5. Zadatak

Prvo rješenje.

Nacrtajmo prizmu kojoj je gornja baza sukladna osnovki piramide, kao na slici. Povucimo dužinu $\overline{VP_3}$ paralelnu s $\overline{CP_2}$.

Tražimo kut $\varphi$ u trokutu $VP_1P_3$. Izračunajmo duljine stranica u tom trokutu:

$$\begin{aligned}|VP_1|&=\sqrt{1+(\sqrt2)^2}=\sqrt3,\\|VP_3|&=|CP_2|=\frac{\sqrt3}2\cdot2\sqrt2=\sqrt6,\\|P_1P_3|&=\sqrt{|P_1P_2|^2+|P_2P_3|^2}=\sqrt{(\sqrt2)^2+1}=\sqrt3.\end{aligned}$$

Prema poučku o kosinusu, vrijedi

$$\cos\varphi=\frac{3+6-3}{2\cdot\sqrt3\cdot\sqrt6}=\frac6{6\sqrt2}=\frac{\sqrt2}2$$

pa je $\varphi=45^\circ$.

Drugo rješenje.

Za kosinus traženog kuta vrijedi

$$\cos\varphi=\frac{\overrightarrow{VP_1}\cdot\overrightarrow{CP_2}}{|\overrightarrow{VP_1}|\cdot|\overrightarrow{CP_2}|}$$

Duljine vektora u nazivniku su

$$\begin{aligned}|\overrightarrow{VP_1}|&=|VP_1|=\sqrt{1^2+(\sqrt2)^2}=\sqrt3,\\|\overrightarrow{CP_2}|&=|CP_2|=2\sqrt2\cdot\frac{\sqrt3}2=\sqrt6.\end{aligned}$$

Prikažimo te vektore ovako:

$$\begin{aligned}\overrightarrow{VP_1}&=\overrightarrow{VC}+\frac12\overrightarrow{CB},\\\overrightarrow{CP_2}&=\overrightarrow{CA}+\frac12\overrightarrow{AB}=\frac12(\overrightarrow{CA}+\overrightarrow{CB}).\end{aligned}$$

Sada je

$$\begin{aligned}\overrightarrow{VP_1}\cdot\overrightarrow{CP_2}&=\frac12\left(\overrightarrow{VC}\cdot\overrightarrow{CA}+\overrightarrow{VC}\cdot\overrightarrow{CB}+\frac12\overrightarrow{CB}\cdot\overrightarrow{CA}+\frac12|\overrightarrow{CB}|^2\right)\\&=\frac12\left(0+0+\frac12\overrightarrow{CB}\cdot\overrightarrow{CA}+\frac12\cdot8\right)=\frac14\overrightarrow{CB}\cdot\overrightarrow{CA}+2.\end{aligned}$$

Sad računamo

$$\begin{aligned}\overrightarrow{CB}\cdot\overrightarrow{CA}&=|\overrightarrow{CB}|\cdot|\overrightarrow{CA}|\cos\angle ACB\\&=2\sqrt2\cdot2\sqrt2\cdot\cos60^\circ=8\cdot\frac12=4.\end{aligned}$$

Zato je

$$\cos\varphi=\frac3{\sqrt3\cdot\sqrt6}=\frac3{3\sqrt2}=\frac{\sqrt2}2,$$

pa traženi kut $\varphi$ iznosi $45^\circ$.