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Chapter 10 · Vector Algebra

Exercise 10.2 · Q15: Point dividing i^+2j^−k^\hat i+2\hat j-\hat k⁠, −i^+j^+k^-\hat i+\hat j+\hat k in the ratio 2 : 1, internally and externally

Exercise 10.2Section formula · 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 i^+2j^−k^\hat i + 2\hat j - \hat k and −i^+j^+k^-\hat i + \hat j + \hat k respectively, in the ratio 2 : 1

(i) internally    (ii) externally

Exam answer

Here OP→=i^+2j^−k^\overrightarrow{OP} = \hat i + 2\hat j - \hat k⁠, OQ→=−i^+j^+k^\overrightarrow{OQ} = -\hat i + \hat j + \hat k 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=12+1[2(−i^+j^+k^)+(i^+2j^−k^)]=−i^+4j^+k^3\begin{aligned} \overrightarrow{OR} &= \frac{m\,\overrightarrow{OQ} + n\,\overrightarrow{OP}}{m + n} \\ &= \frac{1}{2 + 1}\big[2(-\hat i + \hat j + \hat k) \\ &\quad + (\hat i + 2\hat j - \hat k)\big] \\ &= \frac{-\hat i + 4\hat j + \hat k}{3} \end{aligned}

∴ OR→=−13i^+43j^+13k^\therefore\ \overrightarrow{OR} = -\frac13\hat i + \frac43\hat j + \frac13\hat k⁠.

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

OR→=m OQ→−n OP→m−n=12−1[2(−i^+j^+k^)−(i^+2j^−k^)]=−3i^+0j^+3k^\begin{aligned} \overrightarrow{OR} &= \frac{m\,\overrightarrow{OQ} - n\,\overrightarrow{OP}}{m - n} \\ &= \frac{1}{2 - 1}\big[2(-\hat i + \hat j + \hat k) \\ &\quad - (\hat i + 2\hat j - \hat k)\big] \\ &= -3\hat i + 0\hat j + 3\hat k \end{aligned}

∴ OR→=−3i^+3k^\therefore\ \overrightarrow{OR} = -3\hat i + 3\hat k⁠.

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 the point R that 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 m=2m = 2⁠, P gets n=1n = 1⁠. Then it is substitution and collecting components.

Given, to find, and the plan
Given
OP→=i^+2j^−k^\overrightarrow{OP} = \hat i + 2\hat j - \hat k⁠, OQ→=−i^+j^+k^\overrightarrow{OQ} = -\hat i + \hat j + \hat k⁠; R divides the line joining P and Q in the ratio 2 : 1. As points: P(1, 2, −1-1⁠) and Q(−1-1⁠, 1, 1).
To find
OR→\overrightarrow{OR} when the division is (i) internal, (ii) external.
Plan
For each part: (1) choose the formula with m=2m = 2⁠, n=1n = 1⁠; (2) substitute and collect the components; (3) check the ratio with PR→\overrightarrow{PR} and RQ→\overrightarrow{RQ} (or QR→\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

With a⃗\vec a⁠, b⃗\vec b the position vectors of P and Q: 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} (Section 10.5.3, Cases I and II, Figs 10.16 and 10.17). Lesson: the section formula

3Working in components

Vectors are added, subtracted and multiplied by numbers component by component (Section 10.5.1, (i), (ii) and (iv)). Lesson: components

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, n=1n = 1 and 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}.
    PR is twice RQ, so R is nearer to Q; that is why Q carries the larger weight m=2m = 2⁠. The book writes the formula as mb⃗+na⃗m+n\frac{m\vec b + n\vec a}{m + n} with a⃗=OP→\vec a = \overrightarrow{OP}⁠, b⃗=OQ→\vec b = \overrightarrow{OQ}⁠.
    Section formula, Case I (Section 10.5.3, Fig 10.16). Basics: dividing in m : n

  2. Substitute and collect

    OR→=13[2(−i^+j^+k^)+(i^+2j^−k^)]=13[(−2+1)i^+(2+2)j^+(2−1)k^]=−i^+4j^+k^3=−13i^+43j^+13k^.\begin{aligned} \overrightarrow{OR} &= \tfrac13\big[2(-\hat i + \hat j + \hat k) \\ &\quad + (\hat i + 2\hat j - \hat k)\big] \\ &= \tfrac13\big[(-2 + 1)\hat i + (2 + 2)\hat j \\ &\quad + (2 - 1)\hat k\big] \\ &= \frac{-\hat i + 4\hat j + \hat k}{3} \\ &= -\frac13\hat i + \frac43\hat j + \frac13\hat k. \end{aligned}

    So OR→=−13i^+43j^+13k^\overrightarrow{OR} = -\frac13\hat i + \frac43\hat j + \frac13\hat k.

