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Chapter 10 · Vector Algebra

Exercise 10.3 · Q16: A(1,2,7)A(1,2,7)⁠, B(2,6,3)B(2,6,3)⁠, C(3,10,−1)C(3,10,-1) are collinear

Exercise 10.3Collinear points, magnitudes · Sections 10.5.2 and 10.6.1
The question

Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, −1-1⁠) are collinear.

Exam answer AB→=(2−1)i^+(6−2)j^+(3−7)k^=i^+4j^−4k^\begin{aligned} \overrightarrow{AB} &= (2 - 1)\hat i + (6 - 2)\hat j \\ &\quad + (3 - 7)\hat k \\ &= \hat i + 4\hat j - 4\hat k \end{aligned} BC→=(3−2)i^+(10−6)j^+(−1−3)k^=i^+4j^−4k^\begin{aligned} \overrightarrow{BC} &= (3 - 2)\hat i + (10 - 6)\hat j \\ &\quad + (-1 - 3)\hat k \\ &= \hat i + 4\hat j - 4\hat k \end{aligned} AC→=(3−1)i^+(10−2)j^+(−1−7)k^=2i^+8j^−8k^\begin{aligned} \overrightarrow{AC} &= (3 - 1)\hat i + (10 - 2)\hat j \\ &\quad + (-1 - 7)\hat k \\ &= 2\hat i + 8\hat j - 8\hat k \end{aligned} ∣AB→∣=1+16+16=33∣BC→∣=1+16+16=33∣AC→∣=4+64+64=233\begin{aligned} |\overrightarrow{AB}| &= \sqrt{1 + 16 + 16} = \sqrt{33} \\ |\overrightarrow{BC}| &= \sqrt{1 + 16 + 16} = \sqrt{33} \\ |\overrightarrow{AC}| &= \sqrt{4 + 64 + 64} = 2\sqrt{33} \end{aligned}

∴ ∣AB→∣+∣BC→∣=233=∣AC→∣\therefore\ |\overrightarrow{AB}| + |\overrightarrow{BC}| = 2\sqrt{33} = |\overrightarrow{AC}|

Hence the points A, B and C are collinear.

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

The one idea this problem is built on

For any three points, AC≤AB+BC\text{AC} \le \text{AB} + \text{BC}⁠, with equality only when B lies on the segment AC. So if the two short distances add up exactly to the long one, the three points are on one line.

Given, to show, and the plan
Given
A(1, 2, 7), B(2, 6, 3), C(3, 10, −1-1⁠).
To show
A, B and C lie on one line.
Plan
(1) Find AB→\overrightarrow{AB}⁠, BC→\overrightarrow{BC}⁠, AC→\overrightarrow{AC}⁠. (2) Find their lengths. (3) Compare ∣AB→∣+∣BC→∣|\overrightarrow{AB}| + |\overrightarrow{BC}| with ∣AC→∣|\overrightarrow{AC}|⁠. (4) Confirm a second way, with parallel vectors.

Basics you need

1Vector joining two points

PQ→\overrightarrow{PQ} = (coordinates of Q) −- (coordinates of P) (Section 10.5.2). Basics: three numbers fix a point

2Length from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

3Equality in the triangle inequality

∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec a + \vec b| \le |\vec a| + |\vec b|⁠; when ∣AC→∣=∣AB→∣+∣BC→∣|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|⁠, the points A, B, C are collinear (Example 20, Section 10.6.2, and the Remark after it). Lesson: Examples 13–21

Solution

  1. The three vectors

    AB→=(2−1)i^+(6−2)j^+(3−7)k^=i^+4j^−4k^,BC→=(3−2)i^+(10−6)j^+(−1−3)k^=i^+4j^−4k^,AC→=(3−1)i^+(10−2)j^+(−1−7)k^=2i^+8j^−8k^.\begin{aligned} \overrightarrow{AB} &= (2 - 1)\hat i + (6 - 2)\hat j \\ &\quad + (3 - 7)\hat k = \hat i + 4\hat j - 4\hat k, \\ \overrightarrow{BC} &= (3 - 2)\hat i + (10 - 6)\hat j \\ &\quad + (-1 - 3)\hat k = \hat i + 4\hat j - 4\hat k, \\ \overrightarrow{AC} &= (3 - 1)\hat i + (10 - 2)\hat j \\ &\quad + (-1 - 7)\hat k = 2\hat i + 8\hat j - 8\hat k. \end{aligned}
    Each is end minus start, component by component. Notice already that AB→\overrightarrow{AB} and BC→\overrightarrow{BC} come out equal: going from A to B is the same move as going from B to C.
    Vector joining two points (Section 10.5.2). Basics: three numbers fix a point

