Urgent help needed: |a|^2 = (a_1; a_2; a_3)^2 = a_1^2 + a_2^2 + a_3^2

[endrodi88]

2. K2
I have to prove that its true

 

[stapel]

2. K2
I have to prove that its true

Translating from what Google thinks is probably Hungarian, it appears that the indicated exercise says something along the lines of the following:

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\(\displaystyle \mbox{Prove }\, \big|\,\boldsymbol{a}\,\big|^2\,=\, \left(a_1;\, a_2;\, a_3\right)^2\, =\, a_1^2\, +\, a_2^2\, +\, a_3^2\)

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Is the above correct? Is \(\displaystyle \boldsymbol{a}\) a vector, and \(\displaystyle a_1,\, a_2,\) and \(\displaystyle a_3\) are its components? How does your book define the absolute-value bars in this context?

When you reply, please include a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. Thank you!

 

[pka]

2. K2
I have to prove that its true

There are a few symbols that are near universal, \(\displaystyle |\vec{u}|\) or its cousin \(\displaystyle \|\vec{u}\|\) are. They denote the length of the vector \(\displaystyle \vec{u}\).
Working from from the title that you gave this post, I would comment that it is almost impossible to prove a definition.
Now it is possible that your textbook is pointing to some other facts that you have not given us. Short of being made aware of those, here is my take.
If \(\displaystyle \vec{u}=\left<u_1,u_2,u_3\right>\) then by near universal agreement its length is \(\displaystyle \sqrt{u_1^2+u_2^2+u_3^2}\)
Therefore it follows that \(\displaystyle |\vec{u}|^2=|\left<u_1,u_2,u_3\right>|^2=\left( \sqrt{u_1^2+u_2^2+u_3^2}\right)^2=u_1^2+u_2^2+u_3^2\)
As a matter of interest: \(\displaystyle |\vec{u}|^2=\vec{u}\cdot \vec{u}\).