Question
Line \( L \) is given by the vector equation \( \mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} + s \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} \), where \( s \in \mathbb{R} \).
Line \( M \) is given by the vector equation \( \mathbf{r}_2 = \begin{pmatrix} 9 \\ 9 \\ 11 \end{pmatrix} + t \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} \), where \( t \in \mathbb{R} \).
(a) Show that lines \( L \) and \( M \) intersect at a point \( A \) and find the position vector of \( A \). [5]
(b) Verify that the lines \( L \) and \( M \) both lie in the plane \( \Pi \) given by \( \mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 7 \). [3]
(c) (i) Find the position vector of \( C \), where a line through \( B(-3, 12, 2) \) perpendicular to \( \Pi \) intersects \( \Pi \) at point \( C \). [4]
(ii) Hence, find \( |\overrightarrow{BC}| \). [3]
(d) Find the reflection of the point \( B \) in the plane \( \Pi \). [3]
▶️Answer/Explanation
(a) To find the intersection point \( A \), set \( \mathbf{r}_1 = \mathbf{r}_2 \): \[ \begin{pmatrix} 1 + 2s \\ 2 + 3s \\ -3 + 6s \end{pmatrix} = \begin{pmatrix} 9 + 4t \\ 9 + t \\ 11 + 2t \end{pmatrix} \] Solving the system of equations: \[ 1 + 2s = 9 + 4t \quad (1) \\ 2 + 3s = 9 + t \quad (2) \\ -3 + 6s = 11 + 2t \quad (3) \] From equation (2): \[ t = 3s – 7 \] Substitute \( t = 3s – 7 \) into equation (1): \[ 1 + 2s = 9 + 4(3s – 7) \implies 1 + 2s = 9 + 12s – 28 \implies -10s = -20 \implies s = 2 \] Substitute \( s = 2 \) into \( t = 3s – 7 \): \[ t = 3(2) – 7 = -1 \] Verify with equation (3): \[ -3 + 6(2) = 11 + 2(-1) \implies 9 = 9 \] Thus, the lines intersect at \( A \), and the position vector of \( A \) is: \[ \mathbf{r}_A = \begin{pmatrix} 1 + 2(2) \\ 2 + 3(2) \\ -3 + 6(2) \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \\ 9 \end{pmatrix} \]
(b) To verify that \( L \) and \( M \) lie in the plane \( \Pi \), substitute the parametric equations of \( L \) and \( M \) into the plane equation \( \mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 7 \). For line \( L \): \[ \begin{pmatrix} 1 + 2s \\ 2 + 3s \\ -3 + 6s \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(2 + 3s) – 1(-3 + 6s) = 4 + 6s + 3 – 6s = 7 \] For line \( M \): \[ \begin{pmatrix} 9 + 4t \\ 9 + t \\ 11 + 2t \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(9 + t) – 1(11 + 2t) = 18 + 2t – 11 – 2t = 7 \] Thus, both lines lie in the plane \( \Pi \).
(c)(i) The line through \( B(-3, 12, 2) \) perpendicular to \( \Pi \) has direction vector \( \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \). Its parametric equation is: \[ \mathbf{r} = \begin{pmatrix} -3 \\ 12 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \] Substitute into the plane equation \( \mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 7 \): \[ \begin{pmatrix} -3 \\ 12 + 2\lambda \\ 2 – \lambda \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(12 + 2\lambda) – 1(2 – \lambda) = 24 + 4\lambda – 2 + \lambda = 22 + 5\lambda = 7 \] Solving for \( \lambda \): \[ 22 + 5\lambda = 7 \implies 5\lambda = -15 \implies \lambda = -3 \] Thus, the position vector of \( C \) is: \[ \mathbf{r}_C = \begin{pmatrix} -3 \\ 12 + 2(-3) \\ 2 – (-3) \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 5 \end{pmatrix} \]
(c)(ii) The distance \( |\overrightarrow{BC}| \) is: \[ |\overrightarrow{BC}| = \sqrt{(-3 – (-3))^2 + (6 – 12)^2 + (5 – 2)^2} = \sqrt{0 + 36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71 \]
(d) The reflection of \( B \) in the plane \( \Pi \) is found by extending the line through \( B \) and \( C \) to \( B’ \), such that \( C \) is the midpoint of \( B \) and \( B’ \): \[ \mathbf{r}_{B’} = \mathbf{r}_C + (\mathbf{r}_C – \mathbf{r}_B) = \begin{pmatrix} -3 \\ 6 \\ 5 \end{pmatrix} + \begin{pmatrix} 0 \\ -6 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 8 \end{pmatrix} \] Thus, the reflection of \( B \) is \( B'(-3, 0, 8) \).