    First double every component of OQ→\overrightarrow{OQ}⁠: −2i^+2j^+2k^-2\hat i + 2\hat j + 2\hat k⁠. Then add OP→\overrightarrow{OP} component by component, and finally divide each component by 2+1=32 + 1 = 3⁠.
    Sum and scalar multiple by components (Section 10.5.1, (i) and (iv)). Lesson: components

  3. Check that PR is twice RQ

    PR→=(−13−1)i^+(43−2)j^+(13+1)k^=−43i^−23j^+43k^,RQ→=(−1+13)i^+(1−43)j^+(1−13)k^=−23i^−13j^+23k^.\begin{aligned} \overrightarrow{PR} &= \left(-\tfrac13 - 1\right)\hat i + \left(\tfrac43 - 2\right)\hat j \\ &\quad + \left(\tfrac13 + 1\right)\hat k \\ &= -\tfrac43\hat i - \tfrac23\hat j + \tfrac43\hat k, \\ \overrightarrow{RQ} &= \left(-1 + \tfrac13\right)\hat i + \left(1 - \tfrac43\right)\hat j \\ &\quad + \left(1 - \tfrac13\right)\hat k \\ &= -\tfrac23\hat i - \tfrac13\hat j + \tfrac23\hat k. \end{aligned}

    So PR→=2 RQ→\overrightarrow{PR} = 2\,\overrightarrow{RQ}⁠: R lies between P and Q with PR:RQ=2:1\text{PR} : \text{RQ} = 2 : 1⁠.

    Case I in the book starts from exactly this: R divides PQ internally in the ratio m:nm : n when m RQ→=n PR→m\,\overrightarrow{RQ} = n\,\overrightarrow{PR}⁠. Both pieces point the same way (from P towards Q), and the first is twice the second.
    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. With m=2m = 2⁠, n=1n = 1 and 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)⁠, which is why the plus signs become minus signs. Since PR is longer than QR, R lies beyond Q, on the side away from P.
    Section formula, Case II (Section 10.5.3, Fig 10.17). Lesson: the section formula

  2. Substitute and collect

    OR→=12−1[2(−i^+j^+k^)−(i^+2j^−k^)]=(−2−1)i^+(2−2)j^+(2+1)k^=−3i^+0j^+3k^=−3i^+3k^.\begin{aligned} \overrightarrow{OR} &= \tfrac{1}{2 - 1}\big[2(-\hat i + \hat j + \hat k) \\ &\quad - (\hat i + 2\hat j - \hat k)\big] \\ &= (-2 - 1)\hat i + (2 - 2)\hat j + (2 + 1)\hat k \\ &= -3\hat i + 0\hat j + 3\hat k = -3\hat i + 3\hat k. \end{aligned}

    So OR→=−3i^+3k^\overrightarrow{OR} = -3\hat i + 3\hat k.

    The denominator is 2−1=12 - 1 = 1⁠. The minus sign in front of the bracket changes every sign inside it: −(i^+2j^−k^)=−i^−2j^+k^-(\hat i + 2\hat j - \hat k) = -\hat i - 2\hat j + \hat k⁠. The j^\hat j components cancel, so R lies in the xz-plane.
    Difference by components (Section 10.5.1, (ii)). Lesson: components

  3. Check, and picture both points

    PR→=(−3−1)i^+(0−2)j^+(3+1)k^=−4i^−2j^+4k^,QR→=(−3+1)i^+(0−1)j^+(3−1)k^=−2i^−j^+2k^.\begin{aligned} \overrightarrow{PR} &= (-3 - 1)\hat i + (0 - 2)\hat j + (3 + 1)\hat k \\ &= -4\hat i - 2\hat j + 4\hat k, \\ \overrightarrow{QR} &= (-3 + 1)\hat i + (0 - 1)\hat j + (3 - 1)\hat k \\ &= -2\hat i - \hat j + 2\hat k. \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→=−2i^−j^+2k^=QR→\overrightarrow{PQ} = -2\hat i - \hat j + 2\hat k = \overrightarrow{QR}⁠: Q is the midpoint of PR.

