Passing through two points

Theorem 6.3 (vector form, two points). Suppose a line $L$ passes through two distinct points $A$ and $B$, with position vectors $\bar{a}$ and $\bar{b}$. Let $P$, with position vector $\bar{r}$, be any other point on $L$. Since $A$, $P$ and $B$ are all on the same straight line, the vectors $\overline{AP}$ and $\overline{AB}$ are collinear (parallel), so $\overline{AP} = \lambda \overline{AB}$ for some scalar $\lambda$. Now $\overline{AB} = \bar{b} - \bar{a}$ and $\overline{AP} = \bar{r} - \bar{a}$, so substituting gives $\bar{r} - \bar{a} = \lambda(\bar{b} - \bar{a})$, i.e. $\bar{r} = \bar{a} + \lambda(\bar{b} - \bar{a}).$ Notice this is really Theorem 6.1 in disguise: once we know two points, the vector $\bar{b} - \bar{a}$ plays exactly the role that the given direction vector $\bar{b}$ played there — it is simply the direction we compute by subtracting the two known position vectors. As before, this can also be written without a parameter as $(\bar{r} - \bar{a}) \times (\bar{b} - \bar{a}) = \bar{0}$.

Theorem 6.4 (Cartesian form, two points). In coordinates, if $L$ passes through $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, and $P(x,y,z)$ is any other point of $L$, then the direction ratios of $AP$ are $x-x_1,\ y-y_1,\ z-z_1$, while the direction ratios of the line itself (found from its two known points) are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$. Since these two sets of direction ratios represent the same direction, they are proportional, giving $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}.$ Exactly as in Theorem 6.2, this is Theorem 6.2 with the direction ratios $a,b,c$ replaced by the differences $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ computed from the two given points.

Worked examples.

Ex.(5). Find the vector equation of the line through $A(1,2,3)$ and $B(2,3,4)$. Writing the position vectors as $\bar{a} = \hat{i}+2\hat{j}+3\hat{k}$ and $\bar{b} = 2\hat{i}+3\hat{j}+4\hat{k}$, we first compute $\bar{b}-\bar{a} = (2\hat{i}+3\hat{j}+4\hat{k}) - (\hat{i}+2\hat{j}+3\hat{k}) = \hat{i}+\hat{j}+\hat{k}$. By Theorem 6.3, the equation of the required line is $\bar{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(\hat{i}+\hat{j}+\hat{k}).$ As a check, substituting $\lambda = 1$ should reproduce B's position vector: $(\hat{i}+2\hat{j}+3\hat{k}) + (\hat{i}+\hat{j}+\hat{k}) = 2\hat{i}+3\hat{j}+4\hat{k}$, which is exactly $\bar{b}$, confirming the equation is correct.

Ex.(7). Find the Cartesian equations of the line through $A(1,2,3)$ and $B(2,3,4)$ — the Cartesian counterpart of Ex.(5). Here $(x_1,y_1,z_1) = (1,2,3)$ and $(x_2,y_2,z_2) = (2,3,4)$, so by Theorem 6.4: $\frac{x-1}{2-1} = \frac{y-2}{3-2} = \frac{z-3}{4-3},\quad\text{i.e.}\quad \frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{1},$ which simplifies to the neat form $x - 1 = y - 2 = z - 3$ (since dividing by $1$ changes nothing). One can again check that B's own coordinates $(2,3,4)$ satisfy this: $2-1=1$, $3-2=1$, $4-3=1$ — all equal, so B indeed lies on the line.

Ex.(9) — angle between two lines. Find the angle between the lines $\bar{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}-2\hat{j}+\hat{k})$ and $\bar{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(\hat{i}+2\hat{j}+2\hat{k})$. The angle between two lines is defined as the angle between any pair of vectors along each line, regardless of where the lines actually sit in space — so only the two direction vectors $\bar{b} = 2\hat{i}-2\hat{j}+\hat{k}$ and $\bar{c} = \hat{i}+2\hat{j}+2\hat{k}$ matter here (the fact that both lines happen to pass through the same point in this particular example is incidental to the method). Using the standard dot-product formula for the angle between two vectors, $\cos\theta = \frac{\bar{b}\cdot\bar{c}}{|\bar{b}||\bar{c}|}.$ The dot product is $\bar{b}\cdot\bar{c} = (2)(1) + (-2)(2) + (1)(2) = 2 - 4 + 2 = 0$. Since the dot product is already zero, the magnitudes do not even need to be computed to know the answer: $\cos\theta = 0$, so $\theta = 90^\circ$, meaning the two given lines are perpendicular to each other. (Carrying the magnitudes through for completeness: $|\bar{b}| = \sqrt{4+4+1} = 3$ and $|\bar{c}| = \sqrt{1+4+4} = 3$, so $\cos\theta = 0/(3\times3) = 0$, the same conclusion.)

Ex.(10) — showing two lines intersect. Show that the lines $\bar{r} = (-\hat{i}-3\hat{j}+4\hat{k}) + \lambda(-10\hat{i}-\hat{j}+\hat{k})$ and $\bar{r} = (-10\hat{i}-\hat{j}+\hat{k}) + \mu(-\hat{i}-3\hat{j}+4\hat{k})$ intersect, and find their point of intersection. Two lines given in this general vector/parametric form intersect exactly when there is some choice of the two independent parameters, $\lambda$ on the first line and $\mu$ on the second, that makes the corresponding points coincide. Writing out the general point of each line in components: on the first line it is $(-1-10\lambda)\hat{i} + (-3-\lambda)\hat{j} + (4+\lambda)\hat{k}$, and on the second it is $(-10-\mu)\hat{i} + (-1-3\mu)\hat{j} + (1+4\mu)\hat{k}$. Matching the three components gives three equations: $-1-10\lambda = -10-\mu$, $-3-\lambda = -1-3\mu$, and $4+\lambda = 1+4\mu$, which simplify to $10\lambda - \mu = 9,\qquad \lambda - 3\mu = -2,\qquad \lambda - 4\mu = -3.$ This is three equations in only two unknowns, so in general there is no solution at all — the lines intersect only if this particular system happens to be consistent, which can be checked by testing whether the coefficient determinant of the three equations (treating them as a $3\times3$ system, using the third column as constants) is zero: $\begin{vmatrix}10&-1&9\\1&-3&-2\\1&-4&-3\end{vmatrix} = 10(9-8) + 1(-3+2) + 9(-4+3) = 10 - 1 - 9 = 0.$ Since this determinant is zero, the system is consistent and the lines do intersect. Solving any two of the three equations (say the first two) gives $\lambda = 1,\ \mu = 1$; the third equation can then be used purely as a check, and it is indeed satisfied. Substituting $\lambda = 1$ back into the first line's general point $(-1-10\lambda)\hat{i}+(-3-\lambda)\hat{j}+(4+\lambda)\hat{k}$ gives $-11\hat{i} - 4\hat{j} + 5\hat{k}$, which is the position vector of the point of intersection. …