The planes $p_1$ and $p_2$ have equations \(\mathbf{r} \cdot \begin{pmatrix} 1 \ 1 \ -1 \end{p

Question

The planes $p_1$ and $p_2$ have equations $\mathbf{r} \cdot \begin{pmatrix} 1 \ 1 \ -1 \end{pmatrix} = 0$ and $\mathbf{r} \cdot \begin{pmatrix} 2 \ -1 \ -2 \end{pmatrix} = 6$ respectively. Find a vector equation of $l$, the line of intersection between $p_1$ and $p_2$.[2] Another line, $l_1$, has equation $\mathbf{r} = \begin{pmatrix} 2 \ 0 \ 8 \end{pmatrix} + \mu \begin{pmatrix} 1 \ 1 \ 5 \end{pmatrix}$, where $\mu$ is a real parameter. Use the result in part (i) to show that $p_1$, $p_2$ and $l_1$ have exactly one common point of intersection. (You are not allowed to use any calculator for this part of the question.)[3]

Accepted Answer

(i) $\mathbf r=\begin{pmatrix}2\-2\0\end{pmatrix}+\lambda\begin{pmatrix}1\0\1\end{pmatrix},\ \lambda\in\mathbb R$ (ii) $(0,-2,-2)$

2015 NJC P2 Q3

2015 NJC P2 Q3

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