Position vectors, the vector joining two points, and dividing a segment in a ratio

What to assume they know

• A point's position vector and its magnitude, from the second topic of the first module
• The triangle law and the difference of two vectors, from the first topic of this module
• Multiplication of a vector by a scalar, from the second topic of this module
• Component form, and the componentwise rules for sums and differences, from the previous topic
• Dividing a segment in a given ratio in coordinate geometry, from Class XI
• The converse of the Pythagoras theorem, at the level of recognising a right angle from three side lengths

What they should be able to do

• Derive the vector joining two points from their two position vectors using the triangle law
• Write that vector in components and say which point supplies the subtrahend
• Compute its magnitude and recognise the formula as one already met twice in this chapter
• Distinguish internal from external division of a segment by a third point
• Reproduce the chapter's derivation of the internal case, including the step it marks with a question of its own
• Say which coefficient pairs with which endpoint, and explain why the pairing looks reversed
• State the external case, and say what the chapter does instead of proving it
• Obtain the midpoint as the case where the two parts are equal
• Divide the same segment internally and externally in the same ratio, and compare the two answers
• Prove a triangle right-angled from three side lengths, without a scalar product
• Find the ratio in which a point divides a segment through three collinear points
• Show that a named point is the midpoint of a constructed segment

Where it usually goes wrong

• "The joining vector is the first point's position vector minus the second." It is terminal minus initial. Example 10 fixes the order in words before it computes, and the chapter's exercises say directed from precisely because the order is not recoverable from the letters alone.
• "The ratio numbers pair with the nearer point." They cross. The number describing the part next to the first point multiplies the second point's position vector. This is the error that costs marks, and the chapter offers no warning.
• "External division just means putting a minus sign somewhere." Two signs change, one in the numerator and one in the denominator, and the denominator's change is the one students forget. When the two ratio numbers are equal the external formula divides by zero, which is the correct answer: a point cutting a segment externally in the ratio one to one does not exist.
• "The midpoint formula is a separate result." It is the internal formula with the two ratio numbers equal, and it is stated as a Remark for exactly that reason.
• "A right angle can only be found with a scalar product." Example 12 finds one without, using three side lengths, two pages before the scalar product is defined. The same triangle appears again in the scalar product exercise, where the other route is the fast one.
• "If a worked example's answer loses one of the given vectors, something has gone wrong." Example 11's internal answer does exactly that. The arithmetic cancels; the answer is right.
• "Collinear points cannot be divided in a ratio." They are the only points that can. Miscellaneous Exercise Q8 establishes collinearity first and then reads the ratio off the two joining vectors.
• "The chapter's Why? is rhetorical." It marks a genuine gap in the derivation on Part II p. 352 and there are only three such marks in the whole chapter. Each one is a place where the book is asking the reader to supply a step.
• "The magnitude formula in this section is a new formula." It is the third printing of one formula, with coordinate gaps where the earlier two had coordinates.

Questions to check understanding

• Write the vector between two named points, directed one stated way, and give its magnitude — the form of Example 10 and Miscellaneous Exercise Q2
• Produce the unit vector along the segment between two named points — the form of Exercise 10.2 Q8
• Divide a segment internally in a stated ratio — the form of Example 11 part one and Exercise 10.2 Q15 part one
• Divide the same segment externally in the same ratio, and compare — the form of Example 11 part two
• Find a midpoint from two position vectors — the form of Exercise 10.2 Q16
• Prove a triangle right-angled from three position vectors without using a product — the form of Example 12 and Exercise 10.2 Q17
• Show three points collinear and find the ratio one cuts — the form of Miscellaneous Exercise Q8
• Divide externally and then show a given point is the midpoint of a constructed segment — the form of Miscellaneous Exercise Q9
• Derive the external formula from the internal one

