5.32. Find the volume of a triangular pyramid constructed on vectors $\overline{O A}, \overline{O B}$, and $\overline{O C}$, if $|\overrightarrow{O A}|=5,|\overrightarrow{O B}|=2,|\overrightarrow{O C}|=6, \overline{O A} \cdot \overrightarrow{O B}=0$, $\overline{O A} \cdot \overline{O C}=0, \overline{O B} \cdot \overline{O C}=8$.
5.32. Since $\overline{O A} \cdot \overline{O B}=0$ and $\overline{O A} \cdot \overline{O C}=0$, then $\overline{O A} \perp \overline{O B}$ and $\overline{O A} \perp \overline{O C}$, which means $\overline{O A} \perp(O B C)$, i.e., $O A$ is the height of the pyramid $O A B C$ (Fig. 5.21). Let $\angle B O C=\varphi$; then $S_{\triangle O B C}=\frac{1}{2} O B \cdot O C \sin \varphi$. From the condition $\overline{O B} \cdot \overline{O C}=8$ it follows that
$|\overrightarrow{O B}| \cdot|\overrightarrow{O C}| \cos \varphi=8$, or $2 \cdot 6 \cos \varphi=8$,

from which $\cos \varphi=\frac{2}{3}$. Considering that $0<\varphi<180^{\circ}$,

we find $\sin \varphi=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$. Thus,

$$
S_{\triangle O B C}=\frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{\sqrt{5}}{3}=2 \sqrt{5} \text { (sq. units) }
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-616.jpg?height=355&width=302&top_left_y=107&top_left_x=841)

Fig. 5.21

and finally we get

$$
V_{A O B C}=\frac{1}{3} S_{\triangle O B C} \cdot A O=\frac{1}{3} \cdot 2 \sqrt{5} \cdot 5=\frac{10 \sqrt{5}}{3} \text { (cubic units) }
$$

Answer: $\frac{10 \sqrt{5}}{3}$ cubic units.