Chapter 11 · Three Dimensional Geometry

Exercise 11.1 · Q4: $(2, 3, 4)$ ⁠, $(-1, -2, 1)$ ⁠, $(5, 8, 7)$ are collinear

Exercise 11.1Collinear points by direction ratios · Section 11.2.1

The question

Show that the points $(2, 3, 4)$ ⁠, $(-1, -2, 1)$ ⁠, $(5, 8, 7)$ are collinear.

Exam answer

Let $A(2, 3, 4)$ ⁠, $B(-1, -2, 1)$ and $C(5, 8, 7)$ ⁠.

Direction ratios of $AB$ ⁠:

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

Direction ratios of $BC$ ⁠:

\begin{aligned} &5 + 1,\ 8 + 2,\ 7 - 1 \\ &= 6,\ 10,\ 6 \end{aligned}

\frac{6}{-3} = \frac{10}{-5} = \frac{6}{-3} = -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 puts both pieces on one line.

Given, to find, and the plan

Given: The points $A(2, 3, 4)$ ⁠, $B(-1, -2, 1)$ ⁠, $C(5, 8, 7)$ ⁠.
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

Lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are parallel when $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Section 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

Name the points; direction ratios of $AB$

Let $A(2, 3, 4)$ ⁠, $B(-1, -2, 1)$ ⁠, $C(5, 8, 7)$ ⁠. Then $AB$ has direction ratios
$\begin{aligned} &-1 - 2,\ -2 - 3,\ 1 - 4 \\ &= -3,\ -5,\ -3. \end{aligned}$
Second point minus first point, coordinate by coordinate, gives direction ratios of the line $AB$ ⁠. Naming the points makes it easy to say which piece we mean.

Direction ratios $x_2 - x_1, y_2 - y_1, z_2 - z_1$ ⁠. Lesson: the line through two points
Direction ratios of $BC$
$\begin{aligned} &5 - (-1),\ 8 - (-2),\ 7 - 1 \\ &= 6,\ 10,\ 6. \end{aligned}$
The same rule with $B$ as the first point and $C$ as the second. Subtracting a negative number adds it: $5 - (-1) = 5 + 1 = 6$ ⁠.

Direction ratios from two points. Basics: points in space
Show that they are proportional
$\frac{6}{-3} = -2, \quad \frac{10}{-5} = -2, \quad \frac{6}{-3} = -2.$
All three ratios are equal, so $6, 10, 6 = -2\,(-3, -5, -3)$ ⁠: 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. The same factor must work 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 $B$ ⁠. Hence they are the same line, and $A$ ⁠, $B$ ⁠, $C$ are collinear.

Oblique projection, not to scale. The arrow $A \to B$ has direction ratios $-3, -5, -3$ ⁠; the arrow $B \to C$ ⁠, drawn beside the line, has $6, 10, 6$ ⁠, which is $-2$ times as much. Both lie on one line through $B$ ⁠.

Parallel alone is not enough: two different parallel lines never meet. But through a given point there is only one line in a given direction, so two parallel lines through the same point $B$ coincide. All three points are then on that one line.

A line is fixed by one point and a direction. Basics: a line is a point plus a direction
Check with distances
$\begin{aligned} AB &= \sqrt{9 + 25 + 9} = \sqrt{43}, \\ AC &= \sqrt{3^2 + 5^2 + 3^2} = \sqrt{43}, \\ BC &= \sqrt{36 + 100 + 36} \\ &= \sqrt{172} = 2\sqrt{43}. \end{aligned}$
So $AB + AC = BC$ ⁠: $A$ lies between $B$ and $C$ ⁠, exactly in the middle.

For three points, the longest distance equals the sum of the other two exactly when the points are collinear (otherwise the triangle inequality is strict). Here $\sqrt{172} = \sqrt{4 \times 43} = 2\sqrt{43}$ ⁠. Indeed $A$ is the midpoint of $BC$ ⁠: $\left(\frac{-1 + 5}{2}, \frac{-2 + 8}{2}, \frac{1 + 7}{2}\right) = (2, 3, 4)$ ⁠.

Distance formula in space. Basics: Pythagoras twice

Result

$AB$ and $BC$ have proportional direction ratios $-3, -5, -3$ and $6, 10, 6$ and share the point $B$ ⁠, so the points $(2, 3, 4)$ ⁠, $(-1, -2, 1)$ ⁠, $(5, 8, 7)$ are collinear. In fact $(2, 3, 4)$ is the midpoint of the other two.

(The answer key prints no answer for this "show that" question.)

Check it with numbers

$\sqrt{43} \approx 6.557439$ and $\sqrt{172} \approx 13.114877$ ⁠, so $AB + AC \approx 6.557439 + 6.557439$ $= 13.114878$ ⁠, equal to $BC$ up to rounding in the last place.

Walking from $B$ by $\frac12$ of $\overrightarrow{BC}$ ⁠: $(-1 + 3, -2 + 5, 1 + 3) = (2, 3, 4) = A$ ⁠.

Common mistakes

Stopping at "parallel". Proportional direction ratios only show $AB \parallel BC$ ⁠; the common point $B$ is what makes the points collinear.
Sign slips such as $5 - (-1) = 4$ ⁠. Subtracting a negative adds: $5 + 1 = 6$ ⁠.
Checking only two of the three ratios. All three must give the same factor.

What to take away

To show $A$ ⁠, $B$ ⁠, $C$ collinear: direction ratios of $AB$ and $BC$ proportional, plus the common point $B$ ⁠.
A negative factor of proportionality still means parallel; it only reverses the direction.
Distances give an independent check: the longest equals the sum of the other two.

Every fact used, and where it comes from

Fact	Used in	Source
Direction ratios of $PQ$ ⁠: $x_2 - x_1, y_2 - y_1, z_2 - z_1$	Steps 1, 2	Note in Section 11.2.1 (Lesson)
Parallel lines: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$	Step 3	Section 11.4, condition (ii) (Lesson)
Parallel lines with a common point coincide	Step 4	Same argument as Example 5 (Example 5)
Distance formula in space	Step 5	Section 11.2.1 (Basics)

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Exercise 11.1 · Q4: (2,3,4)(2, 3, 4)(2,3,4)⁠, (−1,−2,1)(-1, -2, 1)(−1,−2,1)⁠, (5,8,7)(5, 8, 7)(5,8,7) are collinear

Basics you need

Solution

Name the points; direction ratios of ABABAB

Direction ratios of BCBCBC

Show that they are proportional

Use the common point BBB

Check with distances

Every fact used, and where it comes from

Exercise 11.1 · Q4: $(2, 3, 4)$ ⁠, $(-1, -2, 1)$ ⁠, $(5, 8, 7)$ are collinear

Name the points; direction ratios of $AB$

Direction ratios of $BC$

Use the common point $B$