Intersections and distances in three dimensionsEdexcel A-Level Further Maths: Revision notes

Write the line as $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ and the plane as $n_1x+n_2y+n_3z=k$ (or $\mathbf{r}\cdot\mathbf{n}=k$). Substitute the general point $(a_1+td_1,\,a_2+td_2,\,a_3+td_3)$ into the plane equation to get a linear equation in $t$.

• If $\mathbf{d}\cdot\mathbf{n}\neq0$ there is exactly one intersection: solve for $t$, then substitute back for the point.
• If $\mathbf{d}\cdot\mathbf{n}=0$ the line is parallel to the plane. Then either $\mathbf{a}\cdot\mathbf{n}=k$ (the line lies in the plane) or $\mathbf{a}\cdot\mathbf{n}\neq k$ (no intersection). Example: $\mathbf{r}=(1,0,2)+t(1,2,-1)$ and $2x-y+3z=14$ give $2(1+t)-2t+3(2-t)=8-3t=14$, so $t=-2$ and the point is $(-1,-4,4)$.

The perpendicular distance from $(\alpha,\beta,\gamma)$ to the plane $n_1x+n_2y+n_3z+d=0$ is $\frac{|n_1\alpha+n_2\beta+n_3\gamma+d|}{\sqrt{n_1^2+n_2^2+n_3^2}}.$ Rearrange the plane so that everything is on one side first: $2x-2y+z=7$ becomes $2x-2y+z-7=0$, so $d=-7$. For $A(3,-1,4)$ the numerator is $|6+2+4-7|=5$ and the denominator $\sqrt{4+4+1}=3$, giving $\frac53$. Why it works: the shortest route from the point to the plane runs along the normal $\mathbf{n}$, so the distance is the length of the projection of (point $-$ any point on the plane) onto $\hat{\mathbf{n}}$.

The foot of the perpendicular from $A$ to a plane lies on the line through $A$ with direction $\mathbf{n}$: $\mathbf{r}=\mathbf{a}+\mu\mathbf{n}$. Substitute into the plane to find $\mu$ (this is the line–plane intersection again). For $A(3,-1,4)$ and $2x-2y+z=7$: $2(3+2\mu)-2(-1-2\mu)+(4+\mu)=12+9\mu=7$, so $\mu=-\frac59$ and the foot is $\left(\frac{17}{9},\frac19,\frac{31}{9}\right)$. The reflection $A'$ of $A$ in the plane is as far again on the other side: $\overrightarrow{OA'}=2\overrightarrow{OF}-\overrightarrow{OA}$.

Let the line be $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ and the point be $P$. Two methods:

1. Foot of the perpendicular. Take a general point $F=\mathbf{a}+\lambda\mathbf{d}$, form $\overrightarrow{PF}$ and solve $\overrightarrow{PF}\cdot\mathbf{d}=0$. The distance is $|\overrightarrow{PF}|$.
2. Cross product. $\text{distance}=\frac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}$, because $|\overrightarrow{AP}\times\mathbf{d}|=|\overrightarrow{AP}||\mathbf{d}|\sin\theta$. Example: $l:\ \mathbf{r}=(1,2,0)+\lambda(2,1,-2)$ and $P(4,2,0)$. $\overrightarrow{PF}\cdot\mathbf{d}=9\lambda-6=0$ gives $\lambda=\frac23$, $\overrightarrow{PF}=\left(-\frac53,\frac23,-\frac43\right)$ and the distance is $\sqrt5$.

Parallel lines: take any point on one line and use the point-to-line distance to the other. Skew lines $\mathbf{r}=\mathbf{a}_1+s\mathbf{d}_1$ and $\mathbf{r}=\mathbf{a}_2+t\mathbf{d}_2$: they do not meet and are not parallel. The shortest distance is along their common perpendicular, direction $\mathbf{n}=\mathbf{d}_1\times\mathbf{d}_2$: $\text{distance}=\frac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot\mathbf{n}|}{|\mathbf{n}|}.$ Example: $\mathbf{d}_1=(1,1,0)$, $\mathbf{d}_2=(0,1,1)$ give $\mathbf{n}=(1,-1,1)$. With $\mathbf{a}_2-\mathbf{a}_1=(-1,-2,3)$ the distance is $\frac{4}{\sqrt3}$. To show lines are skew, equate two components, substitute into the third and find a contradiction, then note the directions are not parallel.

• Line and plane meet? Substitute the line into the plane.
• Point to plane? $\frac{|n_1\alpha+n_2\beta+n_3\gamma+d|}{\sqrt{n_1^2+n_2^2+n_3^2}}$.
• Point to line? Foot of the perpendicular ($\overrightarrow{PF}\cdot\mathbf{d}=0$) or cross product.
• Line to line? Parallel: point to line. Skew: $\frac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}$. Leave answers in exact surd form unless a decimal is asked for.