Collinearity of three points in 3D

Three points $A, B, C$ in space are collinear if they lie on a common straight line. In 2D we tested this by checking equal slopes. In 3D we have three coordinates, so the test involves ratios of differences in each.

Tests for collinearity

Test 1: Distance test. $A, B, C$ are collinear iff one of the three distances equals the sum of the other two: $|AC| = |AB| + |BC| \quad \text{(if } B \text{ lies between } A \text{ and } C\text{).}$

This is reliable but requires computing three square roots.

Test 2: Section-ratio test. $A, B, C$ are collinear iff there exists $\lambda$ such that $B = \lambda A + (1 - \lambda) C, \text{ i.e., } B \text{ divides } AC \text{ in ratio } (1 - \lambda) : \lambda.$

Equivalently, $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel , proportional in all three coordinates.

Test 3: Direction-ratio test (cleanest). Compute $\frac{x_B - x_A}{x_C - x_A} = \frac{y_B - y_A}{y_C - y_A} = \frac{z_B - z_A}{z_C - z_A}.$

If all three ratios are equal (with the same value $\lambda$), then $B = A + \lambda(C - A)$, which is the equation of $B$ on segment $AC$. So $A, B, C$ are collinear.

(If any denominator is zero, the corresponding numerator must also be zero , equivalent to $A, B, C$ all sharing that coordinate.)

Parametric line preview

Although Class XII handles lines in space rigorously, here is a preview. A line through $A(x_1, y_1, z_1)$ in the direction $(\alpha, \beta, \gamma)$ has parametric equations $x = x_1 + \alpha t, \quad y = y_1 + \beta t, \quad z = z_1 + \gamma t,$ for $t \in \mathbb{R}$. Eliminating $t$: $\frac{x - x_1}{\alpha} = \frac{y - y_1}{\beta} = \frac{z - z_1}{\gamma}.$

Three points $A, B, C$ are collinear iff there is a single common value of $t$ that produces $B$ and $C$ from $A$ , i.e., the direction-ratio test.

Worked examples

Example 1. Are $(1, 2, 3), (2, 4, 5), (3, 6, 7)$ collinear?

Differences from first: $B - A = (1, 2, 2)$, $C - A = (2, 4, 4) = 2(B - A)$. Yes, collinear. $C$ is at parameter $t = 2$ if $B$ is at $t = 1$.

Example 2. Are $(0, 0, 0), (1, 1, 1), (2, 1, 2)$ collinear?

$B - A = (1, 1, 1)$, $C - A = (2, 1, 2)$. Ratios: $2/1 = 2, 1/1 = 1, 2/1 = 2$. Not all equal. Not collinear.

Example 3. Find $k$ such that $(1, -1, 2), (3, 4, -2), (k, 6, -6)$ are collinear.

$B - A = (2, 5, -4)$. $C - A = (k - 1, 7, -8)$. Ratios: $(k-1)/2, 7/5, -8/(-4) = 2$. Need all equal: $7/5 \ne 2$. So no $k$ works? Let me re-examine. For collinearity we need $7/5 = 2$, which is false. So the second and third coordinates are not in the right ratio , these three points cannot be collinear for any $k$.

Adjust: $(1, -1, 2), (3, 4, 0), (k, 9, -2)$. $B - A = (2, 5, -2)$. $C - A = (k - 1, 10, -4) = 2(B - A) \Rightarrow k - 1 = 4 \Rightarrow k = 5$.

Example 4. Show that the points $A(2, -1, 1), B(1, -3, -5), C(3, -4, -4)$ form a right triangle, not a collinear triple.

$AB^2 = 1 + 4 + 36 = 41$. $AC^2 = 1 + 9 + 25 = 35$. $BC^2 = 4 + 1 + 1 = 6$. $AC^2 + BC^2 = 41 = AB^2$. Right angle at $C$.

Example 5. Find a point on the $z$-axis collinear with $(1, 0, 0)$ and $(2, 0, 4)$.

A point on the $z$-axis is $(0, 0, t)$. For collinearity, $(0, 0, t)$ should lie on the line through $(1, 0, 0)$ and $(2, 0, 4)$. Direction: $(1, 0, 4)$. Parametrise: $(1 + s, 0, 4 s)$. Setting equal to $(0, 0, t)$: $s = -1, t = -4$. So $(0, 0, -4)$.

Try it yourself

1. Are $(1, 1, 1), (2, 3, 5), (3, 5, 9)$ collinear?
2. Are $(0, 0, 0), (1, 2, 3), (2, 4, 6)$ collinear?
3. Find $\lambda$ so that $(2, -3, 4), (3, 1, 8), (\lambda, 5, 12)$ are collinear.
4. Find $\mu$ so that $(\mu, 2, 3), (3, 4, 5), (4, 5, 6)$ are collinear.
5. Show that $(0, 7, 10), (-1, 6, 6), (-4, 9, 6)$ form a right-angled isosceles triangle.
6. Find direction ratios of the line through $(1, 2, 3)$ and $(4, 6, 9)$.
7. Find the point on the $x$-axis collinear with $(1, 2, 3)$ and $(2, 4, 6)$.
8. Three collinear points have $x$-coordinates $1, 3, 5$. If the first two have all coordinates $(1, 1, 1)$ and $(3, 5, 7)$, find the third point.
9. Are $(1, 2, 0), (3, 1, -1), (5, 0, -2)$ collinear?
10. The points $A(1, 2, 3), B(5, 6, 7), C(9, 10, 11)$ , collinear?
11. Find the point on the line through $(1, 0, 0)$ and $(0, 1, 1)$ at $t = 2$.
12. Find the equation $(x - 1)/(?) = (y - 0)/(?) = (z - 0)/(?)$ of the line through $(1, 0, 0)$ with direction $(2, 3, 4)$.

Pitfalls / Tricks

• Equal ratios in all three coordinates are required for 3D collinearity (in 2D it was two ratios).
• If a denominator is zero, the corresponding numerator must be zero too.
• Three collinear points lie on a single line , try to imagine this in 3D before computing.
• Insight. The direction-ratio test is essentially saying: "the vector from $A$ to $C$ is a scalar multiple of the vector from $A$ to $B$." This is the vector form you will meet in Class XII.