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Chapter 10 · Vector Algebra

Miscellaneous · Q9: Dividing 2a⃗+b⃗2\vec a+\vec b⁠, a⃗−3b⃗\vec a-3\vec b externally 1 : 2; P is the midpoint of RQ

Miscellaneous ExerciseSection formula, external division · Section 10.5.3
The question

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a⃗+b⃗)(2\vec a + \vec b) and (a⃗−3b⃗)(\vec a - 3\vec b) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

Exam answer

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

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

OR→=m OQ→−n OP→m−n=1(a⃗−3b⃗)−2(2a⃗+b⃗)1−2=a⃗−3b⃗−4a⃗−2b⃗−1=−3a⃗−5b⃗−1\begin{aligned} \overrightarrow{OR} &= \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n} \\ &= \frac{1(\vec a - 3\vec b) - 2(2\vec a + \vec b)}{1 - 2} \\ &= \frac{\vec a - 3\vec b - 4\vec a - 2\vec b}{-1} \\ &= \frac{-3\vec a - 5\vec b}{-1} \end{aligned}

∴ OR→=3a⃗+5b⃗\therefore\ \overrightarrow{OR} = 3\vec a + 5\vec b⁠.

The position vector of the midpoint of RQ is

OR→+OQ→2=(3a⃗+5b⃗)+(a⃗−3b⃗)2=4a⃗+2b⃗2=2a⃗+b⃗=OP→\begin{aligned} &\frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2} \\ &= \frac{(3\vec a + 5\vec b) + (\vec a - 3\vec b)}{2} \\ &= \frac{4\vec a + 2\vec b}{2} = 2\vec a + \vec b \\ &= \overrightarrow{OP} \end{aligned}

∴\therefore P is the midpoint of the line segment RQ. Hence proved.

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

The one idea this problem is built on

The section formula works with any position vectors, even ones written in terms of other vectors a⃗\vec a and b⃗\vec b⁠: substitute and simplify as in ordinary algebra. For external division in the ratio m:nm : n⁠, OR→=m OQ→−n OP→m−n\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n}⁠. The midpoint is the special case "average of the two ends".

Given, to find and show, and the plan
Given
OP→=2a⃗+b⃗\overrightarrow{OP} = 2\vec a + \vec b⁠, OQ→=a⃗−3b⃗\overrightarrow{OQ} = \vec a - 3\vec b⁠; R divides the line joining P and Q externally in the ratio 1 : 2.
To find
OR→\overrightarrow{OR}⁠.
To show
P is the midpoint of RQ.
Plan
(1) Write the external section formula with m=1m = 1⁠, n=2n = 2⁠. (2) Substitute and simplify. (3) Find the midpoint of RQ and compare with OP→\overrightarrow{OP}⁠. (4) Check the ratio and the picture.

Basics you need

1External division

R divides PQ externally in the ratio m:nm : n when R is on the line PQ, outside the segment, with PR:QR=m:n\text{PR} : \text{QR} = m : n⁠. Basics: dividing in m : n

2Section formula and midpoint

External: OR→=m OQ→−n OP→m−n\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n} (m≠nm \ne n⁠); midpoint of PQ: OP→+OQ→2\frac{\overrightarrow{OP} + \overrightarrow{OQ}}{2} (Section 10.5.3). Lesson: the section formula

3Algebra with vectors

Vectors can be expanded and collected like ordinary symbols: 2(2a⃗+b⃗)=4a⃗+2b⃗2(2\vec a + \vec b) = 4\vec a + 2\vec b (distributive laws, Section 10.5.1). Basics: expanding brackets

Solution

  1. Choose the formula

    The ratio 1 : 2 is written in the order P, Q, so m=1m = 1⁠, n=2n = 2⁠, and PR:QR=1:2\text{PR} : \text{QR} = 1 : 2⁠. For external division,

    OR→=m OQ→−n OP→m−n.\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n}.
    PR is half of QR, so R is nearer to P: it lies beyond P, on the side away from Q. Since m≠nm \ne n⁠, the denominator m−n=−1m - n = -1 is not zero and the formula applies.
    Section formula, Case II (Section 10.5.3, Fig 10.17). Basics: dividing in m : n

  2. Substitute and simplify

    OR→=1(a⃗−3b⃗)−2(2a⃗+b⃗)1−2=a⃗−3b⃗−4a⃗−2b⃗−1=−3a⃗−5b⃗−1=3a⃗+5b⃗.\begin{aligned} \overrightarrow{OR} &= \frac{1(\vec a - 3\vec b) - 2(2\vec a + \vec b)}{1 - 2} \\ &= \frac{\vec a - 3\vec b - 4\vec a - 2\vec b}{-1} \\ &= \frac{-3\vec a - 5\vec b}{-1} = 3\vec a + 5\vec b. \end{aligned}

    So OR→=3a⃗+5b⃗\overrightarrow{OR} = 3\vec a + 5\vec b.

    Expand the brackets: −2(2a⃗+b⃗)=−4a⃗−2b⃗-2(2\vec a + \vec b) = -4\vec a - 2\vec b⁠. Collect: a⃗−4a⃗=−3a⃗\vec a - 4\vec a = -3\vec a and −3b⃗−2b⃗=−5b⃗-3\vec b - 2\vec b = -5\vec b⁠. Dividing by −1-1 changes both signs.
    Distributive laws for scalar multiples (Section 10.5.1). The same steps as Example 11, part (ii). Lesson: Examples 11–12

  3. P is the midpoint of RQ

    OR→+OQ→2=(3a⃗+5b⃗)+(a⃗−3b⃗)2=4a⃗+2b⃗2=2a⃗+b⃗=OP→.\begin{aligned} \frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2} &= \frac{(3\vec a + 5\vec b) + (\vec a - 3\vec b)}{2} \\ &= \frac{4\vec a + 2\vec b}{2} = 2\vec a + \vec b = \overrightarrow{OP}. \end{aligned}

    The midpoint of RQ has the position vector of P, so P is the midpoint of RQ.

