Splitting a vector along the axes so the algebra becomes coordinate arithmetic

What to assume they know

• A point's position vector, and its magnitude taken over from the Class XI distance formula, from the second topic of the first module
• Vector addition by the triangle law, from the first topic of this module
• Multiplication of a vector by a scalar, and the unit vector construction, from the previous topic
• The Pythagoras theorem in a right-angled triangle
• Rectangular axes in space and the coordinates of a point, from Class XI
• Dropping a perpendicular from a point onto a plane, from Class XI

What they should be able to do

• Name the three unit vectors along the axes and say which three points define them
• Decompose a position vector into three pieces by two applications of the triangle law
• Distinguish the three scalar components of a vector from its three vector components, and give the third name the chapter offers
• Derive the magnitude of a vector from its components by applying the Pythagoras theorem twice
• Say where else in the chapter that same magnitude formula appears, and how it arrives there
• State the four componentwise rules the chapter lists, and derive each from the laws printed on the following page
• Decide whether two vectors given in components are equal, and solve for unknown components
• Produce two vectors of equal magnitude that are not equal
• Apply the componentwise test for collinearity as three equal ratios
• Handle a two-component vector in a chapter written for three
• Compute the magnitude of a vector given in components, including a case that comes out at one
• Recognise that the three scalar components are the direction ratios met earlier

Where it usually goes wrong

• "Scalar components and vector components are two names for the same three things." One list is three numbers and the other is three vectors. Exercise 10.2 Q5 asks for both and a student who thinks they are the same gives half an answer.
• "The magnitude formula is a definition you have to accept." It is derived on Part II p. 348 from two right triangles. It is also asserted on Part II p. 339 by citing Class XI, which is where most students meet it, and that is why so few know it has a proof in this chapter.
• "The four component rules are new facts to memorise." Three of the four fall straight out of the laws printed on the very next page. Deriving them takes two minutes and turns four memorised rules into one.
• "Two vectors with the same magnitude are the same vector." Example 5 is exactly this counterexample, and Exercise 10.2 Q2 asks the student to produce another. Length is one number; a vector in space is three.
• "To check whether two vectors are equal I compare their magnitudes." You compare three pairs of components. Equal magnitude is necessary and nowhere near sufficient.
• "Collinear vectors have equal components." They have proportional components, in one common ratio. Equal components would make them the same vector.
• "A two-component vector is a different kind of object." It is a three-component vector whose third component is zero. The chapter never says this and uses two-component vectors in at least four places.
• "The three hatted arrows are just notation." They are constructed, on Part II p. 347, as the position vectors of three named points. That construction is what makes them have magnitude one, which is the property every later computation uses.
• "Every unit vector question tells you it is about a unit vector." Exercise 10.2 Q1's third item is a unit vector and the question only asks for its magnitude. Noticing is the whole reward.

Questions to check understanding

• Write a position vector in component form given the coordinates of its point
• Name the scalar components and the vector components of a stated vector, as two separate lists — the form of Exercise 10.2 Q5
• Compute the magnitude of a vector given in components — the form of Exercise 10.2 Q1
• Derive the magnitude formula from the components using Pythagoras twice
• Solve for unknown components that make two vectors equal — the form of Example 4 and Exercise 10.2 Q4
• Add three vectors given in component form — the form of Exercise 10.2 Q6
• Produce two vectors of equal magnitude that are not equal — the form of Example 5 and Exercise 10.2 Q2
• Test two vectors in components for collinearity by comparing three ratios
• Derive any one of the four componentwise rules from the laws on Part II p. 349

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.

