IB Maths AA HL 3.15 Line relationships in 3D Question Bank

METHOD 1
point on line $L_{1}$ has coordinates $(1+4 \lambda, 2-2 \lambda, 3-\lambda)$
attempt to use a different parameter for $L_{2}$$\frac{x-1}{2}=\frac{y}{3}=1-z=\mu$ or $r=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}2 \\ 3 \\ -1\end{array}\right)$
point on line $L_{2}$ has coordinates $(1+2 \mu, 3 \mu, 1-\mu)$
Note: This A1 may be implied by $\boldsymbol{r}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}2 \\ 3 \\ -1\end{array}\right)$.
$1+4 \lambda=1+2 \mu$$2-2 \lambda=3 \mu$$3-\lambda=1-\mu$
any two of the above equations
attempt to solve two simultaneous equations with two parameters
eg $\lambda=0.25, \mu=0.5$ or $\lambda=1.6, \mu=-0.4$ or $\lambda=-2, \mu=-4$
substitute into third equation or solve a different pair of simultaneous equations
obtain contradiction eg $3-0.25 \neq 1-0.5$ or $1+4(1.6) \neq 1+2(-0.4)$ or $2-2(-2) \neq 3(-4)$ (so the lines do not intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skewMETHOD 2
point on line $L_{1}$ has coordinates $(1+4 \lambda, 2-2 \lambda, 3-\lambda)$
attempt to use the equation of $L_{2}$ to generate at least two equations in $\lambda$
if the two lines intersect,
$\frac{(1+4 \lambda)-1}{2}=\frac{2-2 \lambda}{3}\left(\Rightarrow 2 \lambda=\frac{2-2 \lambda}{3}\right)$$\frac{(1+4 \lambda)-1}{2}=1-(3-\lambda)(\Rightarrow 2 \lambda=\lambda-2)$$\frac{2-2 \lambda}{3}=1-(3-\lambda) \Rightarrow\left(\frac{2-2 \lambda}{3}=\lambda-2\right)$
any two of the above equations
attempt to solve at least one equation in $\lambda$
one of $\lambda=\frac{1}{4}, \lambda=-2, \lambda=\frac{8}{5}$ seen
substitute into second equation or solve second equation
obtain contradiction eg $\lambda=\frac{1}{4} \neq-2$ or $2\left(\frac{1}{4}\right) \neq \frac{1}{4}-2$ (so the lines do not
intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skewMETHOD 3
attempt to use a find Cartesian equation for $L_{1}$$\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-3}{-1}$
attempt to isolate one variable in both equations
$L_{1}: z=\frac{1-x}{4}+3=\frac{y-2}{2}+3 \quad L_{2}: z=\frac{1-x}{2}+1=\frac{-y}{3}+1 \quad$ OR
$L_{1}: y=\frac{1-x}{2}+2=2(z-3)+2 \quad L_{2}: y=\frac{3(x-1)}{2}=3(1-z) \quad$ OR
$L_{1}: x=1-2(y-2)=1-4(z-3) \quad L_{2}: x=\frac{2 y}{3}+1=1-2(z-1)$
attempt to solve for each of the other two variables
e.g. $\frac{1-x}{2}+1=\frac{1-x}{4}+3$ and $\frac{-y}{3}+1=\frac{y-2}{2}+3$
x=-7, y=-1.2 OR x=2, z=1.4 OR y=1.5, z=5
obtain contradiction eg $z=5 \neq 1.4$ OR $y=1.5 \neq-1.2$ OR $x=2 \neq-7$
(so the lines do not intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skew