Cho tứ diện ABCD có \(AB \bot CD,AC \bot BD\). Góc giữa hai véc tơ \(\overrightarrow {AD} \) và \(\overrightarrow {BC} \) là

Vì \(AB \bot CD\) và \(AC \bot BD\) nên ta suy ra

\(\begin{array}{l}
\overrightarrow {AD} .\overrightarrow {BC} = \left( {\overrightarrow {AB} + \overrightarrow {BD} } \right).\left( {\overrightarrow {BD} + \overrightarrow {DC} } \right)\\
= \overrightarrow {AB} .\overrightarrow {BD} + \overrightarrow {AB} .\overrightarrow {DC} + {\overrightarrow {BD} ^2} + \overrightarrow {BD} .\overrightarrow {DC} \\
= \overrightarrow {AB} .\overrightarrow {BD} + 0 + {\overrightarrow {BD} ^2} + \overrightarrow {BD} .\overrightarrow {DC} \\
= \left( {\overrightarrow {AC} + \overrightarrow {CB} } \right).\overrightarrow {BD} + {\overrightarrow {BD} ^2} + \overrightarrow {BD} .\overrightarrow {DC} \\
= \overrightarrow {AC} .\overrightarrow {BD} + \overrightarrow {CB} .\overrightarrow {BD} + {\overrightarrow {BD} ^2} + \overrightarrow {BD} .\overrightarrow {DC} \\
= 0 + \overrightarrow {CB} .\overrightarrow {BD} + {\overrightarrow {BD} ^2} + \overrightarrow {BD} .\overrightarrow {DC} \\
= \left( {\overrightarrow {CB} .\overrightarrow {BD} + \overrightarrow {BD} .\overrightarrow {DC} } \right) + {\overrightarrow {BD} ^2}\\
= \left( {\overrightarrow {CB} + \overrightarrow {DC} } \right).\overrightarrow {BD} + {\overrightarrow {BD} ^2}\\
= \overrightarrow {DB} .\overrightarrow {BD} + {\overrightarrow {BD} ^2}\\
= – {\overrightarrow {BD} ^2} + {\overrightarrow {BD} ^2} = 0.
\end{array}\)

Suy ra \(\overrightarrow {AD} \bot \overrightarrow {BC} \Rightarrow \left( {\overrightarrow {AD} ,\overrightarrow {BC} } \right) = 90^\circ .\)