One point plus one direction fixes a line, in vector form and in the symmetric Cartesian form the parameter eliminates to

What to assume they know

• Position vectors, and the vector from one point to another as a difference of position vectors, from Chapter 10 of this volume
• Scalar multiplication of a vector, and what a negative multiplier does
• Two vectors being parallel exactly when one is a multiple of the other
• The unit vectors along the three axes, and writing a vector by components
• Equating the components of two equal vectors
• Making one variable the subject of a linear equation, and eliminating a variable between equations
• Direction ratios and direction cosines, the fact that any proportional triple serves, and the fact that either triple may sit beneath the fractions — the previous module
• Reading a fraction whose denominator is zero as a warning rather than a value
• The cross product as a vector at right angles to two given vectors, from Chapter 10 of this volume

What they should be able to do

• State the two conditions the chapter says are each enough to fix a line, and say which of the two the chapter goes on to develop
• Set up the vector equation from a point and a direction, deriving it rather than quoting it
• Read the equation in both directions: every point of the line satisfies it, and every value of the parameter delivers a point of the line
• Identify the components of the direction vector as a set of direction ratios, and go back the other way
• Distinguish a component of the direction vector from the length of that vector, as the chapter's own remark warns
• Write the vector equation given a point and a direction vector in components
• Obtain a direction at right angles to two given directions, and use it to build a line
• Show that changing the fixed point to another point of the same line, or scaling the direction vector, leaves the line unchanged
• Recognise which results about lines the Summary states that the body never derives
• Split one vector equation into three scalar equations by comparing components
• State what the three parametric equations say about how the coordinates move together
• Eliminate the parameter and explain why the three resulting quotients must all be equal
• Read a printed symmetric equation backwards, extracting a point and a set of direction ratios from it
• Handle a sign correctly when a coordinate of the anchor point is negative
• Convert between the vector form and the Cartesian form in both directions
• Write the alternative form in which direction cosines sit under the fractions, and say what changes and what does not
• Say what a zero denominator means, and write the equation of a line for which two of the three denominators vanish
• Produce the symmetric equation for a line drawn parallel to a printed one, through a stated point

Where it usually goes wrong

• "The direction vector has to reach from the fixed point to the moving point." It has to be parallel to the line, nothing more. Fig 11.3 draws it off to one side on purpose. Students who believe otherwise think the equation only works for one particular length of direction vector.
• "A line has one equation." It has endlessly many in either form, because any point of the line will serve as the anchor and any non-zero multiple of the direction will serve as the direction. Two students can hand in equations that share not one symbol and both be right, which is why marking this topic by pattern-matching fails. Substituting a second point of the line back in is the check that settles it.
• "The parameter is a coordinate." It is a dial. Turning it moves the point along the line; its value is not a length unless the direction vector happens to have length one, which the chapter never requires.
• "Only positive values of the parameter give points of the line." Negative values run back the other way and zero returns the anchor point. The chapter says the multiplier is a real number and means it.
• "The letter under the modulus bars is the same as the letter in the middle of the components." The chapter's own Remark exists to stop this. One is a component; the other is a length.
• "Two points is the other case, so it must be somewhere in the chapter." The chapter announces it on Part II p. 381 and never returns to it. The result is in the Summary and nowhere else. A student who goes looking for the derivation will not find one, and should build it instead: the difference of the two position vectors is a direction for the line, and then this topic's own result applies.
• "A vector at right angles to two lines has to be found by solving two equations." It can be, and the cross product does it in one step. The chapter uses the cross product for exactly this purpose later on; Miscellaneous Exercise Q5 expects it here.
• "Each fraction gets its own parameter." All three quotients equal one common value, and that value is the parameter. Letting them differ describes a region of space rather than a line, and it is the error that produces three unrelated answers from one question.
• "The numbers on top are the point." They are the point with its signs reversed. A minus in the numerator means a positive coordinate; a plus means a negative one. Example 6 and Exercise 11.2 Q7 both turn on this.
• "The denominators are the direction cosines." They are direction ratios, and usually not normalised. The chapter's Note offers the cosine version as an alternative; the Summary prints only that version, which makes this worse rather than better.
• "A zero denominator means the equation is wrong." It means that coordinate never changes. The line is confined to a plane where that coordinate is fixed. Miscellaneous Exercise Q2 has two such denominators and the chapter offers no guidance at all.
• "I can clear the denominators and get a single equation." Clearing gives two independent equations, not one, because a line in space is the meeting of two conditions. One equation in three coordinates does not describe a line.
• "The parameter disappears in the elimination." It survives as the common value of the three quotients, which is how you get a point back out of a printed symmetric equation without redoing the algebra.
• "The Cartesian form is the real equation and the vector form is a shortcut." They are two representations of the same object, and the chapter derives the second from the first. Which one is convenient depends on the question: angles and distances come out of the vector form with less writing, while checking whether a stated point lies on the line is quicker in the Cartesian one.
• "The chapter will do planes next, so a line and a plane are coming." They are not. §11.3 is followed by §11.4 on the angle between two lines, and the chapter ends at §11.5.2.