• Three points, three arrows, and Fig 10.13 (§10.5.1, Part II p. 347). The section opens by naming three points, one on each axis, each one unit from the origin; observes that the three position vectors of those points have magnitude one; and names those three the unit vectors along the axes, giving each a hat. Read off the printed page: Fig 10.13 is a small axis frame with the three points lettered and their coordinates written out, and the three hatted arrows drawn short and close to the origin. The construction is worth slowing down on: the chapter does not postulate the three unit vectors, it builds them out of three named points, and that is the only place in the chapter where a piece of notation is given a construction rather than a stipulation.
• Fig 10.14 and the decomposition (§10.5.1, Part II p. 347). A point in space, the perpendicular from it down to the horizontal coordinate plane, and the foot of that perpendicular. Read off the printed page: the figure is a box with the origin at one corner and the point at the far corner, the foot marked on the floor of the box, and three edges at the origin labelled with the three vector pieces. Verified as two applications of one law: the triangle law is used once in the floor of the box to reach the foot, and once vertically to climb from the foot to the point. Section 2 should build the box one edge at a time and say "triangle law" out loud at both steps, because the whole of the rest of the chapter's arithmetic rests on those two moves.
• The three names (§10.5.1, Part II p. 348). The three plain numbers are the scalar components; the three pieces they multiply are the vector components; and the chapter then adds that the three numbers are sometimes called rectangular components as well. Verified as a naming hazard: two of the three names differ by one word and denote different kinds of object — one is a triple of numbers, the other a triple of vectors — and an exercise asks for both at once. Show the two lists side by side and never.
• Pythagoras, twice (§10.5.1, Part II p. 348). The chapter says outright that the length of a vector in component form is got by applying the Pythagoras theorem twice, and does it: once in the right triangle lying in the floor of the box, to reach the foot, and once in the upright right triangle, to reach the point. Verified, and worth naming: the word theorem occurs exactly once in this entire chapter, on this page, naming this result — confirmed by grep across all thirty-nine extracted pages and on the page image of each. The chapter proves nothing else it calls a theorem and states no theorem of its own.
• The same formula, nine pages earlier (§10.2, Part II p. 339). The magnitude of a position vector as the square root of the summed squared coordinates is printed twice in this chapter. On Part II p. 339 it is taken over from the Class XI distance formula by citation, with no argument; on Part II p. 348 it is derived, from the box and two right triangles. Verified, and the chapter never connects them: the second is a proof of the first and is not presented as one. Section 5 should show both pages and say so. The second topic of the first module carries the earlier statement; this one carries the proof.
• The four componentwise rules (§10.5.1, Part II p. 348). Two vectors arrive in component form, and four facts are then listed: their sum is got by adding corresponding components; their difference by subtracting them; they are equal exactly when all three pairs of components agree; and a scalar multiple is got by multiplying each component. Verified as unargued: none of the four is derived, or even commented on, and eleven printed lines carry all four. This is the payoff of the entire module — every later computation in the chapter is one of these four — and it is delivered as a list.
• The argument, on the next page (Part II p. 349). The three distributive laws printed at the head of Part II p. 349 are exactly what is needed to derive the four rules on Part II p. 348. Verified: spreading a scalar across a sum of vectors gives the scalar-multiple rule; collecting like terms in the three axis unit vectors gives the sum and difference rules; and the equality rule is the statement that the three axis unit vectors cannot cancel each other, which the chapter never states in any form. Section 7 should run the derivation the chapter skipped, and should flag the equality rule as the one that still has no printed justification. The scaling topic carries the three laws themselves; this topic carries what they are for.
• Example 4 (Part II p. 349). Two vectors in component form are set equal and three unknowns come out. Verified: matching the three pairs of components in turn gives two, two and one. The example exists to make the equality rule concrete and it takes one line.
• Example 5 (Part II p. 350). Two vectors with components one and two, and two and one. Verified: both have magnitude root five, and they are not equal because their components differ. This is the same question Exercise 10.2 Q2 asks in the abstract, done here with numbers. That exercise item itself is carried by the types-of-vectors topic, which needs it for the definition of equality; this topic keeps only Example 5, where the verdict is reached by comparing components. Section 9 is the shortest section in the topic and the one students most often get wrong in the other direction — assuming that unequal vectors must have unequal lengths.
• Remark (i), the componentwise half (Part II p. 349). Two vectors are collinear exactly when their corresponding components stand in one common ratio; the chapter writes the chain of three equal ratios out and identifies the common value with the scalar multiplier. Verified: this is the coordinate translation of the criterion the previous topic states geometrically, and the two halves of Remark (i) are printed as one paragraph. The exercise that drills it, Exercise 10.2 Q11, is carried by the previous topic; this topic carries the criterion and the previous one names it and hands it here.
• Remark (ii) (Part II p. 349). The three scalar components are also called the direction ratios of the vector. Verified as a closing of the loop: the phrase direction ratios was defined on Part II p. 340 as the length times each direction cosine, and this Remark quietly says that the triple so defined is the triple of components. It is one sentence and it joins two sections eleven pages apart. Section 12 should spend thirty seconds on it and hand the consequences to the second topic of the first module, which carries them.
• Remark (iii) (Part II p. 349). The triple of direction cosines assembled into a vector is the unit vector in that direction. Named here for completeness only; it belongs to the direction cosine topic, which carries it.
• Exercise 10.2 Q1 (Part II p. 354). Three magnitudes to compute. Verified: the first vector's three components are all one, so its magnitude is root three; the second's are two, minus seven and minus three, so root sixty-two; and the third's are one over root three three times over, with a minus on the last, so its magnitude is exactly one. The third is a unit vector and the question does not say so — a thirty-second payoff to close section 12 on. And there is a printed slip in the first vector: its third term carries no hat, while the other two vectors in the same line have hats throughout. Confirmed on the printed page; see the note below.
• Exercise 10.2 Q4, Q5 and Q6 (Part II p. 354). Solve for two unknowns making two planar vectors equal; list both kinds of component for the arrow running between two planar points; and add three vectors in component form. Verified: Q4 gives two and three. Q5's vector has components minus seven and six, so its scalar components are minus seven and six and its vector components are minus seven along the first axis and six along the second — the question asks for both lists and marks are lost by giving one. Q6's sum has first component zero, second minus four and third minus one, so the sum is minus four along the second axis and minus one along the third, with nothing along the first.
• Example 26, the chapter's only planar sweep (Part II p. 369, with Fig 10.28). The first of the Miscellaneous Examples asks for every unit vector lying in the coordinate plane, and answers it by putting cosine of an angle on the first component and sine of the same angle on the second. Verified: those two components square and sum to one for every angle, so each such vector does have length one, and as the angle sweeps once around, the tip traces the unit circle the example names — which is why the answer is a whole family rather than a single vector. This is the one place in the chapter where components are read off an angle rather than off a pair of points, and it is the bridge back to the direction-cosine topic: in two dimensions the two components are the two direction cosines. The example sits ten pages past this topic's own material and is easy to miss; it belongs here, and section 11 should end on it. No other brief teaches Example 26 or uses Fig 10.28. Two of them cite the example, but only for the word it uses for a sense of rotation, and both say so.
• Miscellaneous Exercise Q1 (Part II p. 372). The exercise twin of Example 26: write down the unit vector in the coordinate plane that makes a sixth of a right angle with the positive first axis. Verified: the two components are root three over two and one half, and their squares do sum to one. It is the first item of the Miscellaneous Exercise and it is answered by a formula printed three pages earlier, which is worth pointing out — the Miscellaneous Exercise is not uniformly harder than the sectional ones, and a student who has read Example 26 can answer this in one line. No other brief uses it.
• Two dimensions, without comment (Part II pp. 350, 354 and 369). Verified as a pattern: Example 7 works with a vector having two components, Exercise 10.2 Q4 and Q5 are both planar, and Example 26 on Part II p. 369 is set entirely in the coordinate plane. The chapter builds everything in three dimensions and then drops to two whenever it suits, with no sentence anywhere saying that the two-dimensional case is the three-dimensional one with the third component zero. Section 11 should say it. It is thirty seconds and it removes a standing source of confusion.
• Summary bullet eleven (Part II pp. 374–375). Two vectors in component form, and then their sum, a scalar multiple, their scalar product and their vector product, all in components. Verified: the first two of the four belong to this topic and are the compressed form of two of the four rules from Part II p. 348. The difference rule and the equality rule are not in the Summary at all. Confirmed on the page image of all three Summary pages.

