A product that returns a number, and the angle you can extract from it

What to assume they know

• Vectors, magnitude, and the fact that a magnitude is never negative, from the first module
• Component form and the three axis unit vectors, from the third topic of the second module
• Vector addition, its two properties, and multiplication by a scalar, from the first two topics of the second module
• The vector joining two points, from the fourth topic of the second module
• Cosine on the interval from zero to a straight angle, and its sign either side of a right angle
• Composition of functions and pointwise multiplication of functions, from Class XI, at the level of recognising that both exist

What they should be able to do

• Explain the analogy the chapter uses to justify defining two products
• State Definition 2 and identify which of its three ingredients is the new one
• Say what the definition cannot reach, and state the convention that covers it
• Use the sign of the product to decide whether an angle is acute, right or obtuse
• Write perpendicularity as a single equation and use it both ways
• Reproduce the nine products of the three axis unit vectors from two observations
• Extract the angle between two vectors from the product and the two magnitudes
• State the two properties of the scalar product and say which section's Property numbers they are
• Derive the component formula from the definition, using both properties and one observation
• Prove the two inequalities the chapter names, and say which one is used in the other
• Turn the equality case of the second inequality into a test for collinearity
• Decide what can be concluded from a product being zero, and what cannot

Where it usually goes wrong

• "The scalar product of two vectors is a vector." It is a number, and the chapter's very first observation says so before anything else. Every later error in this section is downstream of forgetting it.
• "A product of zero means the two vectors are perpendicular." Only if both are nonzero. Exercise 10.3 Q12 is built on exactly this gap, and its answer is that nothing can be concluded. Exercise 10.3 Q14 asks the student to produce the counterexample directly.
• "The angle between two vectors could be measured either way round." The definition restricts it to the range from zero to a straight angle, which picks one of the two angles at the meeting point. Without the restriction the cosine would not be determined.
• "A negative answer means I have made a mistake." It means the angle is obtuse. Example 14 produces one on purpose and it is the first negative answer in the section.
• "The nine products of the three hats are nine facts to memorise." They are two observations applied nine times: each hat has magnitude one, and different hats are perpendicular.
• "Property 1 in this section is the Property 1 I learnt for addition." It is not. The chapter reuses both labels. Say the section number every time.
• "The scalar in Property 2 has to sit in a particular place." It may sit outside or on either factor. The printed chain says so, but it says one of the three things twice and uses two different letters for the scalar in one statement, so read it carefully.
• "The Cauchy inequality and the triangle inequality are unrelated." The proof of the second cites the first by name at the step where it is needed. They are one argument in two halves.
• "Collinearity has to be checked with components." The Remark on Part II p. 361 checks it with three magnitudes, and Example 21 does exactly that. Both routes work; the magnitude route is the one the chapter chose here.
• "A product of two vectors with itself has no meaning." A vector against itself gives the square of its magnitude, which is Observation 3's particular case and is used in almost every worked example from Example 17 onwards.

Questions to check understanding

• Compute a scalar product from two magnitudes and an angle, and from two vectors in components
• Recover the angle at which two vectors meet, both given in components — the form of Example 14 and Exercise 10.3 Q2
• Show two vectors perpendicular — the form of Example 15 and Exercise 10.3 Q11
• Find a scalar making one vector perpendicular to another — the form of Exercise 10.3 Q10
• Find a magnitude from a product involving sums and differences — the form of Examples 17 and 18 and Exercise 10.3 Q9
• Expand a product of two combinations using both properties — the form of Exercise 10.3 Q7
• State and prove the bound on the size of a scalar product, and use it to prove the bound on the magnitude of a sum
• Decide what follows from a product being zero, and what does not — the form of Exercise 10.3 Q12 and Q14
• Show three points collinear from three magnitudes — the form of Example 21 and Exercise 10.3 Q16
• Find the angle between two unit vectors whose sum is a unit vector — the form of Miscellaneous Exercise Q17

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.

