IBDP Maths AHL 3.15- Different types of lines lines AA HL Paper 1- Exam Style Questions- New Syllabus
Question
Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):
• AHL 3.14: Vector equation of a line in three dimensions — vector equations of \( L_1 \) and \( L_2 \)
• AHL 3.15: Intersecting lines; points of intersection — solving for \( \lambda \), \( \mu \), and \( b \)
▶️ Answer/Explanation
Step 1 – Condition for perpendicular lines:
Direction vectors are:
\( \mathbf{d}_1 = \begin{pmatrix} 0 \\ a \\ 1 \end{pmatrix} \),
\( \mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \).
Since \( L_1 \) and \( L_2 \) are perpendicular, \( \mathbf{d}_1 \cdot \mathbf{d}_2 = 0 \):
\( 0(1) + a(2) + 1(3) = 2a + 3 = 0 \)
\( a = -\frac{3}{2} \)
Step 2 – Equating position vectors at intersection:
Set \( \mathbf{r} \) from \( L_1 \) equal to \( \mathbf{r} \) from \( L_2 \) for some \( \lambda \) and \( \mu \):
\( \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ a \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ -b \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \)
Substitute \( a = -\frac{3}{2} \):
\( \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ -\frac{3}{2} \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ -b \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \)
Step 3 – Solving for \( \mu \), \( \lambda \), and \( b \):
From \( x \)-components: \( 4 = 1 + \mu \Rightarrow \mu = 3 \).
From \( y \)-components: \( -\frac{3}{2}\lambda = 6 \Rightarrow \lambda = -4 \).
From \( z \)-components: \( -1 + \lambda = -b + 3\mu \Rightarrow -5 = -b + 9 \Rightarrow b = 14 \).
Final values:
\( \boxed{a = -\frac{3}{2},\quad b = 14} \)