Applications of vectors to 3-D geometryEdexcel A-Level Further Maths: Revision notes

A line through the point with position vector $\mathbf{a}$ with direction $\mathbf{b}$ has vector equation $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}.$ A line through two points $A$ and $B$ has direction $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$. Writing $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ and eliminating $\lambda$ gives the Cartesian equations $\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}.$ For $\mathbf{r}=(\mathbf{i}-2\mathbf{j}+3\mathbf{k})+\lambda(2\mathbf{i}+\mathbf{j}-2\mathbf{k})$ the Cartesian form is $\frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{-2}$. To test whether a point is on the line, find $\lambda$ from one coordinate and check it gives the other two: $(5,0,-1)$ has $\lambda=2$ in all three.

The vector $\mathbf{r}-\mathbf{a}$ joins the fixed point $A$ to a general point on the line, and it is parallel to $\mathbf{b}$. Two vectors are parallel exactly when their vector product is zero, so $(\mathbf{r}-\mathbf{a})\times\mathbf{b}=\mathbf{0}$ is another equation of the same line. To change from $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$, subtract $\mathbf{a}$ and take the vector product with $\mathbf{b}$ on both sides, using $\mathbf{b}\times\mathbf{b}=\mathbf{0}$. For the line above: $\left(\mathbf{r}-(\mathbf{i}-2\mathbf{j}+3\mathbf{k})\right)\times(2\mathbf{i}+\mathbf{j}-2\mathbf{k})=\mathbf{0}$.

The components $l:m:n$ of any vector along a line are direction ratios: the line with direction $2\mathbf{i}+\mathbf{j}-2\mathbf{k}$ has ratios $2:1:-2$, and so does $4\mathbf{i}+2\mathbf{j}-4\mathbf{k}$. The direction cosines are the cosines of the angles $\alpha,\beta,\gamma$ the line makes with the $x$-, $y$- and $z$-axes: $\cos\alpha=\frac{l}{\sqrt{l^2+m^2+n^2}},\ \cos\beta=\frac{m}{\sqrt{l^2+m^2+n^2}},\ \cos\gamma=\frac{n}{\sqrt{l^2+m^2+n^2}}.$ They are the components of the unit vector along the line, so $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. For ratios $2:1:-2$ the magnitude is $3$, so the direction cosines are $\frac23,\frac13,-\frac23$. For $3:-4:12$ the magnitude is $13$, so the line makes an angle $\cos^{-1}\frac{12}{13}=22.6^\circ$ with the $z$-axis.

A plane through the point with position vector $\mathbf{a}$ and perpendicular to a normal $\mathbf{n}$ has equation $\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}.$ Writing $\mathbf{n}=n_1\mathbf{i}+n_2\mathbf{j}+n_3\mathbf{k}$ gives the Cartesian form $n_1x+n_2y+n_3z=d$, with $d=\mathbf{a}\cdot\mathbf{n}$. Example: through $A(1,2,3)$ with $\mathbf{n}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}$: $d=2-2+6=6$, so $2x-y+2z=6$. A plane can also be written $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ for two non-parallel directions $\mathbf{b}$ and $\mathbf{c}$ in the plane. The normal is then $\mathbf{b}\times\mathbf{c}$.

To find where a line meets a plane, write a general point of the line in terms of $\lambda$, substitute it into the plane's equation, solve for $\lambda$, and substitute back. If the line has direction $\mathbf{d}$ and the plane has normal $\mathbf{n}$, the acute angle $\theta$ between the line and the plane satisfies $\sin\theta=\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|},$ because $\mathbf{d}\cdot\mathbf{n}$ gives the angle with the normal, which is $90^\circ-\theta$. Example: $\mathbf{d}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$, $\mathbf{n}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}$: $\sin\theta=\frac{2}{3\sqrt6}$, so $\theta=15.8^\circ$. The angle between two planes is the angle between their normals: $\cos\phi=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$.

The distance from the point $(x_1,y_1,z_1)$ to the plane $n_1x+n_2y+n_3z=d$ is $\frac{|n_1x_1+n_2y_1+n_3z_1-d|}{\sqrt{n_1^2+n_2^2+n_3^2}}.$ For $(4,3,-1)$ and $2x-y+2z=6$: $\frac{|8-3-2-6|}{3}=1$. To find the foot of the perpendicular, draw the line through the point in the direction $\mathbf{n}$ and find where it meets the plane. The distance from a point $P$ to a line through $A$ with direction $\mathbf{d}$ is $\frac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}.$ For $A(1,-2,0)$, $\mathbf{d}=3\mathbf{i}-4\mathbf{j}+12\mathbf{k}$ and $P(5,1,0)$: $\overrightarrow{AP}\times\mathbf{d}=36\mathbf{i}-48\mathbf{j}-25\mathbf{k}$, which has magnitude $65$, so the distance is $\frac{65}{13}=5$.