Standard +0.3 This is a straightforward 3D vectors question requiring students to set up position vectors, find two vectors meeting at C, then apply the scalar product formula for the angle between vectors. While it involves 3D geometry, the setup is clear, the calculations are routine, and it's a standard textbook application making it slightly easier than average.
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\includegraphics[max width=\textwidth, alt={}, center]{73c0c113-8f35-4e7f-ad5d-604602498b71-2_741_533_1279_808}
The diagram shows a pyramid \(O A B C\) with a horizontal base \(O A B\) where \(O A = 6 \mathrm {~cm} , O B = 8 \mathrm {~cm}\) and angle \(A O B = 90 ^ { \circ }\). The point \(C\) is vertically above \(O\) and \(O C = 10 \mathrm {~cm}\). Unit vectors \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\) are parallel to \(O A , O B\) and \(O C\) as shown.
Use a scalar product to find angle \(A C B\).
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\includegraphics[max width=\textwidth, alt={}, center]{73c0c113-8f35-4e7f-ad5d-604602498b71-2_741_533_1279_808}
The diagram shows a pyramid $O A B C$ with a horizontal base $O A B$ where $O A = 6 \mathrm {~cm} , O B = 8 \mathrm {~cm}$ and angle $A O B = 90 ^ { \circ }$. The point $C$ is vertically above $O$ and $O C = 10 \mathrm {~cm}$. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O B$ and $O C$ as shown.
Use a scalar product to find angle $A C B$.
\hfill \mbox{\textit{CAIE P1 2010 Q5 [6]}}