Intersection of lines and distancesAQA A-Level Further Maths: Revision notes

Write each line in components using a different parameter (say $\lambda$ and $\mu$) and equate the position vectors. Three equations arise, one for each coordinate:

1. Solve any two of them for $\lambda$ and $\mu$.
2. Check the values in the third equation.
3. If it works, the lines intersect: substitute back for the point. If it fails, they do not intersect. Example: $\mathbf r=(1,2,3)+\lambda(1,-1,2)$ and $\mathbf r=(1,-1,4)+\mu(2,1,3)$. Then $1+\lambda=1+2\mu$ and $2-\lambda=-1+\mu$ give $\lambda=2$, $\mu=1$. For $z$: $3+4=7$ and $4+3=7$, so they meet at $(3,0,7)$.

Two lines in 3D can be:

• parallel: direction vectors are multiples of each other (the same line, or distinct);
• intersecting: not parallel, and the three equations are consistent;
• skew: not parallel and no common point. This is possible only in 3D. Check for parallel first (look at the directions), then try to solve. For $\mathbf r=(1,2,3)+\lambda(1,-1,2)$ and $\mathbf r=(1,-1,5)+\mu(2,1,3)$ the $x$ and $y$ equations again give $\lambda=2$, $\mu=1$, but now $z=7$ on one line and $8$ on the other, so the lines are skew.

The shortest distance from a point $P$ to a line is the length of the perpendicular from $P$ to the line. Let $F$ be the foot of the perpendicular.

1. Write $F$ in terms of the parameter.
2. Form $\overrightarrow{PF}$.
3. Set $\overrightarrow{PF}\cdot\mathbf d=0$ and solve for the parameter.
4. The distance is $|\overrightarrow{PF}|$. Example: $P(4,3,2)$ and $\mathbf r=(1,0,2)+\lambda(1,2,2)$. $F=(1+\lambda,2\lambda,2+2\lambda)$, so $\overrightarrow{PF}=(\lambda-3,2\lambda-3,2\lambda)$. Then $(\lambda-3)+2(2\lambda-3)+2(2\lambda)=9\lambda-9=0$ gives $\lambda=1$, $F=(2,2,4)$ and the distance is $|(-2,-1,2)|=3$. The reflection of $P$ in the line is $2\overrightarrow{OF}-\overrightarrow{OP}$.

For skew lines, take a general point $P$ on the first line (parameter $\lambda$) and $Q$ on the second ($\mu$). The shortest distance is along the common perpendicular, so $\overrightarrow{PQ}\cdot\mathbf d_1=0\quad\text{and}\quad\overrightarrow{PQ}\cdot\mathbf d_2=0.$ Solve these simultaneous equations for $\lambda$ and $\mu$, then find $|\overrightarrow{PQ}|$. Example: $\mathbf r=\lambda(1,2,2)$ and $\mathbf r=(1,-1,2)+\mu(2,1,-2)$ give $3-9\lambda=0$ and $-3+9\mu=0$, so $\lambda=\mu=\frac13$, $\overrightarrow{PQ}=(\frac43,-\frac43,\frac23)$ and the distance is $2$. The shortest distance between two parallel lines is found as for a point and a line.