Quick check on an even numbered problem:
$$\mathrm{Given:}\;\vec{u}=(1,1,1),\, \vec{v}=(2,5,2), \langle \vec{u},\vec{v} \rangle=u_1v_1+2u_2v_2+u_3v_3\
\langle \vec{u},\vec{v} \rangle=2+15+2\implies 19\
\|\vec{u}\|=\sqrt{1+2+1}\implies 2\
\|\vec{v}\|=\sqrt{4+50+4}\implies \sqrt{58}\
d(\vec{u},\vec{v})=\sqrt{(2-1)^2+2(5-1)^2+(2-1)^2}\implies\sqrt{34}$$Correct?

Question

Quick check on an even numbered problem:
$$\mathrm{Given:}\;\vec{u}=(1,1,1),\, \vec{v}=(2,5,2), \langle \vec{u},\vec{v} \rangle=u_1v_1+2u_2v_2+u_3v_3\
\langle \vec{u},\vec{v} \rangle=2+15+2\implies 19\
\|\vec{u}\|=\sqrt{1+2+1}\implies 2\
\|\vec{v}\|=\sqrt{4+50+4}\implies \sqrt{58}\
d(\vec{u},\vec{v})=\sqrt{(2-1)^2+2(5-1)^2+(2-1)^2}\implies\sqrt{34}$$Correct?

amistre64 · Answer

i take it that a dot product?  the top looks like its more of a (uv)^2 setup

amistre64 · Answer

but then that would be u.u + 2.u.v + v.v  tho

amistre64 · Answer

the length of (1,1,1) = sqrt(3), not 2

am i reading this right?

e.mccormick · Answer

The top is a given.  Then I get to find $$\langle \vec{u},\vec{v} \rangle\
\|\vec{u}\|\
\|\vec{v}\|\
d(\vec{u},\vec{v})$$
As for that one, what about th2 2 in the given?  Dosn't that change it to: $\sqrt{1^2+2(1^2)+1^2}$?

amistre64 · Answer

its given that:
$$\langle\vec{u},\vec{v} \rangle=u_1v_1+2u_2v_2+u_3v_3\$$
that is notation for dot product right?

amistre64 · Answer

not sure why a 2 would be in there

e.mccormick · Answer

Ah, and I forgot to add it is an inner product defined in $\mathbb{R}^n$.  So they are defining the inner product as that.

e.mccormick · Answer

The 2 is there because it is given that way, as opposed to the Euclidian inner product which is just dot product.

amistre64 · Answer

then wouldnt:$$(\vec u,\vec v)=1.2+2.1.5+1.2=2+10+2=14$$

e.mccormick · Answer

AH HA!  Yes, I did make a mistake there!  I don't know what I was thinking.

amistre64 · Answer

and the norm is defined as:  sqrt(u,v)

amistre64 · Answer

..or something like that :)

e.mccormick · Answer

Yes, it is basically just the distance formula.

e.mccormick · Answer

But all it is measuring is the distance of a vector to itself, so you assume it is the distance to the origin, so the components of the vector are squared, added, then rooted.

e.mccormick · Answer

If you mean the $d(\vec{u},\vec{v})$ part, that is distance between, not the norm.

e.mccormick · Answer

I think the rest of it is correct.  I double checked the other numbers and the book to make sure I used the right formula.  Thanks for finding the one I made a mistake on.

amistre64 · Answer

youre welcome, ive had alot of practice finding all my own mistakes thru the years :)

e.mccormick · Answer

Oh, I know that feeling.  You make a mistake and because you made it, it is invisible!