- #1

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I think it's neither but i'm not sure and i don't understand why. could someone please explain.

Also, Can a vector of magnitude zero have a nonzero component?

Thanks for your help!

- Thread starter confusedaboutphysics
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- #1

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I think it's neither but i'm not sure and i don't understand why. could someone please explain.

Also, Can a vector of magnitude zero have a nonzero component?

Thanks for your help!

- #2

HallsofIvy

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(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.)

- #3

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- #4

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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

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?

- #5

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

- #6

HallsofIvy

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- #7

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

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