follow to a message I am reading, the magnitude of difference vectors is $vec C = vec S - vec T$. V the displacement vectors $vec S$ and $vec T$ having magnitudes $||S||=3m$ and $||T||=4m$ respectively. And by following the rules just previously stated,the size of the distinction vectors is $||C|| = 3m-4m =-1m$

But is this valid? I assumed that a magnitude have the right to not be negative? can the $-1$ suggest the direction the the vector? Or to be I claimed to neglect the an adverse value and simply take the absolute worth of $|-1|$ to it is in $1$?

The text states there might be much more than one best answer come this problem, so if anyone can show the miscellaneous "correct" answer that would help.

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The dominion for a norm linked to a scalar product is$$||vec C||^2 = ||vec S - vec T||^2 = (vec S - vec T)cdot (vec S - vec T) = ||vec S||^2-2vec Scdot vec T+||vec T||^2,$$so here$$||vec C||^2 = 25m^2-2vec Scdot vec T.$$From Cauchy-Schwartz inequality $|vec Scdot vec T| leq ||vec S||||vec T||=12m^2$ girlfriend get$$||vec C||^2 geq25m^2-24m^2=m^2,$$hence$$||vec C|| geq m.$$The same Cauchy-Schwartz inequality leads additionally to$$||vec C|| leq 7m.$$Both extreme cases correspond to colinear vectors (that is, parallel or anti-parallel).

$egingroup$ Yes: $||vec C||^2 = 25m^2 - 2 vec S cdot vec T leq 25m^2 + 2 |vec S cdot vec T| leq 25m^2 + 24m^2$ (note that the authorize of $vec S cdot vec T$ is arbitrary). $endgroup$
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