Chapter 11 · Three Dimensional Geometry

Example 5: $A(2, 3, -4)$ ⁠, $B(1, -2, 3)$ ⁠, $C(3, 8, -11)$ are collinear

ExamplesCollinear points by direction ratios · Section 11.2.1

The question

Show that the points $A(2, 3, -4)$ ⁠, $B(1, -2, 3)$ and $C(3, 8, -11)$ are collinear.

Exam answer

Direction ratios of $AB$ ⁠:

\begin{aligned} &1 - 2,\ -2 - 3,\ 3 + 4 \\ &= -1,\ -5,\ 7 \end{aligned}

Direction ratios of $BC$ ⁠:

\begin{aligned} &3 - 1,\ 8 + 2,\ -11 - 3 \\ &= 2,\ 10,\ -14 \end{aligned}

\frac{2}{-1} = \frac{10}{-5} = \frac{-14}{7} = -2

So the direction ratios of $AB$ and $BC$ are proportional. Hence $AB \parallel BC$ ⁠.

But the point $B$ is common to $AB$ and $BC$ ⁠.

$\therefore$ $A$ ⁠, $B$ ⁠, $C$ are collinear. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Three points are collinear when the piece $AB$ and the piece $BC$ point the same way (or exactly opposite ways) and share a point. "Point the same way" is tested with direction ratios: the two sets must be proportional. The shared point $B$ then forces the two pieces onto one line.

Given, to find, and the plan

Given: $A(2, 3, -4)$ ⁠, $B(1, -2, 3)$ ⁠, $C(3, 8, -11)$ ⁠.
To show: $A$ ⁠, $B$ ⁠, $C$ lie on one straight line.
Plan: (1) Direction ratios of $AB$ ⁠. (2) Direction ratios of $BC$ ⁠. (3) Show they are proportional, so $AB \parallel BC$ ⁠. (4) Use the common point $B$ ⁠. (5) Check another way, with distances.

Basics you need

1Direction ratios of the line through two points

For $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ ⁠: $x_2 - x_1, y_2 - y_1, z_2 - z_1$ (Note in Section 11.2.1). Lesson: direction cosines and direction ratios

2Parallel lines have proportional direction ratios

Any two sets of direction ratios of one line are proportional, and lines with proportional direction ratios are parallel: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Sections 11.2 and 11.4). Basics: direction cosines

3Distance in space

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ ⁠, used for the check. Basics: distance in 3D

Solution

Direction ratios of $AB$
$\begin{aligned} &1 - 2,\ -2 - 3,\ 3 - (-4) \\ &= -1,\ -5,\ 7. \end{aligned}$
Second point minus first point, coordinate by coordinate, gives direction ratios of the line $AB$ ⁠. The book writes the last one as $3 + 4$ ⁠, which is the same as $3 - (-4)$ ⁠.

Direction ratios $x_2 - x_1, y_2 - y_1, z_2 - z_1$ ⁠. Lesson: the line through two points
Direction ratios of $BC$
$\begin{aligned} &3 - 1,\ 8 - (-2),\ -11 - 3 \\ &= 2,\ 10,\ -14. \end{aligned}$
The same rule with $B$ as the first point and $C$ as the second.

Direction ratios from two points. Basics: points in space
Show that they are proportional
$\frac{2}{-1} = -2, \quad \frac{10}{-5} = -2, \quad \frac{-14}{7} = -2.$
All three ratios are equal, so $2, 10, -14 = -2\,(-1, -5, 7)$ ⁠: the direction ratios of $AB$ and $BC$ are proportional, and $AB \parallel BC$ ⁠.

Proportional direction ratios describe the same direction (the factor $-2$ is negative, so $BC$ runs the opposite way to $AB$ ⁠, which still means parallel). It is essential that the same factor works for all three coordinates; if one ratio came out different, the lines would not be parallel.

Parallel lines: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Section 11.4). Lesson: angle between two lines
Use the common point $B$

The lines $AB$ and $BC$ are parallel and both pass through $B$ ⁠, so they are the same line. Hence $A$ ⁠, $B$ ⁠, $C$ are collinear.

