Can a vector of magnitude zero have a nonzero component?

"displacement" is the straight line distance between the starting point and the ending point. Both batter and ball started at the same point and ended at the same point.


(Strictly speaking, "displacement" is the vector form the starting point to the ending point- but it doesn't make sense to talk about one vector being "greater" than another so I assume you really mean the magnitude of the displacement.)

 

A vector CAN NOT have a non-zero component and a zero magnitude.
Example in a 3D space:
Consider a vector [tex]\vec{V}=(v_1,v_2,v_3)[/tex].
By definition its magnitude is:
[tex]\|\vec{V}\|= \sqrt{\vec{V} \cdot \vec{V}}[/tex]
That is [tex]\|\vec{V}\|=\sqrt{v_1^2+v_2^2+v_3^2}[/tex]
Because the square of any number is always positive, it is clear that:
1- [tex]v_1 \neq 0[/tex] or [tex]v_2 \neq 0[/tex] or [tex]v_3 \neq 0[/tex] leads to [tex]\|\vec{V}\|>0[/tex],
2- [tex]\|\vec{V}\|=0[/tex] implies that [tex]v_1 = v_2 = v_3 = 0[/tex]

Does it make it any clearer?

 

Of course you presume that the ball is struck and caught at the same height.

 

You don't have to. If the ball flies up into orbit, goes around the world, then falls back onto the top of Mount Everest while the batter gets into an airplane, flies to Paris, spends the day looking at the Mona Lisa, then takes a boat through the Suez Canal to India, and climbs Mount Everest just in time to catch the ball, the displacement vector (and its length) of the ball and batter are exactly the same!

 

HallsofIvy: I think that NewScientist meant that eventhough the ball can be assumed to be a point, the batter can not. Therefore, defining the displacement of the batter by that of his center of gravity, the ball might have a slightly different displacement if it was struck and caught at different distances from the batter's center of gravity.

However I guess this kind of consideration is out of the scope of this problem.