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Chapter 10 · Vector Algebra

Example 11: Point dividing 3a⃗−2b⃗3\vec a-2\vec b⁠, a⃗+b⃗\vec a+\vec b in the ratio 2 : 1, internally and externally

ExamplesSection formula · Section 10.5.3
The question

Consider two points P and Q with position vectors OP→=3a⃗−2b⃗\overrightarrow{OP} = 3\vec a - 2\vec b and OQ→=a⃗+b⃗\overrightarrow{OQ} = \vec a + \vec b⁠. Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally.

Exam answer

Here OP→=3a⃗−2b⃗\overrightarrow{OP} = 3\vec a - 2\vec b⁠, OQ→=a⃗+b⃗\overrightarrow{OQ} = \vec a + \vec b⁠, and m:n=2:1m : n = 2 : 1⁠.

(i) R divides PQ internally in the ratio 2 : 1, so

OR→=m OQ→+n OP→m+n=2(a⃗+b⃗)+1(3a⃗−2b⃗)2+1=2a⃗+2b⃗+3a⃗−2b⃗3\begin{aligned} \overrightarrow{OR} &= \frac{m\,\overrightarrow{OQ} + n\,\overrightarrow{OP}}{m + n} \\ &= \frac{2(\vec a + \vec b) + 1(3\vec a - 2\vec b)}{2 + 1} \\ &= \frac{2\vec a + 2\vec b + 3\vec a - 2\vec b}{3} \end{aligned}

∴ OR→=5a⃗3\therefore\ \overrightarrow{OR} = \frac{5\vec a}{3}⁠.

(ii) R divides PQ externally in the ratio 2 : 1, so

OR→=m OQ→−n OP→m−n=2(a⃗+b⃗)−1(3a⃗−2b⃗)2−1=2a⃗+2b⃗−3a⃗+2b⃗\begin{aligned} \overrightarrow{OR} &= \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n} \\ &= \frac{2(\vec a + \vec b) - 1(3\vec a - 2\vec b)}{2 - 1} \\ &= 2\vec a + 2\vec b - 3\vec a + 2\vec b \end{aligned}

∴ OR→=4b⃗−a⃗\therefore\ \overrightarrow{OR} = 4\vec b - \vec a⁠.

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

The one idea this problem is built on

The section formula gives the position vector of a point R on the line PQ from the position vectors of P and Q. If R divides PQ in the ratio m:nm : n⁠,

internally:OR→=m OQ→+n OP→m+n,externally:OR→=m OQ→−n OP→m−n.\begin{aligned} &\text{internally:} \\ &\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} + n\,\overrightarrow{OP}}{m + n}, \\ &\text{externally:} \\ &\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n}. \end{aligned}

Each end is weighted by the part of the ratio next to the other end: Q gets mm⁠, P gets nn⁠. The formula does not care what OP→\overrightarrow{OP} and OQ→\overrightarrow{OQ} are made of, so here we simply substitute 3a⃗−2b⃗3\vec a - 2\vec b and a⃗+b⃗\vec a + \vec b and simplify.

Given, to find, and the plan
Given
OP→=3a⃗−2b⃗\overrightarrow{OP} = 3\vec a - 2\vec b⁠, OQ→=a⃗+b⃗\overrightarrow{OQ} = \vec a + \vec b⁠; R divides the line joining P and Q in the ratio 2 : 1.
To find
OR→\overrightarrow{OR} when the division is (i) internal, (ii) external.
Plan
(1) Read off m=2m = 2⁠, n=1n = 1⁠, with P the first point. (2) Substitute into the internal formula and simplify. (3) Check that PR→=2 RQ→\overrightarrow{PR} = 2\,\overrightarrow{RQ}⁠. (4) Repeat with the external formula and check PR→=2 QR→\overrightarrow{PR} = 2\,\overrightarrow{QR}⁠.

