Find the angles between the bisectors of the coordinate…

Find the angles between the bisectors of the coordinate angles $xOy$, $xOz$ and $yOz$.

Explanation of Method

The angles between the bisectors of the coordinate angles.

Step 1: Direction vectors

The bisector of the (positive) angle $xOy$ has the direction $\vec i+\vec j$, and similarly for the others: \[\vec b_1=\vec i+\vec j,\qquad \vec b_2=\vec i+\vec k,\qquad \vec b_3=\vec j+\vec k,\qquad |\vec b_i|=\sqrt2.\]

Step 2: Angles

\[ \vec b_1\cdot\vec b_2=\vec b_1\cdot\vec b_3=\vec b_2\cdot\vec b_3=1, \]

\[ \cos\varphi=\frac{1}{\sqrt2\cdot\sqrt2}=\frac12. \]

\[\angle(\vec b_1,\vec b_2)=\angle(\vec b_1,\vec b_3)=\angle(\vec b_2,\vec b_3)=60^\circ\]

$\vec b_1$ $\vec b_2$ $\vec b_3$ $O$

$\vec b_1\ \text{bisects }xOy$
$\vec b_2\ \text{bisects }xOz$
$\vec b_3\ \text{bisects }yOz$

The bisectors of the three coordinate angles are the face diagonals of the cube from $O$; each pair makes $60^\circ$.

Result

\[\boxed{\,\angle(\vec b_1,\vec b_2)=\angle(\vec b_1,\vec b_3)=\angle(\vec b_2,\vec b_3)=60^\circ\,}\]

Remark

Note. If the bisector of a neighbouring quadrant is used instead (e.g. $\vec i-\vec j$), the cosine is $\pm\tfrac12$, so the angle is $60^\circ$ or $120^\circ$. The acute angle between the lines is always $60^\circ$.