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Chapter 10 · Vector Algebra

Miscellaneous · Q8: A(1,−2,−8)A(1,-2,-8)⁠, B(5,0,−2)B(5,0,-2)⁠, C(11,3,7)C(11,3,7) collinear; the ratio in which B divides AC

Miscellaneous ExerciseCollinear points, section formula · Sections 10.5.1, 10.5.2 and 10.5.3
The question

Show that the points A(1, −2-2⁠, −8-8⁠), B(5, 0, −2-2⁠) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Exam answer AB→=(5−1)i^+(0+2)j^+(−2+8)k^=4i^+2j^+6k^\begin{aligned} \overrightarrow{AB} &= (5 - 1)\hat i + (0 + 2)\hat j \\ &\quad + (-2 + 8)\hat k \\ &= 4\hat i + 2\hat j + 6\hat k \end{aligned} BC→=(11−5)i^+(3−0)j^+(7+2)k^=6i^+3j^+9k^=32(4i^+2j^+6k^)=32 AB→\begin{aligned} \overrightarrow{BC} &= (11 - 5)\hat i + (3 - 0)\hat j \\ &\quad + (7 + 2)\hat k \\ &= 6\hat i + 3\hat j + 9\hat k \\ &= \frac32(4\hat i + 2\hat j + 6\hat k) \\ &= \frac32\,\overrightarrow{AB} \end{aligned}

So AB→\overrightarrow{AB} and BC→\overrightarrow{BC} are parallel, and they have the common point B.

Hence the points A, B and C are collinear.

As BC→=32AB→\overrightarrow{BC} = \frac32\overrightarrow{AB} with 32>0\frac32 > 0⁠, AB→\overrightarrow{AB} and BC→\overrightarrow{BC} have the same direction, so B lies between A and C, and

∣AB→∣=16+4+36=214,∣BC→∣=36+9+81=314\begin{aligned} |\overrightarrow{AB}| &= \sqrt{16 + 4 + 36} = 2\sqrt{14}, \\ |\overrightarrow{BC}| &= \sqrt{36 + 9 + 81} = 3\sqrt{14} \end{aligned}

∴\therefore AB : BC = 2 : 3, so B divides AC internally in the ratio 2 : 3.

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

The one idea this problem is built on

Three points are on one line exactly when the step from A to B and the step from B to C point along the same line: BC→=λ AB→\overrightarrow{BC} = \lambda\,\overrightarrow{AB}⁠. The same number λ\lambda then tells us where B sits: if λ>0\lambda > 0⁠, B is between A and C and AB:BC=1:λ\text{AB} : \text{BC} = 1 : \lambda⁠.

Given, to show and find, and the plan
Given
A(1, −2-2⁠, −8-8⁠), B(5, 0, −2-2⁠), C(11, 3, 7).
To show
A, B, C lie on one line.
To find
The ratio in which B divides AC.
Plan
(1) Find AB→\overrightarrow{AB} and BC→\overrightarrow{BC}⁠. (2) Show one is a multiple of the other. (3) Read off the ratio from the multiple and the lengths. (4) Confirm with the section formula.

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

2Collinear vectors

Two vectors are collinear if and only if one is a scalar multiple of the other, b⃗=λa⃗\vec b = \lambda\vec a (Section 10.5.1, Remarks (i)). Lesson: components

3Dividing a segment in a ratio

B divides AC internally in the ratio m:nm : n when B is between A and C and AB:BC=m:n\text{AB} : \text{BC} = m : n⁠; then OB→=m OC→+n OA→m+n\overrightarrow{OB} = \frac{m\,\overrightarrow{OC} + n\,\overrightarrow{OA}}{m + n} (Section 10.5.3). Basics: dividing in m : n

Solution

  1. The two steps

    AB→=(5−1)i^+(0+2)j^+(−2+8)k^=4i^+2j^+6k^,BC→=(11−5)i^+(3−0)j^+(7+2)k^=6i^+3j^+9k^.\begin{aligned} \overrightarrow{AB} &= (5 - 1)\hat i + (0 + 2)\hat j + (-2 + 8)\hat k \\ &= 4\hat i + 2\hat j + 6\hat k, \\ \overrightarrow{BC} &= (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k \\ &= 6\hat i + 3\hat j + 9\hat k. \end{aligned}
    End minus start, component by component. Watch the double negatives: 0−(−2)=20 - (-2) = 2⁠, −2−(−8)=6-2 - (-8) = 6 and 7−(−2)=97 - (-2) = 9⁠.
    Vector joining two points (Section 10.5.2). Basics: three numbers fix a point

  2. One is a multiple of the other

    The components of BC→\overrightarrow{BC} and AB→\overrightarrow{AB} are in the same ratio: 64=32=96=32\frac64 = \frac32 = \frac96 = \frac32⁠. So

    BC→=32 AB→.\overrightarrow{BC} = \tfrac32\,\overrightarrow{AB}.

    The vectors are parallel and both pass through B, so A, B and C are collinear.

