Is My Solution to the Orthonormalization Process Correct?

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is that correct??
 
I'm not going to go through all of that calculation! You should be able to check for your self if they are independent (if so then since there are three vectors, yes, they form a basis for the subspace) and if they are orthonormal.

I will make two comments. You do not "divide by the normal". You "normalize" a vector by dividing by its length. And it is generally simpler to normalize each vector (divide by its length) as you calculate each vector, not at the end.
 
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so first i divide each vector by its length
and then i do the orthogonalization process
did i wrote the orthogonalization formula correctly
??
 
Last edited:
transgalactic said:
so first i divide each vector by its length
and then i do the orthogonalization process
did i wrote the orthogonalization formula correctly
??

After you have found a vector perpendicular to all the previous vectors, divide by its length. That will simplify the rest of the calculation.

Again, check it yourself. Do all the vectors have length 1 (is the dot product of any vector with itself equal to 1)? Are all the vector orthogonal to all the others (is the dot product of two different vectors 0)?
 
ok after,
whats the right formula for the orthogonolazation process

did i used the correct formula??
did i put the vectors in the right place?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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