Squaring of vectors in absolute value

by M. next
Tags: absolute, squaring, vectors
 P: 378 Is |$\vec{a}$+$\vec{b}$|$^{2}$ equal to the same thing as ($\vec{a}$+$\vec{b}$)$^{2}$? And when is it equal to √(a$^{2}$+b$^{2}$)? Thanks.
Mentor
P: 21,286
 Quote by M. next Is |$\vec{a}$+$\vec{b}$|$^{2}$ equal to the same thing as ($\vec{a}$+$\vec{b}$)$^{2}$?
They're the same, assuming the implied multiplication in the expression on the right is the dot product. Otherwise, multiplication of one vector by another is not defined (with the exception of the cross product).
 Quote by M. next And when is it equal to √(a$^{2}$+b$^{2}$)?
Tip: You don't need so many tex or itex tags. Your squared vector sum can be written like this:
$(\vec{a} +\vec{b})^2$
Or instead of the itex tags, you can use ## delimiters at the front and back.
 P: 378 Thank you for the reply and the tip. But about the √(a^2+b^2)?? My second part of the question?
P: 32
Squaring of vectors in absolute value

 Quote by M. next Thank you for the reply and the tip. But about the √(a^2+b^2)?? My second part of the question?
That is merely the magnitude of both vectors. Assuming that's what you mean? You were a little unclear on the second part. Think of magnitude as the size or length of those two vectors.
P: 1,315
 Quote by M. next Thank you for the reply and the tip. But about the √(a^2+b^2)?? My second part of the question?
$\sqrt{a^2+b^2}$ is the magnitude of $\vec{a}±\vec{b}$,where $\vec{a}$ and $\vec{b}$ are orthogonal (perpendicular) vectors.
 P: 378 Okay. Thank you, yes, it is exactly what I meant.

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