Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
FM June 2021 p13 q06
4272
The lines \(l_1\) and \(l_2\) have equations \(\mathbf{r} = -\mathbf{i} - 2\mathbf{j} + \mathbf{k} + s(2\mathbf{i} - 3\mathbf{j})\) and \(\mathbf{r} = 3\mathbf{i} - 2\mathbf{k} + t(3\mathbf{i} - \mathbf{j} + 3\mathbf{k})\) respectively.
The plane \(\Pi_1\) contains \(l_1\) and the point \(P\) with position vector \(-2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\).
(a) Find an equation of \(\Pi_1\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}\).
(b) The plane \(\Pi_2\) contains \(l_2\) and is parallel to \(l_1\). Find an equation of \(\Pi_2\), giving your answer in the form \(ax + by + cz = d\).
(c) Find the acute angle between \(\Pi_1\) and \(\Pi_2\).
(d) The point \(Q\) is such that \(\overrightarrow{OQ} = -5\overrightarrow{OP}\). Find the position vector of the foot of the perpendicular from the point \(Q\) to \(\Pi_2\).
Solution
(a) The direction vector of \(l_1\) is \(2\mathbf{i} - 3\mathbf{j}\). The vector from \(P\) to a point on \(l_1\) is \(\mathbf{i} - 3\mathbf{k}\). Thus, the equation of \(\Pi_1\) is \(\mathbf{r} = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda (2\mathbf{i} - 3\mathbf{j}) + \mu (\mathbf{i} - 3\mathbf{k})\).
(b) The normal vector to \(\Pi_2\) is \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 0 \\ 3 & -1 & 3 \end{vmatrix} = -9\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}\). Using the point \((3, 0, -2)\) on \(\Pi_2\), the equation is \(9x + 6y - 7z = 41\).
(c) The normal vectors to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 9 \\ 6 \\ -7 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\) respectively. The dot product is \(32\), and the magnitudes are \(\sqrt{166}\) and \(\sqrt{14}\). Thus, \(\cos \alpha = \frac{32}{\sqrt{14 \times 166}}\), giving \(\alpha = 48.4^\circ\) or \(0.845 \text{ rad}\).
(d) \(\overrightarrow{OQ} = -5\overrightarrow{OP} = \begin{pmatrix} 10 \\ 10 \\ -20 \end{pmatrix}\). The equation of \(\Pi_2\) is \(9x + 6y - 7z = 41\). Substituting \(\mathbf{OF} = \begin{pmatrix} 10 + 9t \\ 10 + 6t \\ -20 - 7t \end{pmatrix}\) into the plane equation gives \(t = -\frac{3}{2}\), leading to \(\mathbf{OF} = \left( -\frac{7}{2}, 1, -\frac{19}{2} \right)\).