  2. Their lengths

    ∣AB→∣=1+16+16=33,∣BC→∣=1+16+16=33,∣AC→∣=4+64+64=132=233.\begin{aligned} |\overrightarrow{AB}| &= \sqrt{1 + 16 + 16} = \sqrt{33}, \\ |\overrightarrow{BC}| &= \sqrt{1 + 16 + 16} = \sqrt{33}, \\ |\overrightarrow{AC}| &= \sqrt{4 + 64 + 64} \\ &= \sqrt{132} = 2\sqrt{33}. \end{aligned}
    Square the components, add, take the root. (−4)2=16(-4)^2 = 16 and (−8)2=64(-8)^2 = 64⁠. Since 132=4×33132 = 4 \times 33⁠, 132=4 33=233\sqrt{132} = \sqrt4\,\sqrt{33} = 2\sqrt{33}⁠, which makes the comparison in the next step exact.
    Magnitude from components (Section 10.5.1). Basics: use Pythagoras twice

  3. Compare the distances

    ∣AB→∣+∣BC→∣=33+33=233=∣AC→∣.\begin{aligned} &|\overrightarrow{AB}| + |\overrightarrow{BC}| = \sqrt{33} + \sqrt{33} \\ &= 2\sqrt{33} = |\overrightarrow{AC}|. \end{aligned}

    The two shorter distances add up exactly to the longest one, so A, B and C are collinear, with B between A and C.

    A(1, 2, 7) B(2, 6, 3) C(3, 10, −1) √33 √33 |AC| = 2√33 ≈ 11.49
    A, B and C on one line, drawn true to scale: AB and BC each have length 33\sqrt{33}⁠, and AC has length 2332\sqrt{33}⁠. B is the midpoint of AC.
    If A, B, C were the corners of a genuine triangle, the side AC would be strictly shorter than the path A to B to C (the triangle inequality). Equality leaves no room for a bend at B, so B lies on the segment AC.
    Triangle inequality and the Remark after Example 20; the same method as Example 21. Lesson: Examples 13–21

  4. A second way: parallel vectors through B

    AB→=BC→\overrightarrow{AB} = \overrightarrow{BC}⁠, so the lines AB and BC have the same direction, and both contain the point B. Two lines through the same point with the same direction are the same line, so A, B, C are collinear. Also

    AC→=2i^+8j^−8k^=2 AB→,\overrightarrow{AC} = 2\hat i + 8\hat j - 8\hat k = 2\,\overrightarrow{AB},

    and the midpoint of AC is (1+32,2+102,7−12)=(2,6,3)\left(\frac{1 + 3}{2}, \frac{2 + 10}{2}, \frac{7 - 1}{2}\right) = (2, 6, 3)⁠, which is B.

    Two vectors are collinear exactly when one is a scalar multiple of the other. Being parallel alone is not enough for points: the shared point B is what puts all three on one line. The midpoint fact matches the picture: B is halfway along AC.
    Collinear vectors, b⃗=λa⃗\vec b = \lambda\vec a (Remarks (i), Section 10.5.1); midpoint (Section 10.5.3). Lesson: the section formula
Result

∣AB→∣+∣BC→∣=33+33=233=∣AC→∣|\overrightarrow{AB}| + |\overrightarrow{BC}| = \sqrt{33} + \sqrt{33} = 2\sqrt{33} = |\overrightarrow{AC}|⁠, so A, B and C are collinear (B is the midpoint of AC). Hence proved.

(Textbook answer: the answer key has no entry for this "show that" question.)

Check it with numbers

33≈5.744563\sqrt{33} \approx 5.744563 and 132≈11.489125\sqrt{132} \approx 11.489125⁠. The ratio is exact: 13233=13233=4=2\frac{\sqrt{132}}{\sqrt{33}} = \sqrt{\frac{132}{33}} = \sqrt4 = 2⁠, so AC is exactly twice AB.

Component ratios of AC→\overrightarrow{AC} to AB→\overrightarrow{AB}⁠: 21=84=−8−4=2\frac21 = \frac84 = \frac{-8}{-4} = 2⁠, all the same, as collinear vectors need.

Common mistakes
  1. Using AB→+BC→=AC→\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} as proof. That holds for any three points (triangle law); it is the lengths that must add.
  2. Simplifying 132\sqrt{132} wrongly. 132=233\sqrt{132} = 2\sqrt{33}⁠, not 4334\sqrt{33} or 66\sqrt{66}⁠.
  3. Sign slips. −1−3=−4-1 - 3 = -4 and −1−7=−8-1 - 7 = -8⁠.
What to take away
  1. Three points are collinear when the two shorter distances add up to the longest.
  2. Equivalently, AC→=λ AB→\overrightarrow{AC} = \lambda\,\overrightarrow{AB} for some scalar λ\lambda⁠: parallel vectors through a common point.
  3. Keep surds exact (132=233\sqrt{132} = 2\sqrt{33}⁠) so that the comparison is exact.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsStep 1Section 10.5.2 (Basics: three numbers fix a point)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 2Section 10.5.1 (Basics: use Pythagoras twice)
Triangle inequality; equality means collinearStep 3Example 20 and the Remark after it (Lesson: Examples 13–21)
Collinear vectors: b⃗=λa⃗\vec b = \lambda\vec aStep 4, CheckRemarks (i), Section 10.5.1
Midpoint of a segmentStep 4Section 10.5.3, Remark (Lesson: the section formula)
Triangle law of additionCommon mistakesSection 10.4 (Lesson: adding arrows, head to tail)
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