Define vector that forms equal angles to 3 other vectors

Question

How should I define a vector, that has equal angles to vectors $\vec{i}, \vec{i} + \vec{j}$ and $\vec{i} + \vec{j} + \vec{k}$?

After looking at the problem in a graphical way, I tried taking average from $\vec{i}$ and $\vec{i} + \vec{j} + \vec{k}$, modifying the $\vec{i}$ vector to be of same length than the $\vec{i} + \vec{j} + \vec{k}$. Unfortunately the answer does not seem to be correct.

Answer 1

In general, if you have $3$ vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ and you want to find a vector $\vec{v}$ making equal angles to them. You first construct the unit vectors for those vectors $\hat{v}_1, \hat{v}_2, \hat{v}_3$ and then solve for the equation:

$$\hat{v}_1 \cdot \vec{v} = \hat{v}_2 \cdot \vec{v} = \hat{v}_3 \cdot \vec{v} = \cos\theta\tag{*1}$$ where $\theta$ is the common angle.

This is easy if you have heard of dual basis. Given any $3$ linear independent vectors $\vec{u}_1, \vec{u}_2, \vec{u}_3$, a dual basis of them is another $3$ vectors $\vec{w}_1, \vec{w}_2, \vec{w}_3$ such that

$$\vec{u}_i \cdot \vec{w}_j = \begin{cases}1, & i = j\\ 0, & i \ne j\end{cases}$$

It is not hard to word out the explicit form of $\vec{w}_i$:

$$\vec{w}_1 = \frac{\vec{u}_2 \times \vec{u}_3}{\vec{u}_1\cdot (\vec{u}_2 \times \vec{u}_3)},\quad \vec{w}_2 = \frac{\vec{u}_3 \times \vec{u}_1}{\vec{u}_1\cdot (\vec{u}_2 \times \vec{u}_3)}\quad\text{ and }\quad \vec{w}_3 = \frac{\vec{u}_1 \times \vec{u}_2}{\vec{u}_1\cdot (\vec{u}_2 \times \vec{u}_3)} $$ The nice thing of dual basis is if your know the dot products of a vector $\vec{u}$ with $\vec{u}_i$, you immediate obtain a decomposition of $\vec{u}$ in terms of $\vec{w}_i$:

$$\left\{\begin{align} \vec{u}_1 \cdot \vec{u} &= \lambda_1\\ \vec{u}_2 \cdot \vec{u} &= \lambda_2\\ \vec{u}_3 \cdot \vec{u} &= \lambda_3\\ \end{align}\right. \quad\implies\quad \vec{u} = \lambda_1 \vec{w}_1 + \lambda_2\vec{w}_2 + \lambda_3\vec{w}_3 $$ Apply this to $(*1)$ and notice we don't care about the overall length of $\vec{v}$, we find

$$\begin{align}\vec{v} &\propto \hat{v}_1 \times \hat{v}_2 + \hat{v}_2 \times \hat{v}_3 + \hat{v}_3 \times \hat{v}_1\\ &= \frac{1}{\sqrt{2}}(1,0,0)\times(1,1,0) + \frac{1}{\sqrt{6}}(1,1,0)\times(1,1,1) + \frac{1}{\sqrt{3}}(1,1,1)\times(1,0,0)\\ &= (0,0,\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},0) + (0, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})\\ &\propto (1,\sqrt{2}-1,\sqrt{3}-\sqrt{2}) \end{align} $$

Answer 2

The length is, of course, irrelevant to the angles. Take the vector to be ai+ bj+ ck. The angle it makes with i is $cos^{-1}\left(\frac{a}{\sqrt{a^2+ b^2+ c^2}}\right)$. The angle it makes with i+ j is $cos^{-1}\left(\frac{a+ b}{\sqrt{2}/sqrt{a^2+ b^2+ c^2}}\right)$. The angle it makes with i+ j+ k is $cos^{-1}\left(\frac{a+ b}{\sqrt{3}sqrt{a^2+ b^2+ c^2}}\right)$. Setting those equal gives you two equations. Since the length is irrelevant solving for two of a, b, and c in terms of the third is sufficient.