    Two points with the same position vector (from the same origin O) are the same point. The midpoint formula is the section formula with m=n=1m = n = 1⁠.
    Midpoint, the Remark after Case II (Section 10.5.3). Lesson: the section formula

  4. Check the ratio, and the picture

    PR→=(3a⃗+5b⃗)−(2a⃗+b⃗)=a⃗+4b⃗,QR→=(3a⃗+5b⃗)−(a⃗−3b⃗)=2a⃗+8b⃗.\begin{aligned} \overrightarrow{PR} &= (3\vec a + 5\vec b) - (2\vec a + \vec b) = \vec a + 4\vec b, \\ \overrightarrow{QR} &= (3\vec a + 5\vec b) - (\vec a - 3\vec b) = 2\vec a + 8\vec b. \end{aligned}

    So QR→=2 PR→\overrightarrow{QR} = 2\,\overrightarrow{PR}⁠: both point the same way from P and from Q, and PR:QR=1:2\text{PR} : \text{QR} = 1 : 2⁠, as required. Also PQ→=−a⃗−4b⃗=−PR→\overrightarrow{PQ} = -\vec a - 4\vec b = -\overrightarrow{PR}⁠, so R and Q are equal distances on opposite sides of P.

    R (3a + 5b) P (2a + b) Q (a − 3b) RP = d PQ = d PR = d QR = 2d
    The order of the points on the line, true to scale with one unit d=∣a⃗+4b⃗∣d = |\vec a + 4\vec b|⁠: RP=PQ=d\text{RP} = \text{PQ} = d and RQ=2d\text{RQ} = 2d⁠. R lies beyond P, with PR:QR=1:2\text{PR} : \text{QR} = 1 : 2⁠.
    This works for every a⃗\vec a⁠, b⃗\vec b with a⃗+4b⃗≠0⃗\vec a + 4\vec b \ne \vec 0 (otherwise P and Q would coincide and there would be no line PQ). The distances RP, PQ and QR are all multiples of the same length d=∣a⃗+4b⃗∣d = |\vec a + 4\vec b|⁠, so the picture is exact whatever a⃗\vec a and b⃗\vec b are.
    PR→=OR→−OP→\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} (Section 10.5.2); see also Exercise 10.2 Q15. Lesson: put it all together
Result

OR→=3a⃗+5b⃗\overrightarrow{OR} = 3\vec a + 5\vec b⁠. The midpoint of RQ is (3a⃗+5b⃗)+(a⃗−3b⃗)2=2a⃗+b⃗=OP→\frac{(3\vec a + 5\vec b) + (\vec a - 3\vec b)}{2} = 2\vec a + \vec b = \overrightarrow{OP}⁠, so P is the midpoint of RQ. Hence proved.

(Textbook answer: 3a⃗+5b⃗3\vec a + 5\vec b⁠.)

Check it with numbers

Take the sample a⃗=i^\vec a = \hat i⁠, b⃗=j^\vec b = \hat j⁠. Then P is (2, 1, 0), Q is (1, −3-3⁠, 0) and the formula gives R = (3, 5, 0).

Coordinates: 1(1)−2(2)1−2=3\frac{1(1) - 2(2)}{1 - 2} = 3 and 1(−3)−2(1)1−2=5\frac{1(-3) - 2(1)}{1 - 2} = 5⁠. The midpoint of RQ is (3+12,5−32,0)=(2,1,0)\left(\frac{3 + 1}{2}, \frac{5 - 3}{2}, 0\right) = (2, 1, 0)⁠, which is P.

Distances: PR=1+16=17≈4.123106\text{PR} = \sqrt{1 + 16} = \sqrt{17} \approx 4.123106 and QR=4+64=217≈8.246211\text{QR} = \sqrt{4 + 64} = 2\sqrt{17} \approx 8.246211⁠, in the ratio 1 : 2.

Common mistakes
  1. Using the internal formula. 1 OQ→+2 OP→3=5a⃗−b⃗3\frac{1\,\overrightarrow{OQ} + 2\,\overrightarrow{OP}}{3} = \frac{5\vec a - \vec b}{3} is the point between P and Q, not the external one.
  2. Swapping the weights. m=1m = 1 goes with Q and n=2n = 2 with P.
  3. Losing a sign. −2(2a⃗+b⃗)=−4a⃗−2b⃗-2(2\vec a + \vec b) = -4\vec a - 2\vec b⁠, and the denominator 1−2=−11 - 2 = -1 flips every sign.
What to take away
  1. External division: OR→=m OQ→−n OP→m−n\overrightarrow{OR} = \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n}⁠.
  2. The section formula works with position vectors written in terms of a⃗\vec a and b⃗\vec b⁠.
  3. To show a point is a midpoint, compare its position vector with the average of the two ends.

Every fact used, and where it comes from

FactUsed inSource
Section formula, external divisionSteps 1, 2Section 10.5.3, Case II, Fig 10.17 (Lesson: the section formula)
Distributive laws for scalar multiplesSteps 2, 3Section 10.5.1 (Basics: expanding brackets)
Midpoint OP→+OQ→2\frac{\overrightarrow{OP} + \overrightarrow{OQ}}{2}Step 3Section 10.5.3, Remark
Vector joining two pointsStep 4Section 10.5.2
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q9 (Lesson: put it all together)
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