2022 TMJC P1 Q8
2022 TMJC P1 Q8
11 marks
Prelims
The plane \({{p}_{1}}\) has cartesian equation \(x+z=3\). The plane \({{p}_{2}}\) is perpendicular to \({{p}_{1}}\) and contains the line \({{l}_{1}}\) with equation \(\frac{x+2}{5}=\frac{y+1}{2}=\frac{3-z}{3}\).
- Show that the cartesian equation of the plane \({{p}_{2}}\) is \(-x+4y+z=1\).[2]
- Find a vector equation of the line \({{l}_{2}}\) given that \({{p}_{1}}\) and \({{p}_{2}}\) intersect at \({{l}_{2}}\).[2]
- It is given that the point \(B\) with coordinates \(\left( 0,4,3 \right)\) is on \({{p}_{1}}\) and the perpendicular distance from \(B\) to \({{p}_{2}}\) is \(k\). Find the position vector of the foot of perpendicular from \(B\) to \({{p}_{2}}\) and deduce the value of \(k\).[4]
- Hence, find the vector equations of the lines in \({{p}_{1}}\) such that the perpendicular distance from each line to \({{p}_{2}}\) is \(k\).[3]
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Answer:(ii) \({\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}, \; \alpha \in \mathbb{R}}\) (iii) \({\overrightarrow{OF} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} ; \; k = 3\sqrt{2}}\) (iv) \({\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix} + \beta \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}, \; \beta \in \mathbb{R}}\) OR \({\mathbf{r} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} + \gamma \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}, \; \gamma \in \mathbb{R}}\)
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