• The opening analogy (§10.6, Part II p. 355). Before defining anything, the chapter argues that a second product is even wanted: two numbers multiply to a number and two matrices to a matrix, but two functions can be multiplied in two different ways — value by value, and by feeding one into the other — and so, it says, two vectors are multiplied in two ways as well. Verified as the chapter's own framing and worth keeping intact: it is the only place in the chapter where a definition is motivated by an analogy rather than by a picture or a stipulation, and it prepares a student for a second product arriving seven pages later with a completely different output type. Section 1 is sixty seconds and it earns the whole module.
• Definition 2 and Fig 10.19 (§10.6.1, Part II pp. 355–356). For two nonzero vectors, the product is the first magnitude times the second times the cosine of the angle between them, with the angle running from zero to a straight angle. Read off the printed page: Fig 10.19 is two arrows from a common point with the angle between them marked, and nothing else. Verified as the load-bearing clause: the range on the angle is what makes the product well defined, because two arrows meeting at a point subtend two angles and only one of them lies in that range. The chapter states the range and never explains it. Say it in one sentence.
• The excluded case (§10.6.1, Part II p. 356). If either vector is the zero vector the angle is not defined, so the chapter defines the product to be zero by fiat. Verified as a convention rather than a consequence: nothing forces the value; it is chosen so that the component formula and the distributive property keep working. The vector product's Definition 3 on Part II p. 363 has exactly the same structure — a definition for nonzero vectors and a separate convention for the zero case — and the two should be taught with the same words. Section 3 should say so and hand the parallel to the last topic of this module.
• Observations 1 to 7 (§10.6.1, Part II p. 356). Seven numbered items: the answer is a number; two nonzero vectors are perpendicular exactly when the answer is zero; a zero angle gives the product of the magnitudes and, in particular, a vector against itself gives the square of its magnitude; a straight angle gives the negative of the product of the magnitudes; the nine products of the three axis unit vectors are one on the diagonal and zero off it; the angle can be recovered from the product and the two magnitudes; and the product is commutative, marked with a parenthetical question. Observation 4 is defective and a reviewer should see it: its second sentence, introduced as a particular case, restates its own general statement word for word, where the parallel sentence in Observation 3 gives a genuine particular case. Confirmed on the printed page. Teach Observation 4 with the particular case it should have had — a vector against its own negative — and say the printed line is a repetition.
• The nine products of three hats (Observation 5, Part II p. 356). Verified as a consequence, not a new fact: the diagonal three follow from Observation 3 because each hat has magnitude one, and the off-diagonal three follow from Observation 2 because the axes are perpendicular. Six products are printed and three more follow from commutativity. This is the single most-used fact in the remaining eighteen pages of the chapter and it takes two observations to produce. Section 6 should build the three-by-three table and then never redraw it.
• The angle, extracted (Observation 6, Part II p. 356). The cosine of the angle is the product over the two magnitudes, and the angle itself is the inverse cosine of that. Note for the illustrator: the chapter prints the same quotient twice on one line, once with the multiplication dot set as a centred dot and once as a full stop on the baseline. Cosmetic; set it consistently.
• The two properties (§10.6.1, Part II pp. 356–357). Property 1 is distributivity over addition; Property 2 says a scalar may be moved freely between the two factors and the outside. Property 2 carries two printed defects and both were confirmed on the printed page: the scalar is introduced in the sentence as the letter ell and then used in the formula as a Greek lambda, and the first two members of the four-term chain are identical to each other, so the chain has three distinct statements and four slots. The intended chain is plainly the scalar outside, the scalar on the first factor, and the scalar on the second. Say the property correctly and do not show the printed line.
• The Property numbers are reused (Part II pp. 344, 345, 356, 357, 365). The labels Property 1 and Property 2 name vector addition's commutativity and associativity in §10.4 and then name these two entirely different statements here. Verified by grep across all thirty-nine extracted pages: the word occurs six times in the chapter and there is exactly one Property 3, for the vector product. Always say which section a Property number belongs to. The last topic of this module carries a consequence of this collision that is worse than an ambiguity.
• The component formula, derived (§10.6.1, Part II p. 357). Two vectors in component form are multiplied out: the two properties expand the product into nine terms, and Observation 5 kills six of them and reduces the other three, so the answer is the sum of the three products of corresponding components. Verified as the longest derivation in the chapter: six printed lines, and every step is justified in the margin. The chapter's own margin note names the two properties and the observation used. Section 9 should run all six lines; this is the one place in the chapter where a student can watch a definition turn into an algorithm.
• Examples 13, 14 and 15 (Part II pp. 358–359). Recover an angle from two magnitudes and a product; recover an angle from two vectors in components; and show that the sum and the difference of two given vectors are perpendicular. Verified: in Example 13 the cosine is one half, so the angle is a third of a half-turn. In Example 14 the product is minus one and both magnitudes are root three, so the cosine is minus one third and the angle is obtuse — the first negative answer in the section, and worth pausing on. In Example 15 the sum has components six, two and minus eight, the difference four, minus four and two, and their product is twenty-four minus eight minus sixteen, which is zero.
• Examples 17 and 18 (Part II pp. 359–360). Find the magnitude of a difference from two magnitudes and a product; and find a magnitude from a product of a sum and a difference involving a unit vector. Verified: in Example 17 the squared magnitude is four minus eight plus nine, so the answer is root five. In Example 18 the square of the unknown magnitude minus one is eight, so the magnitude is three, and the chapter says explicitly that the negative root is rejected because a magnitude is not negative — the boxed Note from Part II p. 339 being spent for the second time.
