Exam-Style Problem

Nov 2022 p31 q11

2242

In the diagram, \(OABCD\) is a solid figure in which \(OA = OB = 4\) units and \(OD = 3\) units. The edge \(OD\) is vertical, \(DC\) is parallel to \(OB\) and \(DC = 1\) unit. The base, \(OAB\), is horizontal and angle \(AOB = 90^\circ\). Unit vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are parallel to \(OA, OB\) and \(OD\) respectively. The midpoint of \(AB\) is \(M\) and the point \(N\) on \(BC\) is such that \(CN = 2NB\).

Express vectors \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).
Calculate the angle in degrees between the directions of \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\).
Show that the length of the perpendicular from \(M\) to \(ON\) is \(\sqrt{\frac{22}{5}}\).

Solution

(a) To find \(\overrightarrow{MD}\), we first find \(\overrightarrow{OM}\) which is the midpoint of \(AB\). Since \(OA = 4\mathbf{i}\) and \(OB = 4\mathbf{j}\), \(OM = 2\mathbf{i} + 2\mathbf{j}\). Thus, \(\overrightarrow{MD} = \overrightarrow{OD} - \overrightarrow{OM} = 3\mathbf{k} - (2\mathbf{i} + 2\mathbf{j}) = -2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\).

For \(\overrightarrow{ON}\), since \(N\) divides \(BC\) in the ratio 2:1, \(\overrightarrow{ON} = \overrightarrow{OB} + \frac{2}{3}(\overrightarrow{OC} - \overrightarrow{OB}) = 4\mathbf{j} + \frac{2}{3}(\mathbf{i} + \mathbf{k} - 4\mathbf{j}) = 3\mathbf{j} + \mathbf{k}\).

(b) The angle \(\theta\) between \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\) is given by \(\cos \theta = \frac{\overrightarrow{MD} \cdot \overrightarrow{ON}}{|\overrightarrow{MD}| |\overrightarrow{ON}|}\).

\(\overrightarrow{MD} \cdot \overrightarrow{ON} = (-2)(0) + (-2)(3) + (3)(1) = -6 + 3 = -3\).

\(|\overrightarrow{MD}| = \sqrt{(-2)^2 + (-2)^2 + 3^2} = \sqrt{17}\), \(|\overrightarrow{ON}| = \sqrt{3^2 + 1^2} = \sqrt{10}\).

\(\cos \theta = \frac{-3}{\sqrt{17} \cdot \sqrt{10}}\).

\(\theta = \cos^{-1}\left(\frac{-3}{\sqrt{170}}\right) \approx 103.3^\circ\).

(c) To find the perpendicular distance from \(M\) to \(ON\), consider the projection of \(\overrightarrow{OM}\) on \(\overrightarrow{ON}\).

\(\text{Projection of } \overrightarrow{OM} \text{ on } \overrightarrow{ON} = \frac{\overrightarrow{OM} \cdot \overrightarrow{ON}}{|\overrightarrow{ON}|^2} \overrightarrow{ON}\).

\(\overrightarrow{OM} \cdot \overrightarrow{ON} = (2\mathbf{i} + 2\mathbf{j}) \cdot (3\mathbf{j} + \mathbf{k}) = 6\).

\(\text{Projection length} = \frac{6}{10} = \frac{3}{5}\).

\(\text{Perpendicular distance} = \sqrt{|\overrightarrow{OM}|^2 - \left(\frac{3}{5}\right)^2 |\overrightarrow{ON}|^2} = \sqrt{8 - \frac{9}{5}} = \sqrt{\frac{22}{5}}\).