Figures to have open

• A redraw of Fig 10.13 (Part II p. 347): a small axis frame with the three named points, their coordinate triples written out, and the three hatted arrows drawn from the origin. The chapter's own drawing.
• A redraw of Fig 10.14 (Part II p. 347) built in stages for section 2: the box outline, then the foot of the perpendicular, then the floor triangle closing, then the upright triangle closing. The staging is added here; the chapter prints the finished box only.
• A reuse of that same finished box for section 4, with the two right triangles shaded in turn and their hypotenuses labelled, so the double application of Pythagoras is carried by one drawing rather than two.
• A four-row table for section 6, built with the repo's DataTable component, carrying the four rules as the chapter lists them, with a fifth column added in section 7 for the law each one follows from and the word "none" against the equality rule.
• A two-column comparison for section 5 showing the same magnitude formula on two different pages with its two different justifications. The pairing is added here.
• A redraw of Fig 10.28 (Part II p. 369) for the close of section 11: the unit circle, one arrow from the centre to a point on it, the angle marked at the centre, and the two components dropped to the axes and labelled as cosine and sine. The chapter draws the circle and the arrow; the two dropped components and their labels are the explanation's addition, and they are what makes the link back to direction cosines visible.
• No figure is needed for sections 8, 9, 10 or 12; the chapter prints none for any of that material and the kit shapes above carry it.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 10 "Vector Algebra", §10.5.1 Components of a vector, Part II pp. 347–348, with Fig 10.13 and Fig 10.14
• The magnitude by two applications of the Pythagoras theorem, and the four componentwise rules, Part II p. 348
• The three distributive laws and Remarks (i), (ii) and (iii), Part II p. 349
• The magnitude of a position vector as cited from Class XI, §10.2, Part II p. 339; direction ratios as defined, §10.2, Part II p. 340
• Examples 4 and 5, Part II pp. 349–350
• Exercise 10.2, questions 1, 4, 5 and 6, Part II p. 354; question 19, Part II p. 355; question 2 is named only as Example 5's abstract twin and is carried by the types-of-vectors topic
• Example 26 and Fig 10.28, Miscellaneous Examples, Part II p. 369; Miscellaneous Exercise, question 1, Part II p. 372
• Summary, the component-form bullet, Part II pp. 374–375