Questions to check understanding

• Write the equation in vector form for a line through a stated point, parallel to a stated vector — the form of Example 6 and of Exercise 11.2 Q4
• Given a point as a position vector and a direction as a vector, produce the equation in both forms — the form of Exercise 11.2 Q5
• Read direction ratios off a vector equation, and build a direction vector from stated direction ratios
• Decide whether a stated point lies on a line given in vector form, and produce the parameter value if it does
• Find a direction at right angles to two stated lines and write the equation of a line through a given point with that direction — the form of Miscellaneous Exercise Q5
• Build the equation of the line through two stated points, given only the result the body of the chapter derives
• Convert a vector equation to Cartesian form, showing the elimination — the form of Example 6
• Convert a printed Cartesian equation to vector form, flipping the signs correctly — the form of Exercise 11.2 Q7
• Write the equation of a line through a stated point parallel to a printed line — the form of Exercise 11.2 Q6
• Read a point and a direction-ratio triple off a printed symmetric equation
• Decide whether a stated point lies on a line given in Cartesian form
• Write both forms for a line whose direction has one or more zero entries — the form of Miscellaneous Exercise Q2
• Rewrite a Cartesian equation with direction cosines beneath the fractions
• Show that two differently written equations, in either form and with different anchors, describe one and the same line

