Summing Unit Vectors

Since each unit vector is of magnitude 1, the largest magnitude that the sum of nine unit vectors could possibly be is 9, but the given vector has a magnitude of 10, so no choice of nine unit vectors will achieve $\langle 6, 8 \rangle$ .

Let's name this vector $\langle 6, 8 \rangle$ as $\vec{v}$ and suppose for purposes of contradiction that there is some collection of nine unit vectors $\{\vec{u_i}\}_{i=1}^9$ such that $\vec{v} = \displaystyle{\sum_{i=1}^9\vec{u_i}}$ . Then surely the magnitude of both $\vec{v}$ and that sum must be equal. Applying the triangle inequality yields:

$\|\vec{v}\| = \left\| \displaystyle{\sum_{i=1}^9 \vec{u_i}}\right\| \leq \displaystyle{\sum_{i=1}^9 \|\vec{u_i}\|}$

Since the vectors $\vec{u_i}$ are unit vectors, the sum on the right evaluates to 9, but $\vec{v}$ itself has magnitude $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ . So we have $10 \leq 9$ , contradiction. It follows that $\vec{v}$ can't be exrpressed as a sum of nine unit vectors $\square$ .