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• §10.5.2 and Fig 10.15 (Part II p. 351). Two points are named with their coordinates; both are joined to the origin; the triangle law is applied to the resulting triangle; and the joining vector falls out as the second position vector minus the first. Read off the printed page: Fig 10.15 is an axis frame with the three hatted axis arrows drawn at the origin, the two points marked with their coordinate triples, both joined to the origin by dashed lines, and the joining arrow drawn solid between them. The whole section is one subtraction, and it is the hinge of the chapter: every later worked example that starts from named points passes through this line, and it is nine lines long.
• Terminal minus initial (§10.5.2, Part II p. 351, and Example 10, Part II p. 352). Substituting the component forms and collecting terms gives the three coordinate gaps, each taken in the order terminal minus initial. Example 10 then makes the ordering explicit, naming which of two given points is the start and which the finish before computing anything. Verified: the vector from a point with coordinates two, three and zero to a point with coordinates minus one, minus two and minus four has components minus three, minus five and minus four. Reversing the two points negates all three components, which is the same observation the direction cosine topic makes about Exercise 10.2 Q13, and the two videos should use the same words for it.
• The magnitude of the joining vector (§10.5.2, Part II p. 351). The square root of the summed squared coordinate gaps. Verified as the chapter's third printing of one formula: it appears on Part II p. 339 as the Class XI distance formula applied to a position vector, on Part II p. 348 as a result derived from two right triangles, and here as the same expression with coordinate gaps in place of coordinates. The chapter never says the three are one formula. Section 3 should show all three lines together; it costs thirty seconds and it makes the rest of the chapter's arithmetic feel like one idea instead of three.
• §10.5.3, Case I, and Fig 10.16 (Part II p. 352). A third point on the segment between two given ones. The chapter fixes the ratio by an equation between two scaled segments, then writes each of those two segments as a difference of position vectors, substitutes, and rearranges. Read off the printed page: Fig 10.16 puts the origin at the left with the two given points and the cutting point forming a triangle to its right, the two parts of the divided segment labelled with the two ratio letters, and three arrows drawn from the origin.
• The question mark inside the proof (Part II p. 352). At the step where the two scaled differences are set equal, the chapter prints a parenthetical Why? and moves on. Verified as one of exactly three such marks in the chapter — the others are on Part II p. 356 and Part II p. 360 — confirmed by grep across all thirty-nine extracted pages. It is not decoration: the step needs the definition of the ratio from two lines earlier, and a student who does not go back gets nothing from the derivation. Section 5 should answer the chapter's question out loud.
• The cross pairing (Part II p. 352). The resulting formula weights the second point's position vector by the first ratio number and the first point's by the second. Verified, and it is the single most common source of error in this section: the number describing the part nearer the first point multiplies the position vector of the second point. The chapter states the formula and never remarks on the crossing. Section 6 should make it visible by drawing the two parts and the two coefficients on one segment and joining them with crossing lines.
• §10.5.3, Case II, and Fig 10.17 (Part II p. 353). The external case. The chapter states the result and explicitly declines to derive it, handing the verification over to the reader instead. Verified as the only place in the chapter where it does this — confirmed by grep across all thirty-nine extracted pages. The formula differs from the internal one in exactly two signs. Read off the printed page: Fig 10.17 puts the cutting point beyond the second given point, with the two labelled parts being the long stretch from the first given point and the short overshoot past the second. Section 7 should do the derivation the book skipped, because it is the same four lines with one sign changed and it makes the two formulas one formula.
• The midpoint Remark (Part II p. 353). Setting the two ratio numbers equal in the internal formula gives the average of the two position vectors. Verified: the two numbers cancel top and bottom, leaving one half of the sum. One line, and it is the most-used consequence of the whole section.
• Example 11 (Part II p. 353). Two points given by position vectors built from two other vectors, divided in the ratio two to one, internally and externally. Verified: the internal answer collapses to five thirds of the first building vector, with the second cancelling out entirely; the external answer is four times the second minus the first. The cancellation in the internal case is worth pausing on — it is not a coincidence to be explained, it is arithmetic, and a student who expects the answer to look symmetrical will assume they have made a mistake.
• Example 12 (Part II pp. 353–354). Three points, given by position vectors, are shown to make a triangle with a right angle in it. The solution computes the three side vectors, and then proves the right angle by adding two squared lengths and matching the third. Verified: the three squared lengths are forty-one, six and thirty-five, and six plus thirty-five is forty-one. The scalar product does not exist yet — it arrives on Part II p. 355, two pages later — so the chapter is forced to use the converse of Pythagoras, and it does so without comment. Section 10 should say what is happening: the same fact will be provable in one line in two pages' time, and seeing both routes is worth more than either.
• The same triangle, three times (Part II pp. 353, 355 and 362). Verified by comparing the three printed triples: Example 12, Exercise 10.2 Q17 and Exercise 10.3 Q17 all set the same three position vectors as the vertices of a triangle with a right angle. Example 12 and Exercise 10.3 Q17 list them in the same order; Exercise 10.2 Q17 relabels the three cyclically. The chapter never observes the repetition, and the third appearance sits in the scalar product exercise where a different and shorter method is available. Section 11 is a ninety-second payoff and it is the cleanest bridge this module has into the next.
• Exercise 10.2 Q8, Q15 and Q16 (Part II pp. 354–355). A unit vector along the segment between two named points; the internal and external division of a segment in the ratio two to one; and a midpoint. Verified: in Q8 the joining vector has all three components equal to three, so its magnitude is three root three and the unit vector has one over root three in each place. In Q15 the internal answer has components minus one, four and one, all over three, and the external answer has components minus three, zero and three. In Q16 the midpoint has coordinates three, two and one. Q15 is Example 11 with numbers instead of letters and the two should be taught as one item.
• Miscellaneous Exercise Q2, Q8 and Q9 (Part II p. 372). Both the three plain numbers and the length, for the arrow between two general points; a ratio in which a middle point divides a segment through three collinear points; and an external division followed by a midpoint claim. Verified: Q2 asks the student to restate §10.5.2 in full generality and nothing else. In Q8 the two joining vectors have components four, two, six and six, three, nine, the second being one and a half times the first, so the three points are collinear and the middle one cuts the segment in the ratio two to three. In Q9 the external division in the ratio one to two gives three times the first building vector plus five times the second, and averaging that with the second given point returns the first — which is the midpoint claim. Q9 is the only item in the chapter that runs the section formula and the midpoint Remark against each other, and it is the right closing exercise for this topic.
• Summary bullet eight (Part II p. 374). The internal and external division formulas, printed together, with the internal one written in the reverse term order from the running text on Part II p. 352. Verified as the same formula: the two terms are simply added the other way round. Note it in passing so a student comparing the two pages does not think they disagree. There is no Summary bullet for the vector joining two points and none for the midpoint — confirmed on the page image of all three Summary pages — even though Miscellaneous Exercise Q2 asks for the first and Exercise 10.2 Q16 for the second.

