21M.1.AHL.TZ2.8

pestleMathematicsAIHLPaper 121M· ahl-3-11-vector-equation-of-a-line-in-2d-and-3dsource ↗

Two lines $L_{1}$ and $L_{2}$ are given by the following equations, where $p \in ℝ$ .

$L_{1} : r = (\begin{matrix} 2 \\ p + 9 \\ - 3 \end{matrix}) + λ (\begin{matrix} p \\ 2 p \\ 4 \end{matrix})$

$L_{2} : r = (\begin{matrix} 14 \\ 7 \\ p + 12 \end{matrix}) + μ (\begin{matrix} p + 4 \\ 4 \\ - 7 \end{matrix})$

It is known that $L_{1}$ and $L_{2}$ are perpendicular.

Find the possible value(s) for $p$ .

[3]

In the case that $p < 0$ , determine whether the lines intersect.

[4]

Markscheme / solution

setting a dot product of the direction vectors equal to zero (M1)

$(\begin{matrix} p \\ 2 p \\ 4 \end{matrix}) \cdot (\begin{matrix} p + 4 \\ 4 \\ - 7 \end{matrix}) = 0$

$p (p + 4) + 8 p - 28 = 0$ (A1)

$p^{2} + 12 p - 28 = 0$

$(p + 14) (p - 2) = 0$

$p = - 14, p = 2$ A1

[3 marks]

$p = - 14 \Rightarrow$

$L_{1} : r = (\begin{matrix} 2 \\ - 5 \\ - 3 \end{matrix}) + λ (\begin{matrix} - 14 \\ - 28 \\ 4 \end{matrix})$

$L_{2} : r = (\begin{matrix} 14 \\ 7 \\ - 2 \end{matrix}) + μ (\begin{matrix} - 10 \\ 4 \\ - 7 \end{matrix})$

a common point would satisfy the equations

$2 - 14 λ = 14 - 10 μ$

$- 5 - 28 λ = 7 + 4 μ$ (M1)

$- 3 + 4 λ = - 2 - 7 μ$

METHOD 1

solving the first two equations simultaneously

$λ = - \frac{1}{2}, μ = \frac{1}{2}$ A1

substitute into the third equation: M1

$- 3 + 4 (- \frac{1}{2}) \neq - 2 + \frac{1}{2} (- 7)$

so lines do not intersect. R1

Note: Accept equivalent methods based on the order in which the equations are considered.

METHOD 2

attempting to solve the equations using a GDC M1

GDC indicates no solution A1

so lines do not intersect R1

[4 marks]

Examiners’ report

[N/A]