    Parallel vectors alone would allow two separate parallel lines. Here the line through A and B and the line through B and C share the point B and have the same direction, so they are the same line.
    Collinear vectors: equal component ratios (Section 10.5.1, Remarks (i)); the same method as Example 21 and Exercise 10.3 Q16. Lesson: components

  3. The ratio

    Since 32>0\frac32 > 0⁠, BC→\overrightarrow{BC} points the same way as AB→\overrightarrow{AB}⁠: going from A to B and on to C, we keep moving forward, so B is between A and C. The lengths are

    ∣AB→∣=16+4+36=56=214,∣BC→∣=36+9+81=126=314.\begin{aligned} |\overrightarrow{AB}| &= \sqrt{16 + 4 + 36} \\ &= \sqrt{56} = 2\sqrt{14}, \\ |\overrightarrow{BC}| &= \sqrt{36 + 9 + 81} \\ &= \sqrt{126} = 3\sqrt{14}. \end{aligned}

    So AB:BC=2:3\text{AB} : \text{BC} = 2 : 3⁠, and B divides AC internally in the ratio 2 : 3.

    A(1, −2, −8) B(5, 0, −2) C(11, 3, 7) AB = 2√14 BC = 3√14 AB : BC = 2 : 3
    A, B and C on one line, drawn true to scale: AB=214\text{AB} = 2\sqrt{14}⁠, BC=314\text{BC} = 3\sqrt{14} and AC=514\text{AC} = 5\sqrt{14}⁠, so B divides AC internally in the ratio 2 : 3.
    ∣BC→∣=32∣AB→∣|\overrightarrow{BC}| = \frac32|\overrightarrow{AB}| gives the ratio directly; the surds just confirm it: 56=4×1456 = 4 \times 14 and 126=9×14126 = 9 \times 14⁠. Had the multiple been negative, B would lie outside the segment AC and the division would be external.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5); internal division (Section 10.5.3). Basics: dividing in m : n

  4. Confirm with the section formula

    The point dividing AC internally in the ratio 2 : 3 has position vector

    2 OC→+3 OA→2+3=(22+3)i^+(6−6)j^+(14−24)k^5=5i^+0j^−2k^=OB→.\begin{aligned} &\frac{2\,\overrightarrow{OC} + 3\,\overrightarrow{OA}}{2 + 3} \\ &= \frac{(22 + 3)\hat i + (6 - 6)\hat j + (14 - 24)\hat k}{5} \\ &= 5\hat i + 0\hat j - 2\hat k = \overrightarrow{OB}. \end{aligned}
    The weight 2 goes with C (the end away from the "2" part AB) and 3 with A. Landing exactly on B(5, 0, −2-2⁠) confirms both the collinearity and the ratio at once.
    Section formula, internal division (Section 10.5.3); see Exercise 10.2 Q15. Lesson: the section formula
Result

BC→=32 AB→\overrightarrow{BC} = \frac32\,\overrightarrow{AB} with the common point B, so A, B and C are collinear. B divides AC internally in the ratio 2 : 3.

(Textbook answer: 2 : 3.)

Check it with numbers

14≈3.741657\sqrt{14} \approx 3.741657⁠, so AB≈7.483315\text{AB} \approx 7.483315⁠, BC≈11.224972\text{BC} \approx 11.224972 and AC=514≈18.708287\text{AC} = 5\sqrt{14} \approx 18.708287⁠; AC→=10i^+5j^+15k^\overrightarrow{AC} = 10\hat i + 5\hat j + 15\hat k has length 350=514\sqrt{350} = 5\sqrt{14}⁠, equal to AB + BC.

The section formula coordinate by coordinate: 2(11)+3(1)5=5\frac{2(11) + 3(1)}{5} = 5⁠, 2(3)+3(−2)5=0\frac{2(3) + 3(-2)}{5} = 0⁠, 2(7)+3(−8)5=−2\frac{2(7) + 3(-8)}{5} = -2⁠.

Common mistakes
  1. Writing the ratio upside down. B divides AC in AB : BC = 2 : 3, not 3 : 2.
  2. Sign slips with negative coordinates. −2−(−8)=6-2 - (-8) = 6⁠, not −10-10⁠.
  3. Stopping at "parallel". Parallel vectors need a common point before the points can be called collinear.
What to take away
  1. A, B, C are collinear when BC→=λ AB→\overrightarrow{BC} = \lambda\,\overrightarrow{AB} (common point B).
  2. If λ>0\lambda > 0⁠, B divides AC internally in the ratio 1:λ1 : \lambda⁠; here 1:32=2:31 : \frac32 = 2 : 3⁠.
  3. The section formula is a quick check of any ratio.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsStep 1Section 10.5.2 (Basics: three numbers fix a point)
Collinear vectors: b⃗=λa⃗\vec b = \lambda\vec a⁠, equal component ratiosStep 2Section 10.5.1, Remarks (i)
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; direction of λa⃗\lambda\vec aStep 3Section 10.5
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 3Section 10.5.1 (Basics: use Pythagoras twice)
Internal division and the section formulaSteps 3, 4Section 10.5.3, Case I (Lesson: the section formula)
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q8 (Lesson: put it all together)
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