Find Magnitude of Vector a: Solve Ax2 + Ay2 + Az2

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SUMMARY

To find the magnitude of a vector a represented as a = axi + ayj + azk, the correct formula is √(ax² + ay² + az²). This accounts for all three dimensions in a 3D coordinate system, where k represents the z-axis. Using Pythagoras' theorem, the magnitude can be visualized as the hypotenuse of a right triangle formed by the components of the vector. Omitting the az component results in an incomplete calculation of the vector's magnitude.

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If you are given vector a = axi + ayj + azk and you want to find the magnitude of a, then would you just do the square root of ax2 + ay2? If so, then what about az? What is azk actually? I'm a little unclear about it being related to a.
 
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You would do the square root of ax2 + ay2 + ay2.
Sorry, I don't know how to make superscripts. Too bad they don't copy.
 
You mean az for the last one?

And to do superscripts, just put "[.SUP] [/SUP.]" without the quotations and periods (caps don't matter). And it's for subscripts.
 
Yes az, and thanks for the tip.
 
Just a quick explanation for that azk term, if you still are unclear about it.

k adds another dimension to the vector. When a vector is only in i,j it's 2D and i,j,k is 3D. See it as drawing it in a xyz-coordinate system instead of a xy one. And as was mentioned earlier, it changes things a little bit when it comes to the magnitude. You wouldn't just do the √(ax2 + ay2), because then you wouldn't take the third dimension into consideration. From there it's not too big of a leap to √(ax2 + ay2 + az2).

You can prove it using Pythagoras' theroem. If you draw a 3D vector in a coordinate system you can figure it out. It can be a bit tricky to see, but it's not impossible.
 
You could do it that way and you'll get the same answer.
It would be interesting to prove that . . .
 
Actually I just realized that the square root of az2 + ay2 is the hypothesis. So I guess it makes sense.
 

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