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 two conditions (§11.3, Part II p. 381). The chapter opens the section by listing two circumstances, either of which settles a line: a point together with a direction, or two points. It then gives §11.3.1 to the first. The second gets no subsection anywhere in the chapter — see the rationalisation note below. Section 1.
• The setup and Fig 11.3 (§11.3.1, Part II pp. 381–382). An anchor point carries a position vector of its own; a direction vector is given; a moving point carries the position vector the equation is about. Read off the page image, the figure puts the origin at the lower left with the three axes, marks the fixed point and the moving point along a line labelled at its right-hand end, draws both position vectors from the origin, and draws the direction vector as a separate arrow set above and parallel to the line, not lying on it. That separation is the figure's best idea: the direction is not a piece of the line, it is a direction the line happens to have.
• The derivation (§11.3.1, Part II p. 382). Three lines. The join from the fixed point to the moving point is parallel to the direction vector, so it is some real multiple of it. That join is also the difference of the two position vectors. Equate the two descriptions and rearrange. Verified: the whole content is the parallel-means-multiple fact from Chapter 10, applied once. The chapter then adds the converse in one sentence — each value of the multiplier hands back a point of the line.
• A parameter symbol that failed to print (Part II p. 382). The chapter's central displayed result — the vector equation itself, numbered one — prints its parameter as a double chevron instead of the Greek letter. Read on the printed page and unambiguous. Every other appearance of the parameter in the chapter, including the parametric equations six lines below and the Summary bullet on Part II p. 392, uses the Greek letter correctly. The one line a student would copy off the page is the one line that is broken. It is worth thirty seconds at the end of section 3, the moment the equation is shown: name it, show the correct symbol, and tell the student their own printed copy has the same mark.
• The Remark on the direction vector's components (Part II p. 382). If the direction vector is written by components, those three numbers are a set of direction ratios for the line; and given direction ratios, that combination of the unit vectors is a vector parallel to it. Both directions are printed, and this is the hinge between this module and the previous one.
• The warning inside the same Remark (Part II p. 382). The chapter closes the Remark by warning against confusing the middle component with the length of the direction vector — the same letter is doing two jobs on one line. Verified as a live hazard: the letter names the y-component in the component form, while the same letter under a modulus names the length of the whole vector. Show both at once, once, and move on.
• Example 6, first half (Part II pp. 382–383). Through the point with coordinates five, two, minus four, parallel to the direction with components three, two, minus eight. Verified: the equation is that point's own position vector plus the parameter times the direction vector. The chapter then expands the right-hand side by components, which is where the second half of the explanation picks the same example up again.
• Exercise 11.2 Q4 (Part II p. 389). Through the point with coordinates one, two, three, parallel to the direction with components three, two, minus two. Verified: the equation is that point's position vector plus the parameter times that direction. A one-line substitution, and the right item to work because nothing in it needs simplifying.
• Miscellaneous Exercise Q5 (Part II p. 391). A line through the point with coordinates one, two, minus four, at right angles to two lines given in Cartesian form with direction ratios three, minus sixteen, seven and three, eight, minus five. Verified: a direction at right angles to both is their cross product, which works out to twenty-four, thirty-six, seventy-two, and every entry carries a factor of twelve, so two, three, six serves just as well. The answer is the named point's own position vector plus the parameter times that tidy direction. Two things worth saying aloud. First, this is the item where the previous module pays off — any proportional triple is as good as any other, so the factor of twelve may simply be dropped. Second, the chapter never states, as a result, that the cross product of two vectors is at right angles to both; it uses the fact later, on Part II p. 386, in one clause. Chapter 10 supplies it.
• Exercise 11.2 Q5 (Part II p. 389). A point given as a position vector and a direction given as a vector, wanted in both forms — so this single item carries both halves of the explanation. Verified: the vector form is the direct substitution; the Cartesian form has denominators one, two and minus one, under x minus two, y plus one and z minus four. Note the printed slip recorded below before showing this item.
• The unnumbered heading (Part II p. 382). The Cartesian material sits under a bold black heading that carries no section number — the chapter's numbered headings jump from §11.3.1 straight to §11.4. There is a second such unnumbered heading later, on Part II p. 387. Worth knowing because a student navigating by section number will not find this material listed anywhere.
• The component split (Part II p. 382). The moving point's position vector, the anchor's position vector and the direction vector are each written by components; substituting into the vector equation and matching the coefficients of the three unit vectors gives three scalar equations, one per coordinate. Verified: the step is legitimate exactly because the three unit vectors are independent — two vectors are equal only when all three pairs of components agree. The chapter does not say why matching is allowed.
• The three parametric equations (Part II p. 382). Each coordinate is its anchor value plus the parameter times the matching direction entry. Verified: the same parameter appears in all three, which is the whole content — one dial, three dependent readings. A student who lets the three parameters differ has described a solid region, not a line.
• The elimination (Part II p. 382). Make the parameter the subject of each of the three equations and set the three expressions equal. Verified: each expression is a coordinate gap over the matching direction entry, so the symmetric form is a statement that three particular quotients coincide, and the common value is the parameter itself. The chapter drops the parameter at this point and never mentions that it survives as the common value; that is a useful thirty seconds, because it explains why the form works and how to get a point back out of it.
• The boxed Note (Part II p. 382). The same shape with the three direction cosines beneath the fractions instead of direction ratios. Verified: both are correct, since cosines are themselves a set of ratios; what changes is that the common value of the three quotients now equals the actual distance travelled from the anchor, because the direction has length one. The chapter states the form and does not draw that consequence. It is an added remark and should be flagged as such.
• Example 6, second half (Part II pp. 382–383). The chapter takes the vector equation it has just built, expands the right side by components, and eliminates. Verified: through the point with coordinates five, two, minus four, with direction three, two, minus eight, the three fractions are x minus five over three, y minus two over two, and z plus four over minus eight. Note the third numerator: the point's coordinate is negative, so the equation carries a plus. That sign flip is the single most common arithmetic slip in this topic and this example contains it.
• Exercise 11.2 Q6 (Part II p. 389). Through the point with coordinates minus two, four, minus five, parallel to a printed line whose denominators are three, five, six. Verified: the direction is read straight off the printed line's denominators, and the answer is x plus two over three, y minus four over five, z plus five over six. Two of the three numerators carry a plus because two of the point's coordinates are negative. This is the cleanest exercise item in the chapter for section 10.
• Exercise 11.2 Q7 (Part II p. 389). A printed Cartesian equation, wanted in vector form. Verified: the anchor point has coordinates five, minus four, six — read off by flipping the sign of each constant in the numerators — and the direction has components three, seven, two. The vector equation is that point's position vector plus the parameter times that direction. The sign flip is the whole exercise, and a student who reads the anchor's second coordinate as plus four has made the error this item exists to catch.
• Miscellaneous Exercise Q2 (Part II p. 390). A line parallel to the x-axis through the origin. Verified, and this is the sharpest item in the chapter for section 11: the direction is one, zero, zero and the anchor is the origin, so the vector form is simply the parameter times the first unit vector. Written in the symmetric shape, two of the three denominators are zero, and the equation has to be read as the pair of conditions that the second and third coordinates both vanish while the first runs free. The chapter never discusses a zero denominator anywhere. Checked against the page image of every folio from 381 to 393. Leaving a fraction with a zero underneath unexplained is worse than not showing it.
• The Summary bullets (Part II p. 392). Three belong to this topic, and all three are now in one video. One gives the vector equation from a point and a direction — the body's own result. One gives the equation from two points as the first position vector plus the parameter times the difference of the two. That second one is not derived, not stated and not used anywhere in the body of the chapter. The third gives the Cartesian form with direction cosines beneath the fractions. Verified against the body: the Summary prints only the cosine version, while the body prints the ratio version as its main result and the cosine version in a Note. A student revising from the Summary alone will think the denominators must be normalised. Make it a beat.

