Calculating Magnitude and Direction Cosines of a Vector in Physics

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To find the magnitude of vector A with components 4, 6, and 3, use the formula |A| = sqrt(x^2 + y^2 + z^2), which results in a magnitude of 7. The direction cosines can be calculated by dividing each component by the magnitude, giving cosines for the angles that vector A makes with the coordinate axes. The discussion clarifies that the formula for magnitude should be applied correctly, emphasizing that the correct expression is |A| = sqrt(x^2 + y^2 + z^2) rather than including z^3. The conversation highlights that these calculations are mathematical in nature, not purely physics-related. Understanding these concepts is essential for mastering vector analysis in physics.
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I just started physics this year and having a difficult time, lol.

anyways, my question is:

suppose I was given a vector A, and A has x, y, z components of 4,6,3

How would I find its magnitude of A and the cosines of the angles that makes B?
 
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It's like finding the length of a diagonal of a rectangle of length, breadth and depth of 4, 6, and 3 respectively, by the Pythagoras's theorem. And what's B?
 
i'm sorry, I meant what are the angles that make vector A, sorry.
 
also, I'm just wondering why I can't apply the formula Magnitude A = sqrt (x^2, y^2, z^2)
 
I just want to know if I use the |A| = sqrt (x^2 + y^2 + z^3)
 
or the rectangular length like bel has mentioned above.
 
tdusffx said:
I just want to know if I use the |A| = sqrt (x^2 + y^2 + z^3)

Yep, you can. :smile:
 
hmm, ok thank you guys for the help.
 

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