Finding angle in 110-dimensional Euclidean space

A vector in 110-dimensional Euclidean space $\mathbf{R}^{110}$ makes equal acute angles with the positive directions of the 110 coordinate axes. Approximately what is that angle in degrees?

This is intended to be a challenge question on my homework, however I have no idea how to even begin, thanks in advance

• Can you determine the angle between a vector and the positive first coordinate axis? – Daniel Fischer Mar 1 '16 at 22:22
• The dot product is still a valid operation in such dimensions, as a hint the vector you have can be represented as an 110x1 vector of just ones – Triatticus Mar 1 '16 at 22:23
• From what the key says, the answer is 84.528750, I still dont know how to figure it out – EconDude Mar 1 '16 at 22:26

In $\mathbb{R^2}$, the vector $<1, 1>$ accomplishes this by making an angle of $45^\circ$ with both the positive-x and the positive-y axis.

This is because this vector dotted with any (and all) of the basis vectors gives a dot product of 1. So, we can calculate $cos(\theta) = \frac{1}{\sqrt{2}}$

In $\mathbb{R^3}$, the vector $<1, 1, 1>$ accomplishes this by making an angle of $54.74^\circ$ with the positive-x, positive-y, and positive-z axis.

This is because, similarly to above, we can calculate $cos(\theta) = \frac{1}{\sqrt{3}}$

Can you take it from here?

• Ohh I see thanks! So the dot product will always result in 1 if we follow $$\cos\theta = \frac{ \mv{a}\cdot \mv{b}}{||\mv{a}||||\mv{b}||}$$ the numerator will always be one, but how can you determine that the bottom will be $${sqrt{110}}$$ – EconDude Mar 1 '16 at 22:37
• The denominator is the product of the length of the vectors. Your basis vector has length, by definition, equal to 1. The other vector is just a vector of ones. In $\mathbb{R^2}$, the vector had two 1s, so its length was $\sqrt{2}$. In $\mathbb{R^{110}}$, the vector will have 110 1s, so its length will be... – anonymouse Mar 1 '16 at 22:39
• That makes perfect sense thanks for the help! – EconDude Mar 1 '16 at 22:41