Lines and planes in three dimensionsEdexcel International A Level Further Maths: Revision notes

The line through the point with position vector $\mathbf{a}$ in the direction $\mathbf{b}$ has vector equation $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$. The same line can be written $(\mathbf{r}-\mathbf{a})\times\mathbf{b}=\mathbf{0}$, because $\mathbf{r}-\mathbf{a}$ is parallel to $\mathbf{b}$ exactly when their vector product is zero. Eliminating $\lambda$ gives the cartesian form $\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}$. Example: $(\mathbf{r}-(\mathbf{i}+2\mathbf{j}-\mathbf{k}))\times(2\mathbf{i}-\mathbf{j}+3\mathbf{k})=\mathbf{0}$ is the line through $(1,2,-1)$ with direction $(2,-1,3)$, that is $\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{3}$. It meets $y=0$ where $\lambda=2$, at $(5,0,5)$.

A plane through the point $\mathbf{a}$ with normal $\mathbf{n}$ has equation $\mathbf{r}\cdot\mathbf{n}=p$, where $p=\mathbf{a}\cdot\mathbf{n}$. With $\mathbf{n}=\alpha\mathbf{i}+\beta\mathbf{j}+\gamma\mathbf{k}$, the cartesian form is $\alpha x+\beta y+\gamma z=p$. Example: $\mathbf{r}\cdot(2\mathbf{i}-\mathbf{j}+2\mathbf{k})=6$ is $2x-y+2z=6$. The parametric (vector) form is $\mathbf{r}=\mathbf{a}+s\mathbf{b}+t\mathbf{c}$, where $\mathbf{b}$ and $\mathbf{c}$ are two non-parallel vectors in the plane. To convert, take $\mathbf{n}=\mathbf{b}\times\mathbf{c}$ and $p=\mathbf{a}\cdot\mathbf{n}$. In the other direction, choose a point on the plane and two non-parallel vectors perpendicular to $\mathbf{n}$.

The shortest distance from the point $(x_1,y_1,z_1)$ to the plane $\alpha x+\beta y+\gamma z=p$ is $d=\frac{|\alpha x_1+\beta y_1+\gamma z_1-p|}{\sqrt{\alpha^2+\beta^2+\gamma^2}}.$ It is measured along the normal. For $A(4,1,3)$ and $2x-y+2z=6$: $2(4)-1+2(3)=13$, so $d=\frac{|13-6|}{3}=\frac73$. For the vector form $\mathbf{r}\cdot\mathbf{n}=p$ the same result is $\frac{|\mathbf{a}\cdot\mathbf{n}-p|}{|\mathbf{n}|}$.

Two non-parallel planes meet in a line. That line lies in both planes, so its direction is perpendicular to both normals: $\mathbf{d}=\mathbf{n}_1\times\mathbf{n}_2$. Find a point on both planes by setting one coordinate (often $z=0$) and solving the two remaining equations. Then write $\mathbf{r}=\mathbf{p}+\lambda\mathbf{d}$. Example: $x+2y-z=4$ and $2x-y+3z=3$. $\mathbf{n}_1\times\mathbf{n}_2=5\mathbf{i}-5\mathbf{j}-5\mathbf{k}$. With $z=0$, $x=2$, $y=1$. The line is $\mathbf{r}=2\mathbf{i}+\mathbf{j}+\lambda(\mathbf{i}-\mathbf{j}-\mathbf{k})$.

Skew lines are not parallel and do not meet. For $\mathbf{r}=\mathbf{a}_1+\lambda\mathbf{b}_1$ and $\mathbf{r}=\mathbf{a}_2+\mu\mathbf{b}_2$, the common perpendicular has direction $\mathbf{b}_1\times\mathbf{b}_2$ and $d=\frac{\left|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)\right|}{\left|\mathbf{b}_1\times\mathbf{b}_2\right|}.$ This is also the distance from a point on one line to the plane containing the other line and parallel to the first. To show lines are skew: equate two components to solve for $\lambda$ and $\mu$, show the third component fails, and check the directions are not parallel. Example: $(1,0,2)+\lambda(1,2,2)$ and $(2,-1,4)+\mu(2,1,-2)$ have $\mathbf{b}_1\times\mathbf{b}_2=(-6,6,-3)$, $|\mathbf{b}_1\times\mathbf{b}_2|=9$, $(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)=-18$, so $d=2$.