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Nov 2008 p1 q4
2254
The diagram shows a semicircular prism with a horizontal rectangular base \(ABCD\). The vertical ends \(AED\) and \(BFC\) are semicircles of radius 6 cm. The length of the prism is 20 cm. The mid-point of \(AD\) is the origin \(O\), the mid-point of \(BC\) is \(M\) and the mid-point of \(DC\) is \(N\). The points \(E\) and \(F\) are the highest points of the semicircular ends of the prism. The point \(P\) lies on \(EF\) such that \(EP = 8\) cm.
Unit vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are parallel to \(OD, OM\) and \(OE\) respectively.
(i) Express each of the vectors \(\overrightarrow{PA}\) and \(\overrightarrow{PN}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).
(ii) Use a scalar product to calculate angle \(APN\).
Solution
(i) To find \(\overrightarrow{PA}\), note that \(P\) is 8 cm along \(EF\) from \(E\), so \(P\) has coordinates \((8, 0, 6)\). \(A\) is at \((0, 0, 0)\). Thus, \(\overrightarrow{PA} = (0 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = -8\mathbf{i} - 6\mathbf{k}\).
For \(\overrightarrow{PN}\), \(N\) is at \((20, 0, 0)\). Thus, \(\overrightarrow{PN} = (20 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = 12\mathbf{i} - 6\mathbf{k}\).
(ii) The dot product \(\overrightarrow{PA} \cdot \overrightarrow{PN} = (-8)(12) + (0)(0) + (-6)(-6) = -96 + 36 = -60\).
The magnitudes are \(|\overrightarrow{PA}| = \sqrt{(-8)^2 + 0^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\).