    P R (i) Q R (ii) 2 1 PR = 6 QR = 3 |PQ| = 3
    The line through P and Q drawn true to scale, one scale throughout: ∣PQ→∣=3|\overrightarrow{PQ}| = 3⁠. The internal point R (i) is 2 from P and 1 from Q; the external point R (ii) is 6 from P and 3 from Q, beyond Q.
    For external division both PR→\overrightarrow{PR} and QR→\overrightarrow{QR} point from an end towards R, the same way, and the first is twice the second. 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). The same pattern as Example 11. Lesson: Examples 11–12
Result

(i) Internally: OR→=−13i^+43j^+13k^\overrightarrow{OR} = -\frac13\hat i + \frac43\hat j + \frac13\hat k⁠. (ii) Externally: OR→=−3i^+3k^\overrightarrow{OR} = -3\hat i + 3\hat k⁠.

(Textbook answer: (i) −13i^+43j^+13k^-\frac13\hat i + \frac43\hat j + \frac13\hat k⁠; (ii) −3i^+3k^-3\hat i + 3\hat k⁠.)

Check it with numbers

(i) R is about (−0.333333,1.333333,0.333333)(-0.333333, 1.333333, 0.333333)⁠. The coordinate section formula agrees:

(2(−1)+13,2(1)+23,2(1)+(−1)3)=(−13,43,13).\begin{aligned} &\left(\tfrac{2(-1) + 1}{3}, \tfrac{2(1) + 2}{3}, \tfrac{2(1) + (-1)}{3}\right) \\ &= \left(-\tfrac13, \tfrac43, \tfrac13\right). \end{aligned}

Distances: PR=169+49+169=2\text{PR} = \sqrt{\frac{16}{9} + \frac49 + \frac{16}{9}} = 2 and RQ=49+19+49=1\text{RQ} = \sqrt{\frac49 + \frac19 + \frac49} = 1⁠.

(ii) The coordinate formula for external division gives

(2(−1)−11,2(1)−21,2(1)−(−1)1)=(−3,0,3).\begin{aligned} &\left(\tfrac{2(-1) - 1}{1}, \tfrac{2(1) - 2}{1}, \tfrac{2(1) - (-1)}{1}\right) \\ &= (-3, 0, 3). \end{aligned}

Distances: PR=16+4+16=6\text{PR} = \sqrt{16 + 4 + 16} = 6 and QR=4+1+4=3\text{QR} = \sqrt{4 + 1 + 4} = 3⁠, again 2 : 1.

Common mistakes
  1. Swapping the weights. 2 OP→+OQ→3=13i^+53j^−13k^\frac{2\,\overrightarrow{OP} + \overrightarrow{OQ}}{3} = \frac13\hat i + \frac53\hat j - \frac13\hat k is the point with PR:RQ=1:2\text{PR} : \text{RQ} = 1 : 2⁠, nearer to P.
  2. Swapping the order in the external formula. 2 OP→−OQ→=3i^+3j^−3k^2\,\overrightarrow{OP} - \overrightarrow{OQ} = 3\hat i + 3\hat j - 3\hat k lies beyond P, not beyond Q.
  3. Sign slips inside the bracket. In part (ii), −(i^+2j^−k^)-(\hat i + 2\hat j - \hat k) gives +k^+\hat k⁠, so the k^\hat k component is 2+1=32 + 1 = 3⁠, not 1.
What to take away
  1. Internal: m OQ→+n OP→m+n\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: PR:RQ=m:n\text{PR} : \text{RQ} = m : n⁠, so Q gets mm and P gets nn⁠.
  3. Check with PR→=mnRQ→\overrightarrow{PR} = \frac{m}{n}\overrightarrow{RQ} (internal) or PR→=mnQR→\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
Sum, difference and scalar multiple by componentsPart (i), Step 2; Part (ii), Step 2Section 10.5.1, (i), (ii), (iv)
Vector joining two points: terminal minus initialPart (i), Step 3; Part (ii), Step 3Section 10.5.2
The answerResultNCERT answer key, Exercise 10.2 Q15 (Lesson: components, lengths and unit vectors)
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