• Examples 19 and 20, and the Remark (Part II pp. 360–361). The first bounds the size of a product by the product of the magnitudes; the second bounds the magnitude of a sum by the sum of the magnitudes; and a Remark then reads the equality case of the second as forcing three points onto one line. Verified as a four-item chain and the best-structured passage in the chapter: Example 20's proof cites Example 19 by name at the step that needs it, and the Remark then runs Example 20 backwards. Section 10 and section 11 should be taught as one argument in two halves. Note also that Example 19's proof divides by both magnitudes, which is why it has to dispose of the zero case in its first line.
• Example 21 (Part II p. 361). Three points given by position vectors are shown collinear. Verified: the three joining vectors have components three, minus one and minus two; six, minus two and minus four; and nine, minus three and minus six, so their magnitudes are root fourteen, twice root fourteen and three times root fourteen, and the longest equals the other two added together. This is the Remark applied, and it does not use the scalar product at all — it uses only magnitudes. Section 11 should say so: the section formula topic could have set this item and the chapter has put it here because of the Remark above it.
• The boxed Note under Example 21 (Part II p. 361). Although the three joining segments sum to the zero vector, the three points are not the vertices of a triangle. Named here for completeness: it is a counterexample to the converse of a Summary bullet about addition, and the first topic of the second module carries it. Do not spend a section on it twice.
• Exercise 10.3 questions 1, 2 and 5 to 18 (Part II pp. 361–362). Verified in turn, and the ones worth choosing are named: Q1 gives a cosine of one over root two and so a right angle halved. Q2 gives a cosine of five sevenths. Q5 is the one item in the exercise that exercises both halves of the section at once: it hands the student three vectors, each a seventh of a triple built from the same three numbers two, three and six with the signs and the order shuffled, and asks first that each be shown to have length one and then that the three be shown mutually square. Verified: every one of the three triples has squared entries summing to forty-nine, so each seventh-scaled vector has length one; and all three pairwise products come out zero. It is the chapter's only orthonormal triple other than the axis vectors themselves, and it is worth a beat for that reason alone — the same seventh-scaled triple turns up again as the answer to a Miscellaneous item that the vector product topic carries. Q6 is the one item in the exercise whose answer is not tidy — the smaller magnitude comes out as two root fourteen over twenty-one and the larger as sixteen times that first surd over the same denominator — and it is worth showing precisely because the rest are tidy. Q8 forces both magnitudes to one. Q9 gives root thirteen. Q10 gives a multiplier of eight. Q11 is a two-line cancellation. Q12 is the trap of the exercise: the first condition forces the first vector to be the zero vector, after which the second condition holds for every second vector, so nothing whatever can be concluded — and a student who answers "perpendicular" has fallen for it. Q13 gives minus three halves. Q14 asks for a counterexample to a converse and is the scalar product's twin of an item in the next exercise. Q15 gives an inverse cosine of ten over root one hundred and two. Q16's two joining vectors are identical, which settles collinearity in one line. Q17 is the same right-angled triangle the section formula topic already met twice. Q18 is the unit vector condition, answered by the reciprocal of the modulus of the scalar.
• Miscellaneous Examples 27, 28 and 29 (Part II pp. 370–371). Three worked items sitting ten pages past this section, all three of them scalar product problems, and an explanation that stops at Example 21 leaves the section's three best examples on the table. Verified in turn. Example 27 gives four points by position vector and asks for the angle between two of the joining vectors: the first has components one, four and minus one with length three root two, the second has components minus two, minus eight and two with length six root two, their product is minus thirty-six, so the cosine is exactly minus one and the angle is a half-turn. Then the chapter adds a one-line alternative that makes the whole computation unnecessary — the second joining vector is minus twice the first, which settles collinearity outright. Show the long way, then the one line, and say which one a reader should have seen first. Example 28 has three vectors of lengths three, four and five, each one square to whatever the remaining pair adds up to; adding the three conditions kills the cross terms, so the squared length of the sum is nine plus sixteen plus twenty-five and the answer is five root two. Both magnitude triples were read on a checked the printed page, because the text layer of these two pages drops every one of them. Example 29 has three vectors summing to the zero vector with lengths three, four and two, and asks for the sum of the three pairwise products; dotting the condition with each vector in turn and adding gives twice that sum as minus twenty-nine, so the answer is minus twenty-nine halves. Examples 28 and 29 are the same trick run forwards and backwards — one uses vanishing products to find a length, the other uses a vanishing sum to find the products — and they should be taught as a pair. Where they go: Example 27 belongs in section 7, where the angle is extracted, and it is the best available demonstration that the formula returns a straight angle rather than failing; Examples 28 and 29 belong together at the end of section 12, where a vanishing product is being read for what it means.
• Miscellaneous Exercise questions 13, 15, 16 and 17 (Part II pp. 372–373). Verified: Q13 gives a value of one, after squaring away a square root. Q15 is the statement that the squared magnitude of a sum splits exactly when the product vanishes, which is the Pythagoras theorem written in vectors and is worth saying out loud. Q16's answer is the closed interval from zero to a right angle. Q17's answer is two thirds of a half-turn, obtained by squaring the sum of two unit vectors and setting the result to one. Note on Q16: **the item is worded with only when where exactly when is meant**, and under the literal reading more than one option is defensible; teach the intended answer and do not dwell. Question 19 compares the two products and belongs to the last topic of this module, which carries it.
• Summary bullet nine (Part II p. 374). The definition of the scalar product and the formula for the angle, in one bullet. Verified as the whole of what the Summary keeps from §10.6.1: there is no Summary bullet for perpendicularity, none for the nine products of the axis unit vectors, none for either inequality, and none for the two properties. The component formula survives, inside the last bullet on Part II p. 375. Confirmed on the page image of all three Summary pages.

