Problem source

A pyramid has a horizontal rectangular base $ABCD$ and its vertex $V$ is vertically above the centre of the base. The acute angle between the face $AVB$ and the base is $\alpha$, the acute angle between the face $BVC$ and the base is $\beta$ and the obtuse angle between the faces $AVB$ and $BVC$ is $\pi - \theta$.
\begin{questionparts}
\item The edges $AB$ and $BC$ are parallel to the unit vectors $\mathbf{i}$ and $\mathbf{j}$, respectively, and the unit vector $\mathbf{k}$ is vertical. Find a unit vector that is perpendicular to the face $AVB$.
Show that
$$\cos \theta = \cos \alpha \cos \beta.$$
\item The edge $BV$ makes an angle $\phi$ with the base. Show that
$$\cot^2 \phi = \cot^2 \alpha + \cot^2 \beta.$$
Show also that
$$\cos^2 \phi = \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta} \geq \frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$$
and deduce that $\phi < \theta$.
\end{questionparts}

Solution source

\begin{tikzpicture}[scale=2]
    % Define the perspective view coordinates for a square base
    % Using a vanishing point approach for better perspective
    
    % Vanishing point (implicit, not drawn)
    \coordinate (VP1) at (15,4);
    \coordinate (VP2) at (-10,4);
    
    % Base coordinates with proper perspective distortion
    \coordinate (A) at (0, 0) ;     % Bottom left
    \coordinate (B) at ($(A)!0.25!(VP1)$);    % Bottom right
    \coordinate (C) at  ($(A)!0.25!(VP2)$);     % Top right
    \coordinate (D) at ($(B)!0.20!(VP2)$);    % Top left

    \coordinate (M1) at ($(A)!0.5!(C)$);
    \coordinate (M2) at ($(B)!0.5!(D)$);

    \coordinate (X) at ($(M1)!0.5!(M2)$);
    \coordinate (X1) at  ($(C)!0.5!(D)$);
    \coordinate (X2) at ($(A)!0.5!(B)$);

    % \coordinate (R) at (-0.5,2);
    \coordinate (R) at ($(X) + (0, 1.5)$);
    % \coordinate (S) at ($(R)!0.125!(VP1)$);

    \coordinate (Y) at  ($(A)!0.3!(B)$);

    \node[below] at (A) {$A$}; 
    \node[right] at (B) {$B$}; 
    \node[left] at (C) {$D$}; 
    \node[above] at (D) {$C$}; 

    \node[above] at (R) {$V$};


    % \draw[<->, thin, >=latex, gray, dashed] (X) -- (R) node[midway, left, label, black] {$h$};
    \draw[<->, thin, >=latex, gray, dashed] ($(A)!-0.01!(VP1)$) -- ($(C)!-0.01!(VP1)$) node[midway, left, label, black] {$2d$};
    \draw[<->, thin, >=latex, gray, dashed] ($(A)!-0.01!(VP2)$) -- ($(B)!-0.01!(VP2)$) node[midway, below, label, black] {$2b$};
    % \draw[<->, thin, >=latex, gray, dashed] ($(A)!0.01!(VP1)$) -- ($(M1)!0.01!(VP1)$) node[midway, left, label, black] {$p$};
    % \draw[<->, thin, >=latex, gray, dashed] ($(A)!0.01!(VP2)$) -- ($(Y)!0.01!(VP2)$) node[midway, above, label, black] {$p$};
    
    % Draw the base with perspective distortion
    \draw[thick] (A) -- (R) -- (C) -- cycle;
    \draw[thick] (A) -- (R) -- (B) -- cycle;
    % \draw[thick] (R) -- (S) -- (B);
    \draw[dashed] (A) -- (B) -- (D) -- (C) -- cycle;
    \draw[dashed] (D) -- (R) -- (C);
    % \draw[dashed] (R) -- (S);
    \draw [dashed] (R) -- (M1) -- (X);

    \pic [draw, angle radius=0.8cm, "$\beta$"] {angle = X--M1--R};
    % \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = S--M2--X};
    \pic [draw, angle radius=0.8cm, "$\alpha$"] {angle = X--X1--R};
    \pic [draw, angle radius=0.8cm, "$\alpha$"] {angle = R--X2--X};
    \pic [draw, angle radius=0.8cm, "$\phi$"] {angle = R--B--X};

    % \draw[dashed] (R) -- (M1) -- (M2) -- (S);
    \draw[dashed] (R) -- (X) node[pos=0.6, right] {$h$};
    \draw[dashed] (R) -- (X1) -- (X2) -- (R);
    \draw[dashed] (B) -- (X);
    
\end{tikzpicture}

\begin{questionparts}
\item 
Let $A = (0,0,0)$ and then $B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}$ and $V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}$

We also have \begin{align*}
    && \tan \alpha &= \frac{h}{d}\\
    && \tan \beta &= \frac{d}{b} \\
    && \vec{AV} \times \vec{VB} &= \begin{pmatrix}
        b \\ d \\ h
    \end{pmatrix} \times \begin{pmatrix}
        -b \\ d \\ h
    \end{pmatrix} \\
    &&&= \begin{pmatrix}
        0 \\ -2bh \\ 2db
    \end{pmatrix} \\
    &&&= 2b \begin{pmatrix}
        0 \\ -d \tan \alpha \\ d
    \end{pmatrix} \\
    &&&= k \begin{pmatrix}
        0 \\ - \sin \alpha \\ \cos \alpha
    \end{pmatrix}
\end{align*}

similarly for the vector perpendicular to the other face it must be $\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}$


Looking at the angle between these perpendicular (to find the angles between the faces we see:

\begin{align*}
    \begin{pmatrix}
        0 \\ - \sin \alpha \\ \cos \alpha
    \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= 
        \cos \alpha \cos \beta
\end{align*}

But this is also $\pi -$ the angle between the planes, ie $\cos \theta = \cos \alpha \cos \beta$

\item $\,$

\begin{align*}
    && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\
    && \cot^2 \alpha &= \frac{d^2}{h^2} \\
    && \cot^2 \beta &= \frac{b^2}{h^2} \\
    \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha
\end{align*}

\begin{align*}
    && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\
    && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\
    && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\
    && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\
    &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\
    &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\
    &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\
    &&&= \cos^2\phi
\end{align*}

Also notice that \begin{align*}
    && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\
    &&&= 2 \cos \theta  \\
    \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\
    &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\
    &&&> \cos^2 \theta \\
    \Rightarrow && \phi &< \theta
\end{align*}
\end{questionparts}