Stretching by a scalar, and dividing a vector by its own length

What to assume they know

• Vectors, magnitude and direction, and the modulus notation, from the first module
• The zero vector, collinear vectors, the negative of a vector and the unit vector notation, from the third topic of the first module
• Vector addition and its two laws, from the previous topic
• Absolute value of a real number, and the fact that it is never negative
• Rationalising a denominator, and the arithmetic of surds, from Class IX
• Component form of a vector, at the level of reading one — the topic that builds it follows this one

What they should be able to do

• State what the chapter asserts about the product of a vector and a scalar, and distinguish the assertions from the one thing it derives
• Apply the magnitude rule, keeping the modulus bars on the scalar
• Read the five arrows of the chapter's scaling figure, including the two whose minus signs a text search will not find
• Explain what happens when the scalar is minus one, and connect it to the definition given nine pages earlier
• State what scaling does to the zero vector, and quote the boxed Note that covers it
• Restate collinearity as a single equation in one unknown scalar
• List the chapter's three so-called distributive laws and say which of them distributes over what
• Derive the unit vector formula by choosing the scalar to be the reciprocal of the magnitude
• State the condition under which that choice is legal, and say why
• Produce a vector of stated magnitude pointing the way a given vector points
• Produce the one-unit vector along a sum of two given vectors, in the right order of operations
• Recognise a family of exercise items as the same question in different clothes

Where it usually goes wrong

• "Multiplying by a negative scalar makes the magnitude negative." A magnitude is a length and can never be negative; the bars round the scalar in the chapter's own rule are there to prevent exactly this. The sign goes into the direction, which is the only place it can go.
• "Half of a vector points somewhere different from the vector." It does not; every scaled copy is collinear with the original. Only the sign of the scalar can change where it points, and then only by reversing it exactly.
• "The fourth arrow in the scaling figure is another positive multiple." It is minus one half, and it points the other way. The extracted text of that page loses the minus sign, so this error is built into any workflow that reads the book without looking at it.
• "Scaling the zero vector by a large number gives a long vector." It gives the zero vector, whatever the scalar. The chapter's boxed Note on Part II p. 347 says so in one line.
• "All three of the chapter's laws on Part II p. 349 are distributive laws." One of them nests two scalings and involves no addition at all, so nothing is distributed. Another distributes over the addition of numbers. Only the third distributes over the addition of vectors, which is what the heading suggests all three do.
• "To find a unit vector along a sum, take the unit vectors and add them." That gives a different vector, and in general not a unit vector at all. Add first, then normalise. Example 8 is printed in that order and the order is the lesson.
• "To get a vector of magnitude seven, multiply the original by seven." Only if the original already had magnitude one. Normalise first. Example 7's two steps are the whole content of that example.
• "The unit vector formula works for any vector." It excludes the zero vector, and the chapter states the exclusion on the line where it makes the choice. A student who does not carry the condition will divide by zero the first time an exercise hands them a sum that cancels.
• "There is only one answer to a unit-vector question." Miscellaneous Exercise Q5 has two, plus and minus. The question asks for a multiplier, not for a direction, and both signs give a vector of length one.

Questions to check understanding

• Compute the magnitude of a scaled vector from the magnitude of the original, with a negative scalar
• State what a stated scalar does to a drawn vector, in direction and in length
• Decide whether two given vectors are collinear by finding the scalar — the form of Exercise 10.2 Q11
• Judge four statements about collinear vectors and say which fail — the form of Exercise 10.2 Q19
• Produce the unit vector along a given vector — the form of Example 6 and Exercise 10.2 Q7
• Produce a vector of stated magnitude along a given vector — the form of Example 7 and Exercise 10.2 Q10
• Normalise a total of two given vectors — the form of Example 8 and Exercise 10.2 Q9
• Find every multiplier that turns a given vector into a unit vector, keeping both signs — the form of Miscellaneous Exercise Q5
• State the condition under which a vector may be divided by its own magnitude, and say what goes wrong without it

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.