Basics you need

1Dividing in a ratio

"R divides PQ internally in the ratio m:nm : n" means R lies between P and Q with PR:RQ=m:n\text{PR} : \text{RQ} = m : n⁠; "externally" means R lies on the line PQ, outside the segment, with PR:QR=m:n\text{PR} : \text{QR} = m : n⁠. Basics: dividing in m : n

2The section formula

Internal: OR→=mb⃗+na⃗m+n\overrightarrow{OR} = \frac{m\vec b + n\vec a}{m + n}⁠; external: OR→=mb⃗−na⃗m−n\overrightarrow{OR} = \frac{m\vec b - n\vec a}{m - n}⁠, where a⃗\vec a and b⃗\vec b are the position vectors of P and Q (Section 10.5.3, Cases I and II, Figs 10.16 and 10.17). Lesson: the section formula

3Vector joining two points

PR→=OR→−OP→\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP}⁠: terminal minus initial (Section 10.5.2). Lesson: adding arrows, head to tail

Part (i): R divides PQ internally

  1. Read the ratio and choose the formula

    The ratio 2 : 1 is written in the order P, Q, so m=2m = 2 is the part next to P and n=1n = 1 the part next to Q: PR:RQ=2:1\text{PR} : \text{RQ} = 2 : 1⁠. For internal division,

    OR→=m OQ→+n OP→m+n.\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} + n\,\overrightarrow{OP}}{m + n}.
    The book states the formula as mb⃗+na⃗m+n\frac{m\vec b + n\vec a}{m + n}⁠, where its a⃗\vec a and b⃗\vec b are the position vectors of P and Q. In this question the letters a⃗\vec a and b⃗\vec b stand for two other vectors, so we write the formula with OP→\overrightarrow{OP} and OQ→\overrightarrow{OQ} to keep the two meanings apart. Since PR is twice RQ, R is nearer to Q, which is why Q carries the larger weight m=2m = 2⁠.
    Section formula, Case I (Section 10.5.3, Fig 10.16). Basics: dividing in m : n

  2. Substitute and simplify

    OR→=2(a⃗+b⃗)+1(3a⃗−2b⃗)2+1=2a⃗+2b⃗+3a⃗−2b⃗3=5a⃗3.\begin{aligned} \overrightarrow{OR} &= \frac{2(\vec a + \vec b) + 1(3\vec a - 2\vec b)}{2 + 1} \\ &= \frac{2\vec a + 2\vec b + 3\vec a - 2\vec b}{3} \\ &= \frac{5\vec a}{3}. \end{aligned}

    So OR→=53a⃗\overrightarrow{OR} = \frac{5}{3}\vec a.

    Vectors can be multiplied out and collected like algebra: 2a⃗+3a⃗=5a⃗2\vec a + 3\vec a = 5\vec a⁠, and the b⃗\vec b terms cancel because 2b⃗−2b⃗=0⃗2\vec b - 2\vec b = \vec 0⁠. The answer has no b⃗\vec b at all, so R lies on the line through O in the direction of a⃗\vec a⁠.
    Vector addition is commutative and associative, and λ(u⃗+v⃗)=λu⃗+λv⃗\lambda(\vec u + \vec v) = \lambda\vec u + \lambda\vec v (Sections 10.4 and 10.5). Lesson: the section formula

  3. Check that PR is twice RQ

    PR→=53a⃗−(3a⃗−2b⃗)=−43a⃗+2b⃗,RQ→=(a⃗+b⃗)−53a⃗=−23a⃗+b⃗.\begin{aligned} \overrightarrow{PR} &= \tfrac53\vec a - (3\vec a - 2\vec b) = -\tfrac43\vec a + 2\vec b, \\ \overrightarrow{RQ} &= (\vec a + \vec b) - \tfrac53\vec a = -\tfrac23\vec a + \vec b. \end{aligned}

    So PR→=2 RQ→\overrightarrow{PR} = 2\,\overrightarrow{RQ}⁠: the two pieces point the same way and PR is twice as long as RQ. R is between P and Q with PR:RQ=2:1\text{PR} : \text{RQ} = 2 : 1⁠.

    This is the definition the book starts from in Case I: R divides PQ internally in the ratio m:nm : n when m RQ→=n PR→m\,\overrightarrow{RQ} = n\,\overrightarrow{PR}⁠, here 2 RQ→=PR→2\,\overrightarrow{RQ} = \overrightarrow{PR}⁠. Checking it directly confirms both the formula and the algebra.
    PR→=OR→−OP→\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} (Section 10.5.2). Lesson: Examples 11–12

Part (ii): R divides PQ externally

  1. The external formula

    Now R lies on the line PQ outside the segment, with PR:QR=2:1\text{PR} : \text{QR} = 2 : 1⁠. Again m=2m = 2⁠, n=1n = 1⁠, and since m≠nm \ne n⁠,