Figures to have open

• A redraw of Fig 10.19 (Part II p. 356): two arrows from a common point with the angle marked between them. The chapter's own drawing, and the only figure in §10.6.1. Reuse it for sections 2, 4, 5 and 7 with only the angle changing, so four different points ride on one image.
• A three-by-three table for section 6, built with the repo's DataTable component, carrying the nine products of the three axis unit vectors, with the diagonal and the off-diagonal entries visually distinguished and each traced to the observation that supplies it.
• A redraw of Fig 10.21 (Part II p. 360) for section 10: the triangle carrying two vectors and their sum, which is the picture behind the triangle inequality. The chapter's own drawing, and the only figure in the worked-example run.
• A single number line for section 4 with the product's value marked and the three angle regions coloured, so the sign and the angle are read off one scale.
• A three-panel build for section 11: three points, the three joining segments with their magnitudes measured, and the two shorter ones laid end to end against the longest.
• No figure is needed for sections 1, 3, 8, 9 or 12 beyond the kit shapes named above; the chapter prints none for any of that material.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 10 "Vector Algebra", §10.6 Product of Two Vectors, the opening analogy, Part II p. 355
• §10.6.1 Scalar (or dot) product of two vectors, with Definition 2, Fig 10.19 and the zero-vector convention, Part II pp. 355–356
• Observations 1 to 7, Part II p. 356; Property 1, Part II p. 356; Property 2 and the component derivation, Part II p. 357
• Examples 13, 14, 15, 17, 18, 19, 20 and 21 with Fig 10.21, the Remark and the boxed Note, Part II pp. 358–361
• Exercise 10.3, questions 1, 2 and 5 to 18, Part II pp. 361–362
• Miscellaneous Examples 27, 28 and 29, Part II pp. 370–371
• Miscellaneous Exercise, questions 13, 15, 16 and 17, Part II pp. 372–373
• Definition 3 and its zero-vector convention, §10.6.3, Part II p. 363, cited only for the structural parallel
• Summary, the scalar product bullet, Part II p. 374, and the component form bullet, Part II p. 375