Two lines L₁ and L₂ are given by the following equations, where p ∈ ℝ. L₁: r = (2, p+9, -3) + λ(p, 2p, 4)

setting a dot product of the direction vectors equal to zero $\left(\begin{array}{c}p \\ 2p \\ 4\end{array}\right) \cdot\left(\begin{array}{c}p+4 \\ 4 \\ -7\end{array}\right)=0$ $p(p+4)+8p-28=0$ $p^{2}+12p-28=0$ $(p+14)(p-2)=0$ $p=-14, p=2$

$p=-14 \Rightarrow$ $L_{1}: r=\left(\begin{array}{c}2 \\ -5 \\ -3\end{array}\right)+\lambda\left(\begin{array}{c}-14 \\ -28 \\ 4\end{array}\right)$ $L_{2}: r=\left(\begin{array}{c}14 \\ 7 \\ -2\end{array}\right)+\mu\left(\begin{array}{c}-10 \\ 4 \\ -7\end{array}\right)$ a common point would satisfy the equations

$\begin{aligned} & 2-14 \lambda=14-10 \mu \\ & -5-28 \lambda=7+4 \mu \\ & -3+4 \lambda=-2-7 \mu \end{aligned}$

METHOD 1

solving the first two equations simultaneously $\lambda=-\frac{1}{2}, \mu=\frac{1}{2}$ substitute into the third equation: $-3+4\left(-\frac{1}{2}\right) \neq-2+\frac{1}{2}(-7)$ so lines do not intersect. Note: Accept equivalent methods based on the order in which the equations are considered.

METHOD 2

attempting to solve the equations using a GDC