IBDP Maths AHL 3.12 Concept of a vector AA HL Paper 1- Exam Style Questions- New Syllabus
Question
The line $L_1$ is defined by the Cartesian equation:
$\dfrac{x-1}{2} = \dfrac{y+2}{3} = z$
A second line, $L_2$, is defined by the vector equation:
$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ -8 \end{pmatrix} + t \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \quad t \in \mathbb{R}$
Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):
AHL 3.12: Concept of a vector and position vectors — part (a)
AHL 3.15: Intersecting and skew lines; points of intersection — part (b)
▶️ Answer/Explanation
(a)
Set the Cartesian ratios equal to a parameter $\lambda$:
$\dfrac{x-1}{2} = \lambda \implies x = 2\lambda + 1$
$\dfrac{y+2}{3} = \lambda \implies y = 3\lambda – 2$
$z = \lambda \implies z = \lambda$
Expressing this in vector form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$:
$\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$
Result: $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$
(b)
Equate the parametric components of $L_1$ and $L_2$:
$L_1: x = 1 + 2\lambda, \; y = -2 + 3\lambda, \; z = \lambda$
$L_2: x = t, \; y = 4, \; z = -8 + 2t$
From the $y$-components: $-2 + 3\lambda = 4 \implies 3\lambda = 6 \implies \lambda = 2$.
Substitute $\lambda = 2$ into $L_1$ to find the coordinates:
$x = 1 + 2(2) = 5$
$y = -2 + 3(2) = 4$
$z = 2$
Verification with $L_2$:
Using $x = 5 \implies t = 5$.
Check $z$ for $L_2$: $z = -8 + 2(5) = 2$. (Consistent)
Result: $(5, 4, 2)$