Vector equations of linesEdexcel International A Level Maths: Revision notes

A line is fixed by one point on it and a direction. If $A$ has position vector $\mathbf{a}$ and the line is parallel to $\mathbf{b}$, $\mathbf{r}=\mathbf{a}+t\mathbf{b},$ where $t$ is a scalar parameter. Each value of $t$ gives the position vector of one point on the line. $\mathbf{a}$ is a position vector and $\mathbf{b}$ the direction vector; any non-zero multiple of $\mathbf{b}$ gives the same line. Example: $\mathbf{r}=(2\mathbf{i}-\mathbf{j}+3\mathbf{k})+t(3\mathbf{i}+2\mathbf{j}-2\mathbf{k})$. When $t=2$ the point is $(8,3,-1)$.

For points $C$ and $D$ with position vectors $\mathbf{c}$ and $\mathbf{d}$, the direction is $\mathbf{d}-\mathbf{c}$, so $\mathbf{r}=\mathbf{c}+t(\mathbf{d}-\mathbf{c}).$ With $t=0$ you are at $C$ and with $t=1$ at $D$. For $C(1,4,-2)$ and $D(3,3,0)$, $\mathbf{r}=(1,4,-2)+t(2,-1,2)$. To test whether a point lies on a line, equate one coordinate to find $t$, then check the other two. For $E(-3,6,-6)$: $1+2t=-3$ gives $t=-2$, and the $y$ and $z$ values both match, so $E$ is on the line. You can also find where a line meets a coordinate plane by setting that coordinate to zero.

Two lines are parallel if their direction vectors are scalar multiples of each other. $\mathbf{r}=\mathbf{a}+t(2\mathbf{i}-\mathbf{j}+4\mathbf{k})$ is parallel to any line with direction $-4\mathbf{i}+2\mathbf{j}-8\mathbf{k}$, since $-4,2,-8$ are $-2$ times $2,-1,4$. Parallel lines either never meet or are the same line. If they are the same line, a point from one line also lies on the other. Parallel lines never intersect, and they are never skew.

To find whether two lines meet, use different parameters, $s$ and $t$, for the two lines, since the point need not be reached by the same parameter. Equate the $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ components to get three equations in two unknowns. Solve two of them, then check the third. If it holds, the lines intersect and you substitute back to find the point. Example: $\mathbf{r}=(1,1,2)+s(2,1,-1)$ and $\mathbf{r}=(1,4,5)+t(1,-1,-2)$ give $1+2s=1+t$, $1+s=4-t$, $2-s=5-2t$. Then $s=1$, $t=2$, the third equation holds ($1=1$), and the lines meet at $(3,2,1)$.

In three dimensions, two lines that are not parallel may still never meet. They are skew. The test is: the direction vectors are not multiples of each other, and the three component equations are inconsistent (the values of $s$ and $t$ from two equations fail in the third). Example: $\mathbf{r}=(2,1,0)+s(1,2,3)$ and $\mathbf{r}=(-1,0,4)+t(2,-1,1)$. The first two equations give $s=-1$, $t=1$, but the third requires $-3=5$, which is false. So the lines are skew. Summary: parallel directions means parallel (or identical) lines; otherwise a consistent solution means intersecting and an inconsistent one means skew. In a context, skew paths mean the objects never collide.