# Magnitude Of Difference Vectors

According to a text I am reading, the magnitude of difference vectors is $\vec C = \vec S - \vec T$. With 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 magnitude of the difference vectors is $||C|| = 3m-4m =-1m$

But is this valid? I thought that a magnitude can not be negative? Could the $-1$ indicate the direction of the vector? Or am I supposed to ignore the negative value and simply take the absolute value of $|-1|$ to be $1$?

The text says there may be more than one right answer to this problem, so if anyone could show the various "correct" answers that would help.

• Try to think about the maximum and minimum of the vector addition (included substraction) and also imagine about two concentric circles. You will not get a negative result for a vector $\textbf{C}$
– Lugo
Commented Jan 5, 2017 at 6:16
• You can not add/subtract vectors like that. They follow the paralellogram or the triangle law. If I give you two sides of a Triangle as 4m and 3m, would you say that the third side is 7m or would you worry about the orientation of the sides with respect to one another? Commented Jan 5, 2017 at 6:20
• C seems to be the vector that points from T to S. Calling it magnitude feels like a typical bizarre "simplification" that messes up all logical thinking. Just ignore the name and take the normal of the "magnitude vector". Which is not the formula you posted, since S and T as written are just the normals of the vectors (when they have arrows you need to express them wrt a basis). (With normal I meant ||•||)
– Emil
Commented Jan 5, 2017 at 7:07

The rule for a norm associated to a scalar product is $$||\vec C||^2 = ||\vec S - \vec T||^2 = (\vec S - \vec T)\cdot (\vec S - \vec T) = ||\vec S||^2-2\vec S\cdot \vec T+||\vec T||^2,$$ so here $$||\vec C||^2 = 25m^2-2\vec S\cdot \vec T.$$ From Cauchy-Schwartz inequality $|\vec S\cdot \vec T| \leq ||\vec S||||\vec T||=12m^2$ you get $$||\vec C||^2 \geq25m^2-24m^2=m^2,$$ hence $$||\vec C|| \geq m.$$ The same Cauchy-Schwartz inequality leads also to $$||\vec C|| \leq 7m.$$ Both extreme cases correspond to colinear vectors (that is, parallel or anti-parallel).
• 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 sign of $\vec S \cdot \vec T$ is arbitrary).