• What §10.5 asserts (Part II p. 346). Four claims arrive in one paragraph and none is argued: the product of a vector and a scalar is again a vector; it is collinear with the original; its direction agrees with the original when the scalar is positive and reverses when the scalar is negative; and its magnitude is the modulus of the scalar times the original magnitude. Verified as a stipulation: the chapter offers no proof for any of the four and no proof is possible, because they are what defines the operation. Section 1 should say so cleanly. A student who spends the section looking for the missing derivation has been misled by every other section of the chapter, all of which do derive something.
• The magnitude rule (Part II p. 346). The magnitude of the scaled vector is written with modulus bars round the scalar as well as round the vector. Verified as the point of the notation: without the bars round the scalar the statement would be false for negative scalars, since a magnitude can never come out below zero — which is exactly the boxed Note from Part II p. 339 being spent for the first time. Show the two together.
• Fig 10.12 (§10.5, Part II p. 346). Read off the printed page: five arrows, labelled as the vector itself, half of it, twice it, minus half of it and minus twice it. Each carries small tick marks so the lengths can be compared. The first three point up and to the right; **the two negative ones point down and to the left. *The extracted text of this page drops the minus sign on the fourth arrow entirely***, so a reader working from the text layer sees the same label twice and no reversal at all. This is the clearest example in the chapter of why a figure claim needs the printed page. The chapter's own phrase for this figure is worth keeping: it calls it a geometric visualisation, which is precisely what it is doing in place of an argument.
• The scalar minus one (Part II p. 346). Taking the scalar to be minus one gives a vector of the same magnitude pointing the opposite way; the chapter names it the negative of the vector, adds the second name additive inverse, and states that the vector and its negative sum to the zero vector in either order. Verified as a second definition of an object already defined: §10.3 on Part II p. 341 built the same thing geometrically, from a segment with its two letters swapped. The chapter does not say the two are the same. Section 4 should put the two entries side by side; the third topic of module one plants this and hands it here.
• The boxed Note on scaling the zero vector (Part II p. 347). Any scalar times the zero vector is the zero vector. Verified as the edge case that matters: it says scaling can never manufacture a direction out of nothing, which is why the unit vector construction two lines above it has to exclude the zero vector by hand. This is one of six boxed Notes in the chapter.
• Collinearity as one equation (Remark (i), Part II p. 349). Whatever the scalar, the scaled vector is collinear with the original; and conversely two vectors are collinear exactly when one is a nonzero scalar times the other. Verified: the forward direction is one of the four stipulations from Part II p. 346, so the content is entirely in the converse, which the chapter asserts without proof. The same Remark continues into a componentwise test — three equal ratios — and that half belongs to the next topic, which carries it; name it here and hand it over.
• The three distributive laws (Part II p. 349). The chapter introduces them by saying that vector addition and scaling together yield them, and then prints three: the first adds two scalars and multiplies once; the second nests two scalings into one; the third spreads a single scalar across a sum of two vectors. Verified, and this is a real mislabelling: the second involves no addition of any kind, so nothing is being distributed over anything in it — it is the associativity of scaling, filed under a heading that does not describe it. And the first distributes over the addition of numbers, not of vectors, while only the third distributes over the addition of vectors. Section 7 should sort the three; it takes forty seconds and it is the difference between three rules memorised and three rules understood.
• The unit vector derivation (Part II pp. 346–347). Take the scalar to be one over the magnitude of the vector, provided the vector is not the zero vector; then by the magnitude rule the scaled vector has magnitude one. The chapter concludes that this scaled vector is the one of unit length pointing the way the original points, and gives it the hat notation. Verified, and it is the only thing §10.5 derives: every other statement in the section is a stipulation or a figure. Three lines of algebra, and they produce the single most-used object in the remaining twenty-nine pages of the chapter. Section 8 is the centre of this topic and should be paced accordingly.
• The condition (Part II p. 346). The chapter writes the exclusion twice in one line, once as an inequality and once in words, naming the excluded case a null vector. Verified: the exclusion is forced, because the reciprocal of zero is not a number; and the boxed Note on the next page explains what would go wrong if one tried anyway. Section 9 is thirty seconds and it is the part students omit.
• Example 6 (Part II p. 350). Find the unit vector along a vector whose components are two, three and one. Verified: the magnitude is the square root of fourteen, so the answer has components two, three and one, each over the square root of fourteen. This is the derivation of section 8 run once, on numbers.
• Example 7 (Part II p. 350). Find a vector of magnitude seven along a vector whose components are one and minus two. Verified: the magnitude is the square root of five, the unit vector has components one and minus two each over the square root of five, and multiplying by seven gives seven over root five and minus fourteen over root five. The two-step shape is the whole lesson — normalise first, then scale — and reversing the steps is the commonest error.
• Example 8 (Part II pp. 350–351). Normalise a sum: two vectors are given and the one-unit vector along their total is wanted. Verified: the sum has components four, three and minus two, its magnitude is the square root of twenty-nine, and the unit vector is those three components each over the square root of twenty-nine. Note the ordering: the sum is formed first and normalised afterwards, because normalising the two addends separately and adding the results gives a different vector. Section 11 should show the wrong order once, briefly, and say why it fails.
• Exercise 10.2 Q7, Q9 and Q10 (Part II p. 354). A unit vector along a given vector; the same for a total of two given vectors; and a vector of magnitude eight along a given vector. Verified in turn: the first has magnitude root six; the second's sum comes to a vector with components one, zero and one, of magnitude root two, so the unit vector is one over root two along the first and third axes and nothing along the second — the cleanest arithmetic in the exercise; the third has magnitude root thirty, so the answer is eight over root thirty times the given vector. These are Examples 6, 8 and 7 in the same order, with different numbers, which is why section 12 groups them.
• Exercise 10.2 Q11 and Q19 (Part II pp. 354–355). Show two given vectors are collinear; and decide which of four statements about collinear vectors are incorrect. Verified: in Q11 the second vector is minus two times the first, so the collinearity test of Remark (i) is satisfied outright. In Q19 the statement that one vector is a scalar times the other is the correct one; the statement that the two are equal up to sign is incorrect because their magnitudes may differ; the statement that their components are not proportional is incorrect because collinearity is exactly proportionality; and the statement that they share a direction and differ only in magnitude is incorrect because a negative multiplier reverses one of them. Three of the four are incorrect, and the item is worded in the plural, which is worth flagging so a student does not stop after finding one.
• Miscellaneous Exercise Q5, Q6 and Q7 (Part II p. 372). Find the multiplier that turns a stated vector into one of unit length; find a vector of magnitude five running parallel to a stated resultant; and normalise a stated combination of three vectors. Verified: in Q5 the magnitude is the modulus of the unknown times root three, so the unknown is plus or minus one over root three, and both signs are genuine — the negative one gives the opposite unit vector, which is still a unit vector. In Q6 the resultant has components three, one and zero, of magnitude root ten, so the answer is five times that vector over root ten. In Q7 the combination comes to components three, minus three and two, of magnitude root twenty-two, so the unit vector is those three over root twenty-two. All three lean on component arithmetic that the next topic builds; cite them here and say so.
• Summary bullets six and seven (Part II p. 374). Scaling changes a vector's magnitude by the modulus of the scalar and keeps or reverses its direction according to the sign; and the vector over its own magnitude is the unit vector along it. Verified as the whole of what the Summary keeps from §10.5: there is no Summary bullet for the three distributive laws, none for the collinearity criterion, and none for the zero vector under scaling. Confirmed on the page image of all three Summary pages.

