Problem source

Let $\mathbf{n}$ be a vector of unit length and $\Pi$ be the plane through the origin perpendicular to $\mathbf{n}$. For any vector $\mathbf{x}$, the projection of $\mathbf{x}$ onto the plane $\Pi$ is defined to be the vector $\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\,\mathbf{n}$.
 
The vectors $\mathbf{a}$ and $\mathbf{b}$ each have unit length and the angle between them is $\theta$, which satisfies $0 < \theta < \pi$. The vector $\mathbf{m}$ is given by $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.
 
\begin{questionparts}
    \item Show that $\mathbf{m}$ bisects the angle between $\mathbf{a}$ and $\mathbf{b}$.
 
    \item The vector $\mathbf{c}$ also has unit length. The angle between $\mathbf{a}$ and $\mathbf{c}$ is $\alpha$, and the angle between $\mathbf{b}$ and $\mathbf{c}$ is $\beta$. Both angles are acute and non-zero.
 
    Let $\mathbf{a}_1$ and $\mathbf{b}_1$ be the projections of $\mathbf{a}$ and $\mathbf{b}$, respectively, onto the plane through the origin perpendicular to $\mathbf{c}$. Show that $\mathbf{a}_1 \cdot \mathbf{c} = 0$ and, by considering $|\mathbf{a}_1|^2 = \mathbf{a}_1 \cdot \mathbf{a}_1$, show that $|\mathbf{a}_1| = \sin\alpha$.
 
    Show also that the angle $\varphi$ between $\mathbf{a}_1$ and $\mathbf{b}_1$ satisfies
    \[
        \cos\varphi = \frac{\cos\theta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta}.
    \]
 
    \item Let $\mathbf{m}_1$ be the projection of $\mathbf{m}$ onto the plane through the origin perpendicular to $\mathbf{c}$. Show that $\mathbf{m}_1$ bisects the angle between $\mathbf{a}_1$ and $\mathbf{b}_1$ if and only if
    \[
        \alpha = \beta \qquad \text{or} \qquad \cos\theta = \cos(\alpha - \beta).
    \]
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& \cos \angle MOB &= \frac{\mathbf{m} \cdot \mathbf{b}}{|\mathbf{m}||\mathbf{b}|} \\
&&&= \frac{\cos \theta + 1}{2\sqrt{\frac14(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})}} \\
&&&= \frac{\cos \theta + 1}{\sqrt{1+1+2\cos \theta}} \\
&&&= \frac{1 + \cos \theta}{\sqrt{2(1+\cos \theta})} \\
&&&= \frac1{\sqrt{2}} \sqrt{1+\cos \theta} \\
&&&= \cos \tfrac{\theta}{2}
\end{align*}
Since $0 < \theta < \pi$ we must have $\angle MOB = \tfrac{\theta}{2}$ ie it is the angle bisector.

\item The plane through the origin perpendicular to $\mathbf{c}$ has $\mathbf{x} \cdot \mathbf{c} = 0$, so
 \begin{align*}
&& \mathbf{a}_1 \cdot \mathbf{c} &= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot \mathbf{c} \\
&&&= \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} \\
&&&= 0 \\
\\
&& |\mathbf{a}_1|^2 &= \mathbf{a}_1 \cdot \mathbf{a}_1 \\
&&&= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \\
&&&= 1 - 2(\mathbf{a} \cdot \mathbf{c})^2  + \mathbf{a} \cdot \mathbf{c} \\
&&&= (1-\cos^2 \alpha) \\
&&&=  \sin^2 \alpha \\
\Rightarrow && |\mathbf{a}_1| &= \sin \alpha \\
\Rightarrow && |\mathbf{b}_1| &= \sin \beta \tag{changing a and b} \\
\\
&& \cos \phi &= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1}{|\mathbf{a}_1||\mathbf{b}_1|} \\
&&&= \frac{(\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{c})}{\sin \alpha \sin \beta} \\
&&&= \frac{\mathbf{a} \cdot \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})+(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})}{\sin \alpha \sin \beta} \\
&&&= \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta}
\end{align*}

\item Note that $\mathbf{m}_1 = \tfrac12(\mathbf{a}_1 + \mathbf{b}_1)$ either by expanding or by noting that projection is linear
 \begin{align*}
&& \cos \angle M_1OB_1 &= \frac{\mathbf{m}_1 \cdot \mathbf{b}_1}{|\mathbf{m}_1||\mathbf{b}_1|} \\
&&&= \frac{(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{b}_1}{2|\mathbf{m}_1||\mathbf{b}_1|} \\
&&&= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1 + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\
&&&= \frac{|\mathbf{a}_1 || \mathbf{b}_1| \cos \phi + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\
&&&= \frac{|\mathbf{a}_1 |\cos \phi + |\mathbf{b}_1|}{2|\mathbf{m}_1|} \\
&&&= \frac{\sin \alpha \cos \phi + \sin \beta}{2\sin \frac{\theta}{2}} \\
&&&= \frac{\sin \alpha \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \sin \beta}{2\sin \frac{\theta}{2}} \\
&&&= \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} \\
\Rightarrow && \cos \angle M_1OA_1 &=  \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} 
\end{align*}
$M_1$ is a bisector iff these two cosines are equal, ie
\begin{align*}
&& \cos \angle M_1OB_1 &= \cos \angle M_1OA_1  \\
\Leftrightarrow && \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \\
\Leftrightarrow && \cos \theta (\sin \alpha - \sin \beta) &= \cos \alpha \cos \beta(\sin \alpha - \sin \beta) + \sin \alpha \sin \beta (\sin \alpha - \sin \beta) \\
\Leftrightarrow &&0 &= (\sin \alpha - \sin \beta)( \cos \theta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta)) \\
&&&= (\sin \alpha - \sin \beta) (\cos \theta - \cos (\alpha - \beta))
\end{align*}
From which the result immediately follows

\end{questionparts}