Multidimensional Vectors

The fact that the vectors are 2015-dimensional is unnecessary; this applies to vectors of all dimensions. We'll can generalize this and represent $\vec{a}$ as $\langle a_{1},a_{2},\dots,a_{n} \rangle$ and $\vec{b}$ as $\langle b_{1},b_{2},\dots,b_{n} \rangle$ .

Using the definition of dot product of two vectors, $\vec{a} \cdot \vec{b} = \|\vec{a} \| \| \vec{b} \| \cos \theta$

$a_{1} b_{1} + a_{2} b_{2} + \cdots + a_{n} b_{n} = (\sqrt{8})(\sqrt{96})(\frac{\sqrt{3}}{2}) = 24$

Squaring the initial equations using the definition of magnitude yields $a_{1}^2+a_{2}^2+\dots+a_{n}^2 = 8$ and $b_{1}^2+b_{2}^2+\dots+b_{n}^2 = 96$ .

Recognizing that $\|\vec{a} - \vec{b} \|^2 = \| \langle (a_{1}-b_{1}), (a_{2}-b_{2}),\dots, (a_{n}-b_{n}) \rangle \|^2$ is key because that leads to $\|\vec{a}-\vec{b}\|^2 = a_{1}^2-2a_{1}b_{1}+b_{1}^2+a_{2}^2-2a_{2}b_{2}+b_{2}^2+\dots+a_{n}^2-2a_{n}b_{n}+b_{n}^2$ .

This is the sum of the squares of the $a$ and $b$ terms minus twice the sum of the $a_{n}b_{n}$ products, which we found above! Therefore, $\|\vec{a} - \vec{b} \|^2 = 8 + 96 - 2(24) = 56$ .

The vectors $\vec{a}$ , $\vec{b}$ and $\vec{a}-\vec{b}$ form a triangle. We can use the law of cosines to calculate:

$||\vec{a}-\vec{b}||^2=||\vec{a}||^2+||\vec{b}||^2-2||\vec{a}||\cdot||\vec{b}||\cos(30^\circ)$

$=8+96-2\sqrt{8\times 96}\times \frac12\sqrt3$

$=8+96-48$

$=\boxed{56}.$

$\|\vec{a}-\vec{b}\|^2=(\vec{a}-\vec{b}). (\vec{a}-\vec{b})$

$=\vec{a}.\vec{a}+\vec{b}.\vec{b}-2\vec{a}.\vec{b}$

$=8+96-2\sqrt{8}\sqrt{96}\dfrac{\sqrt{3}}{2}$

$=104-\sqrt{8\times 8\times 4\times 3 \times 3}$

$=104-8\times 2\times 3=\boxed{56}$