We're given an inner product defined by
[tex]\langle p,q\rangle=p(-2)q(-2)+p(0)q(0)+p(2)q(2)[/tex]
That is, we multiply the values of [tex]p(x)[/tex] and [tex]q(x)[/tex] at [tex]x=-2,0,2[/tex] and add those products together.
[tex]p(x)=2x^2+6x+1[/tex]
[tex]q(x)=3x^2-5x-6[/tex]
The inner product is
[tex]\langle p,q\rangle=-3\cdot16+1\cdot(-6)+21\cdot(-4)=-138[/tex]
To find the norms [tex]\|p\|[/tex] and [tex]\|q\|[/tex], recall that the dot product of a vector with itself is equal to the square of that vector's norm:
[tex]\langle p,p\rangle=\|p\|^2[/tex]
So we have
[tex]\|p\|=\sqrt{\langle p,p\rangle}=\sqrt{(-3)^2+1^2+21^2}=\sqrt{451}[/tex]
[tex]\|q\|=\sqrt{\langle q,q\rangle}=\sqrt{16^2+(-6)^2+(-4)^2}=2\sqrt{77}[/tex]
Finally, the angle [tex]\theta[/tex] between [tex]p[/tex] and [tex]q[/tex] can be found using the relation
[tex]\langle p,q\rangle=\|p\|\|q\|\cos\theta[/tex]
[tex]\implies\cos\theta=\dfrac{-138}{22\sqrt{287}}\implies\theta\approx1.95\,\mathrm{rad}\approx111.73^\circ[/tex]
