Find angle between two vectors using inner product?

Question

$u=(0,-5), v=(4,3)$ and $\langle u,v \rangle=2u_1v_1+u_2v_2$ The problem says to find the angle $\theta$ between the vectors and round your answer to the two decimal places. I've been trying to do this, but my answer is a bit off. The answer should be $2.06$ radians.
This is what I've been trying
I've gotten $2.18$ radians. I'm not sure about where I'm going wrong?

Answer 1

Salaam!
Let $u=(u_1,u_2)$ and $v=(v_1,v_2)$ be vectors, we have the dot product defined as follows: $$<u,v> = 2u_1 v_1 + u_2 v_2$$.
Also, $$<u,v>= \mid u \mid \mid v \mid cos \theta$$ where: $$\mid u \mid =\sqrt {(u_1^2+u_2^2)}$$ $$ \mid v \mid =\sqrt{ (2v_1^2+v_2^2)}$$ and $\theta$ is the angle between vectors $u$ and $v$. Thus, in your example, we have $$u=(0,-5), v=(4,3)$$ Thus, $$<u,v>=(0 \times 4)+(-5 \times 3)=-15$$ $$\mid u \mid = \sqrt{ 0^2 + (-5)^2} =5$$ $$\mid v \mid = \sqrt{ 2\times 4^2 + 3^2} =\sqrt{41}$$  $$\theta = cos^{-1}(\frac{{-15}}{5 \times \sqrt{41}}) \approx 2.06$$ I hope this helps! please, feel free to ask more questions if you still have any.