Figures to have open

• A redraw of Fig 10.15 (Part II p. 351): an axis frame with the three hatted axis arrows at the origin, two points with their coordinate triples, both joined to the origin, and the joining arrow drawn between them. The chapter's own drawing.
• A redraw of Fig 10.16 (Part II p. 352) for sections 5 and 6, with the two parts of the divided segment labelled and the three arrows from the origin drawn. Reuse the same drawing for section 6 with crossing lines added between the two labels and the two far endpoints; the crossing lines are not in the book.
• A redraw of Fig 10.17 (Part II p. 353) for section 7, with the cutting point beyond the second given point and the two parts labelled as the chapter labels them. Draw it beside the internal figure at the same scale so the two can be compared; the chapter prints them on facing pages and they are never seen together.
• A single segment reused across sections 4, 8 and 9, with the cutting point moving along it and past its end, so that internal, midpoint and external are three positions of one drawing rather than three drawings.
• A three-panel timeline for section 11 carrying the three pages on which the same triangle appears. Not in the book; the chapter draws the triangle no times at all.
• No figure is needed for sections 2, 3, 10 or 12 beyond the kit shapes named above.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 10 "Vector Algebra", §10.5.2 Vector joining two points, Part II p. 351, with Fig 10.15
• §10.5.3 Section formula, Case I with Fig 10.16, Part II p. 352; Case II with Fig 10.17, and the Remark on the midpoint, Part II p. 353
• Examples 10, 11 and 12, Part II pp. 352–354
• Exercise 10.2, questions 8, 15, 16 and 17, Part II pp. 354–355
• Exercise 10.3, question 17, Part II p. 362, cited only for the repetition
• Miscellaneous Exercise, questions 2, 8 and 9, Part II p. 372
• The magnitude of a position vector, §10.2, Part II p. 339, and the same formula derived, §10.5.1, Part II p. 348
• Summary, the section formula bullet, Part II p. 374