Figures to have open

• A redraw of Fig 10.12 (Part II p. 346) for section 3: five arrows on a common baseline, labelled as the vector, one half of it, twice it, minus one half of it and minus twice it, with the last two drawn pointing the opposite way and with the tick marks kept, since the ticks are how the lengths are compared. The chapter's own drawing, laid out on a shared baseline rather than scattered, so the lengths can be read against each other.
• A single reusable arrow for sections 1, 2, 8 and 9, drawn once and then scaled in place, so that four different points are made about one picture.
• A three-row table for section 7, built with the repo's DataTable component, carrying each of the three laws, what is being added in it, and what is being distributed over what. The classification is added here; the chapter supplies only the collective heading.
• A two-panel comparison for section 11: the sum normalised, and the two addends normalised then added, with both results drawn from the same origin and both magnitudes marked. The wrong panel is added here; the chapter shows only the right one.
• No figure is needed for sections 4, 5, 6, 10 or 12 beyond the kit shapes named above; the chapter prints none for any of that material and none is wanted.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 10 "Vector Algebra", §10.5 Multiplication of a Vector by a Scalar, Part II pp. 346–347, with Fig 10.12
• The unit vector construction and the boxed Note on scaling the zero vector, Part II pp. 346–347
• The three distributive laws and Remark (i), Part II p. 349
• Examples 6, 7 and 8, Part II pp. 350–351
• Exercise 10.2, questions 7, 9, 10 and 11, Part II p. 354; question 19, Part II p. 355
• Miscellaneous Exercise, questions 5, 6 and 7, Part II p. 372
• The unit vector entry and the negative of a vector, §10.3, Part II p. 341
• Summary, the scaling bullet and the unit vector bullet, Part II p. 374