Sine and Cosine mixture

Consider the three vectors $\mathbf{v}_1 = ( 2 \cos \alpha, 2 \sin \alpha ) , \mathbf{v}_2 = ( 5 \sin \beta , 5 \cos \beta ), \mathbf{v}_3 = (-8, -15)$ , then

the indicated sum is the square of the magnitude of $\mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3$ , we can choose $\alpha$ and $\beta$ so that $\mathbf{v}_1$ and $\mathbf{v}_2$ point in the opposite direction as $\mathbf{v}_3$ , hence the length of the resultant vector is $\sqrt{(-8)^2 + (-15)^2 } - 2 - 5 = 17 - 7 = 10$ , so that the square of the magnitude is $10^2 = \boxed{100}$ .

Consider three vectors $\mathbf {v_a} = (2\cos \alpha, 2\sin \alpha)$ , $\mathbf {v_b} = (5\sin \beta, 5 \cos \beta)$ , and $\mathbf {v_c} = (8,15)$ . Let $\mathbf {v_d = v_a+v_b}$ . Then $(2\cos \alpha + 5\sin \beta - 8)^2 + (2\sin \alpha + 5\cos \beta -15)^2 = |\mathbf{v_d-v_c}|^2$ . And $|\mathbf{v_d-v_c}|$ (denoted by the dotted line) is minimum when $\alpha = \tan^{-1} \frac 8{15}$ and $\beta = \tan^{-1} \frac {15}8$ . Then we have

$\mathbf{|v_d-v_c| = |v_c-v_d| = |v_c|-|v_d|} = \sqrt{8^2+15^2} - \mathbf{|v_a|-|v_b|} = 17 - 2 - 5 = 10$

Therefore $\mathbf{|v_c-v_d|^2 = 10^2 = \boxed{100}}$ .