Oblique projection, not to scale. The arrow $AB$ (gold) and the arrow $BC$ (blue, drawn beside the line so both can be seen) lie along one line. $BC$ is $-2$ times $AB$ ⁠, and $A$ is halfway between $B$ and $C$ ⁠.

Being parallel alone is not enough: two different parallel lines never meet. Through a given point there is only one line in a given direction, so two parallel lines through the same point $B$ coincide. This "parallel plus a common point" argument is the whole proof.

One and only one line passes through a point in a given direction (Section 11.3). Basics: a line is a point plus a direction
A second check, by distances
$\begin{aligned} AB &= \sqrt{1 + 25 + 49} = \sqrt{75} = 5\sqrt3, \\ AC &= \sqrt{1 + 25 + 49} = 5\sqrt3, \\ BC &= \sqrt{4 + 100 + 196} = 10\sqrt3. \end{aligned}$
So $AB + AC = BC$ ⁠: $A$ lies on the segment $BC$ ⁠, in fact at its midpoint, since the midpoint of $BC$ is
$\Big(\frac{1 + 3}{2}, \frac{-2 + 8}{2}, \frac{3 - 11}{2}\Big) = (2, 3, -4).$
For $AC$ the differences are $3 - 2, 8 - 3, -11 + 4 = 1, 5, -7$ ⁠, the same squares as for $AB$ ⁠. When the two shorter distances add up exactly to the longest, the three points cannot form a triangle (the triangle inequality would be strict), so they lie on one line. This agrees with the picture: the order along the line is $B$ ⁠, $A$ ⁠, $C$ ⁠.

Distance formula in space. Basics: distance in 3D

Result

The direction ratios of $AB$ ( $-1, -5, 7$ ⁠) and of $BC$ ( $2, 10, -14$ ⁠) are proportional, so $AB \parallel BC$ ⁠; since $B$ is common to both, $A$ ⁠, $B$ ⁠, $C$ are collinear. Moreover $A$ is the midpoint of $BC$ ⁠.

(Textbook answer: $A$ ⁠, $B$ ⁠, $C$ are collinear points, by the same argument.)

Check it with numbers

The line through $B$ with direction ratios $-1, -5, 7$ is $(1 - t,\ -2 - 5t,\ 3 + 7t)$ ⁠. At $t = -1$ it gives $(2, 3, -4) = A$ ⁠, and at $t = -2$ it gives $(3, 8, -11) = C$ ⁠. In decimals, $AB = AC \approx 8.660254$ and $BC \approx 17.320508$ ⁠.

Common mistakes

Stopping at "parallel". Proportional direction ratios only give $AB \parallel BC$ ⁠; the common point $B$ is what makes the points collinear.
Thinking a negative factor means "not parallel". $-2$ means $BC$ points the opposite way along the same direction, which is still parallel.
Checking only two of the three ratios. All three quotients must be equal.

What to take away

Collinear = parallel pieces with a common point.
Test "parallel" by dividing matching direction ratios: one common value means proportional.
Distances give a second test: the two shorter lengths add up to the longest.

Every fact used, and where it comes from

Fact	Used in	Source
Direction ratios of the line through two points	Steps 1, 2	Note in Section 11.2.1 (Lesson)
Proportional direction ratios mean parallel lines	Step 3	Sections 11.2 and 11.4 (Lesson)
A line is fixed by one point and a direction	Step 4	Section 11.3 (Basics)
Distance formula; midpoint $\frac{B + C}{2}$	Step 5	Section 11.2.1; section formula, Section 10.5.3 (Basics)

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Example 5: A(2,3,−4)A(2, 3, -4)A(2,3,−4)⁠, B(1,−2,3)B(1, -2, 3)B(1,−2,3)⁠, C(3,8,−11)C(3, 8, -11)C(3,8,−11) are collinear

Basics you need

Solution

Direction ratios of ABABAB

Direction ratios of BCBCBC

Show that they are proportional

Use the common point BBB

A second check, by distances

Every fact used, and where it comes from

Example 5: $A(2, 3, -4)$ ⁠, $B(1, -2, 3)$ ⁠, $C(3, 8, -11)$ are collinear

Direction ratios of $AB$

Direction ratios of $BC$

Use the common point $B$