    OR→=m OQ→−n OP→m−n.\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n}.
    External division is internal division with the ratio m:(−n)m : (-n)⁠: the point lies outside, so one of the two pieces points backwards. That is why the plus signs of the internal formula become minus signs. The formula needs m≠nm \ne n⁠; for m=nm = n no such point exists, because no point outside the segment is equally far from P and Q.
    Section formula, Case II (Section 10.5.3, Fig 10.17). Lesson: the section formula

  2. Substitute and simplify

    OR→=2(a⃗+b⃗)−1(3a⃗−2b⃗)2−1=2a⃗+2b⃗−3a⃗+2b⃗=−a⃗+4b⃗.\begin{aligned} \overrightarrow{OR} &= \frac{2(\vec a + \vec b) - 1(3\vec a - 2\vec b)}{2 - 1} \\ &= 2\vec a + 2\vec b - 3\vec a + 2\vec b \\ &= -\vec a + 4\vec b. \end{aligned}

    So OR→=4b⃗−a⃗\overrightarrow{OR} = 4\vec b - \vec a.

    The denominator is 2−1=12 - 1 = 1⁠. The minus sign in front of the bracket changes both signs inside it: −(3a⃗−2b⃗)=−3a⃗+2b⃗-(3\vec a - 2\vec b) = -3\vec a + 2\vec b⁠. Then 2a⃗−3a⃗=−a⃗2\vec a - 3\vec a = -\vec a and 2b⃗+2b⃗=4b⃗2\vec b + 2\vec b = 4\vec b⁠.
    −(u⃗−v⃗)=−u⃗+v⃗-(\vec u - \vec v) = -\vec u + \vec v (negative of a vector, Section 10.3). Basics: the same algebra rules

  3. Check, and picture both points

    PR→=(4b⃗−a⃗)−(3a⃗−2b⃗)=−4a⃗+6b⃗,QR→=(4b⃗−a⃗)−(a⃗+b⃗)=−2a⃗+3b⃗.\begin{aligned} \overrightarrow{PR} &= (4\vec b - \vec a) - (3\vec a - 2\vec b) \\ &= -4\vec a + 6\vec b, \\ \overrightarrow{QR} &= (4\vec b - \vec a) - (\vec a + \vec b) \\ &= -2\vec a + 3\vec b. \end{aligned}

    So PR→=2 QR→\overrightarrow{PR} = 2\,\overrightarrow{QR}⁠, and R lies beyond Q with PR:QR=2:1\text{PR} : \text{QR} = 2 : 1⁠. Also PQ→=OQ→−OP→=−2a⃗+3b⃗=QR→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = -2\vec a + 3\vec b = \overrightarrow{QR}⁠: R is as far beyond Q as Q is from P.

    x y O P Q R (i) R (ii) sample: a = î, b = ĵ OP = 3a − 2b OQ = a + b (i) OR = 5a/3 PR : RQ = 2 : 1 (ii) OR = 4b − a PR : QR = 2 : 1
    Drawn to scale, one scale on both axes, for the sample choice a⃗=i^\vec a = \hat i⁠, b⃗=j^\vec b = \hat j⁠, so that P is (3, −2-2⁠) and Q is (1, 1). The internal point (i) splits PQ in the ratio 2 : 1 and lies on the line of a⃗\vec a⁠, because OR→=53a⃗\overrightarrow{OR} = \frac53\vec a⁠. The external point (ii) is at (−1,4)(-1, 4)⁠, with Q the midpoint of P and R, so PR=2 QR\text{PR} = 2\,\text{QR}⁠. The answers hold for every a⃗\vec a and b⃗\vec b⁠; only the picture uses the sample.
    For external division the two vectors PR→\overrightarrow{PR} and QR→\overrightarrow{QR} point the same way (both from an end towards R), and the first is twice the second. Because QR→=PQ→\overrightarrow{QR} = \overrightarrow{PQ}⁠, going from P to Q and then the same step again lands on R.
    QR→=OR→−OQ→\overrightarrow{QR} = \overrightarrow{OR} - \overrightarrow{OQ} (Section 10.5.2); collinear vectors u⃗=λv⃗\vec u = \lambda\vec v (Section 10.5). Lesson: Examples 11–12
Result

(i) Internally: OR→=5a⃗3\overrightarrow{OR} = \frac{5\vec a}{3}⁠. (ii) Externally: OR→=4b⃗−a⃗\overrightarrow{OR} = 4\vec b - \vec a⁠.

