and the magnitude does not give the direction of the vector. $\endgroup$ – Mikael Fremling Jan 12 '17 at 18:10

enter image description here

In above Figure move the end point $\:\mathrm{B}\:$ of the vector $\:\mathbf{S}\:$ around a circle of radius $\:|\mathbf{S }|=3\:$. Try to find the length $\:|\mathbf{S}-\mathbf{T}|\:$ of the vector $\:\mathbf{S}-\mathbf{T}\:$ when $\:\mathrm{B}\:$ is on points $\:\mathrm{P_{ii}},\mathrm{P_{iii}},\mathrm{P_{iv}}\:$.

  • $\begingroup$ Thanks, but you reversed the magnitudes of S and T. Can you tell me the name of the program you used? $\endgroup$ – DigiNin Gravy Jan 12 '17 at 18:17
  • $\begingroup$ @DigiNin Gravy : GeoGebra $\endgroup$ – Frobenius Jan 12 '17 at 18:35

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