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- Thread starter moatasim23
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Simon Bridge

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A *is* expressed in terms of x y and z.

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CompuChip

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You can replace them by ##r_1, r_2, r_3##, if that makes you feel any better. In general, if

Also note that though the fact that one is named

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No its not..here at least

- #5

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You can replace them by ##r_1, r_2, r_3##, if that makes you feel any better. In general, ifvis a vector, it is customary to denote its components byv_{1},v_{2},v_{3}. However, ifris the position vector, then (x, y, z) is also quite common.

Also note that though the fact that one is namedrhints that it comes from a physical application in which a position vector is involved, the mathematical identity actually holds for any two vectorsu,v.

If we use r1,r2,r3 then how would the vector operator operator operate on it?Like it didnt in A when we used A1,A2,A3.

- #6

SteamKing

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What are you talking about?

- #7

Simon Bridge

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I'm sorry - the example in your attachment very clearly states thatNo its not..here at least

That means that

- the x component of

- the y component of

- the z component of

Therefore: A is resolved in terms of x, y, and z.

What did you think it meant?

- #8

CompuChip

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If ##\mathbf v = v_1 \mathbf i + v_2 \mathbf j + v_3 \mathbf k## and ##\mathbf u = u_1 \mathbf i + u_2 \mathbf j + u_3 \mathbf k## then

$$\mathbf u \times \mathbf v = (u_2 v_3 - u_3 v_2) \mathbf i + (u_3 v_1 - u_1 v_3) \mathbf j + (u_1 v_3 - u_3 v_1) \mathbf k$$

That's just how the cross product works. It doesn't matter how you call the components. You could replace ##u_1##, ##u_2## and ##u_3## by ##x##, ##y## and ##z## or clubs, spades, hearts or bunny, cow, eagle and the definition would still be the same.

Is it the notation of a vector like##\mathbf v = v_1 \mathbf i + v_2 \mathbf j + v_3 \mathbf k## instead of ##\mathbf v = (v_1, v_2, v_3)## that confuses you?

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