Angle and distance of defined inner vector space

Question

Let the vectors u and v be vectors in R$^4$. The inner product space is defined by

$<u, v> = 2u$₁$v$₁ $+ u$₂$v$₂ $+ u$₃$v$₃ $ + 3u$₄$v$₄

When u $= (2, -2, 4, 2)$ and v = $(4, 1, 4, 2)$ find the angle and distance between u and v.

Now the angle is normally defined by $\arccos$ u ∙ v/||u|| ||v||. Would I still use the dot product answer of $26$ or the $<u, v>$ calculation of $42$?

Then for calculating the norm such as $<u, u>$^1/2 do I sub in u in the place of v which $ = \sqrt40$ or apply with itself $ = \sqrt72$. Which method is correct?

Then how does distance work when it is normally $<u-v, u-v>$^1/2? I got the answer of 3 but I am not confident that is correct. Thanks.

Answer 1

Any inner product $\langle.,.\rangle$ induces a unique norm $\|.\|_{\langle.,.\rangle}$ by $\|u\|_{\langle.,.\rangle}:=\sqrt{\langle u,u\rangle}$. As each inner product and its induced norm satisfy the Cauchy-Schwarz inequality $$|\langle u,v\rangle|\leq\|u\|_{\langle.,.\rangle} \|v\|_{\langle.,.\rangle}$$ one may define the angle $\angle_{\langle.,.\rangle}(u,v)$ between $u$ and $v$ in respect to $\langle.,.\rangle$ via $$\cos\bigl(\angle_{\langle.,.\rangle}(u,v)\bigr)=\frac{\langle u,v\rangle}{\|u\|_{\langle.,.\rangle} \|v\|_{\langle.,.\rangle}}.$$ Hence the angle depends on the chosen inner product, just don't mix it with the dot product. One omits the indices "${}_{\langle.,.\rangle}$" in case the context is not ambiguous.