Figures to have open

• A redraw of Fig 11.3 (Part II p. 382): the axis frame, the anchor point and the moving point on a line labelled at one end, both position vectors drawn from the origin, and the direction vector drawn as a separate arrow parallel to the line and clear of it. Keep that separation; it is the whole visual argument. Carries sections 2 and 3 as one drawing that gains elements.
• A step-by-step version of the same drawing for section 3, with the multiplier running through positive values, zero and negative values so the moving point sweeps the line in both directions.
• A large pair of glyphs to close section 3: the chevron as printed and the Greek letter as intended. Not in the book.
• A single line drawn once for section 7, with three different anchor points marked and three different direction arrows, and the three resulting equations set below in one type size so they read as peers. Not in the book; the chapter makes this point nowhere.
• A dial with three coupled readouts for section 8. An added device; the chapter prints the three parametric equations with no illustration. It should reappear, unchanged, behind section 9 when the dial is eliminated.
• An annotated printed symmetric equation for section 10, with the sign flips drawn as arrows. Not in the book; the chapter never annotates its own form.
• An axis frame for section 11 with a line lying along the x-axis, and the two fixed coordinates marked as the conditions they are. Not in the book, and the only figure in this brief that the chapter would need and does not supply.
• No figure is needed for sections 1, 4, 5, 6 or 12; those are text, notation and arithmetic.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 11 "Three Dimensional Geometry", §11.3 Equation of a Line in Space, the two conditions, Part II p. 381
• §11.3.1, the setup and Fig 11.3, Part II pp. 381–382; the derivation, the converse sentence and the numbered vector equation, Part II p. 382
• The Remark on direction ratios and the warning about the repeated letter, Part II p. 382
• The unnumbered heading deriving the Cartesian form from the vector form, the component split, the parametric equations, the elimination and the boxed Note, all Part II p. 382
• Example 6 in full, both halves, Part II pp. 382–383
• Exercise 11.2 questions 4, 5, 6 and 7, Part II p. 389
• Miscellaneous Exercise questions 2 and 5, Part II pp. 390–391
• Summary, the vector-equation bullet, the two-point bullet and the Cartesian-form bullet, Part II p. 392