(Textbook answer: (i) OR→=2(a⃗+b⃗)+(3a⃗−2b⃗)2+1=5a⃗3\overrightarrow{OR} = \frac{2(\vec a + \vec b) + (3\vec a - 2\vec b)}{2 + 1} = \frac{5\vec a}{3}⁠; (ii) OR→=2(a⃗+b⃗)−(3a⃗−2b⃗)2−1=4b⃗−a⃗\overrightarrow{OR} = \frac{2(\vec a + \vec b) - (3\vec a - 2\vec b)}{2 - 1} = 4\vec b - \vec a⁠.)

Check it with numbers

Take a⃗=i^\vec a = \hat i⁠, b⃗=j^\vec b = \hat j (in the xy-plane): P is (3, −2-2⁠) and Q is (1, 1).

(i) The coordinate section formula gives (2⋅1+1⋅33,2⋅1+1⋅(−2)3)=(53,0)\left(\frac{2 \cdot 1 + 1 \cdot 3}{3}, \frac{2 \cdot 1 + 1 \cdot (-2)}{3}\right) = \left(\frac53, 0\right)⁠, and 53a⃗=53i^\frac53\vec a = \frac53\hat i⁠, the same point, with 53≈1.666667\frac53 \approx 1.666667⁠. Then PR=2313≈2.403701\text{PR} = \frac23\sqrt{13} \approx 2.403701 and RQ=1313≈1.201850\text{RQ} = \frac13\sqrt{13} \approx 1.201850⁠, exactly in the ratio 2 : 1.

(ii) (2⋅1−1⋅31,2⋅1−1⋅(−2)1)=(−1,4)\left(\frac{2 \cdot 1 - 1 \cdot 3}{1}, \frac{2 \cdot 1 - 1 \cdot (-2)}{1}\right) = (-1, 4)⁠, and 4b⃗−a⃗=−i^+4j^4\vec b - \vec a = -\hat i + 4\hat j⁠, the same point. Then PR=213≈7.211103\text{PR} = 2\sqrt{13} \approx 7.211103 and QR=13≈3.605551\text{QR} = \sqrt{13} \approx 3.605551⁠, again exactly 2 : 1.

Common mistakes
  1. Swapping the weights. 2 OP→+OQ→3=7a⃗−3b⃗3\frac{2\,\overrightarrow{OP} + \overrightarrow{OQ}}{3} = \frac{7\vec a - 3\vec b}{3} is the point with PR:RQ=1:2\text{PR} : \text{RQ} = 1 : 2⁠, nearer to P. The weight mm goes with Q.
  2. Mixing up the letters. In the book's formula a⃗\vec a and b⃗\vec b are the position vectors of P and Q; here they are not. Substituting a⃗\vec a for OP→\overrightarrow{OP} gives a wrong answer.
  3. Losing a sign in part (ii). −(3a⃗−2b⃗)=−3a⃗+2b⃗-(3\vec a - 2\vec b) = -3\vec a + 2\vec b⁠, so the b⃗\vec b terms add up to 4b⃗4\vec b instead of cancelling.
What to take away
  1. Internal: OR→=m OQ→+n OP→m+n\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} + n\,\overrightarrow{OP}}{m + n}⁠; external: change both plus signs to minus.
  2. Each end is weighted by the part of the ratio next to the other end.
  3. Check with PR→=mn RQ→\overrightarrow{PR} = \frac{m}{n}\,\overrightarrow{RQ} (internal) or PR→=mn QR→\overrightarrow{PR} = \frac{m}{n}\,\overrightarrow{QR} (external).

Every fact used, and where it comes from

FactUsed inSource
Section formula, internal divisionPart (i), Steps 1, 2Section 10.5.3, Case I, Fig 10.16 (Lesson: the section formula)
Section formula, external divisionPart (ii), Steps 1, 2Section 10.5.3, Case II, Fig 10.17
Multiplying out and collecting vector termsPart (i), Step 2; Part (ii), Step 2Sections 10.4 and 10.5 (Basics: the same algebra rules)
Vector joining two points: terminal minus initialPart (i), Step 3; Part (ii), Step 3Section 10.5.2
Collinear vectors u⃗=λv⃗\vec u = \lambda\vec vPart (ii), Step 3Section 10.5
The book's working and answerResultExample 11, Section 10.5.3 (Lesson: Examples 11–12)
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