Deep Dive: Position Vectors Distance and Foot of Perpendicular

Provenance: written 2026-09-29. Read/checked directly for this document: research/veridian/WMA14-verified-facts.md in full, with a targeted re-grep on "perpendicular", "foot of" and "position vector" (§1 spec text for 7.1–7.7, §4 mark-scheme conventions, §5.7's vectors subsection in full, §6's series/document count, §7's standalone-lesson shortlist which does not include this topic, §8's flagged gaps, and the dedicated lesson-audit-log entry for this exact file); the complete lesson file content/veridian/curriculum/wma14/position-vectors-distance-and-foot-of-perpendicular.ts (all 1,035 lines — both worked chains, all seven MCQs, the trap-taxonomy, the reference card and its own header/flag disclosure); the sibling lesson content/veridian/curriculum/wma14/vectors-scalar-product-and-skew-lines.ts (checked for overlap, since both lessons share spec-7 territory); and all three existing WMA14 technique-hub files — codex-formula-and-method-bank.md §2.10 and §7, forge-exam-technique-guide.md §1 and its foot-of-perpendicular checklist item, and signal-frequency-and-boundaries.md §1 and §3 — to confirm what is already written and avoid restating it. content/veridian/question-bank/wma14.ts exists but contains no citation of "perpendicular", "foot of", "position vector" or "distance" (checked directly by grep), so nothing there overlaps either. Every new numeric example introduced below was independently computed and cross-checked with sympy (exact Matrix/Rational/sqrt arithmetic) before being written in, following this course's standing discipline.

#WMA14 · CODEX DEEP-DIVE — Position Vectors, Distance, and the Foot of a Perpendicular (spec 7.4–7.5)

#What this document adds, and what it deliberately does not repeat

This is a single-topic deep-dive on spec points 7.4 ("Position vectors") and 7.5 ("The distance between two points"), specifically as they combine with 7.6 (vector equations of lines) and 7.7 (the scalar product) to produce this unit's genuinely hardest recurring question type: finding the foot of the perpendicular from an external point to a line, the resulting shortest distance, and the reflection of that point in the line. It does not reproduce the lesson file's own two worked chains, its chain-drill, its method-comparison block, its five foundational MCQs, or its marked-solution — those stay the canonical worked exemplars, and a reader who wants full past-paper-length working, or the foundational vector-arithmetic teaching that precedes the hard content, should go there. It also does not re-derive the frequency finding already stated precisely in signal-frequency-and-boundaries.md §1 (this spec area is "seen in 1 of 7 series checked") — that finding is cited below, not recomputed.

What follows instead is three things this course's existing WMA14 material states informally or applies case-by-case but never writes down in one place: (1) the underlying vector algebra carried one level more formally than a teaching lesson needs to, including a general closed-form solution to the foot-of-perpendicular problem that the lesson solves afresh for each specific scenario rather than stating once; (2) a structural taxonomy of the genuinely distinct cases this technique splits into, each illustrated with a short, fresh micro-example using numbers that appear nowhere else in this course's WMA14 material; and (3) an exam-angle synthesis that reads the one real anchor question this spec area has — October 2021, Q7 — at the level of its full mark-scheme dependency structure and its explicitly credited alternative methods, rather than narrating it part by part again. Section 2.4 in particular states an observation about why the documented wrong equation is solvable that is not, so far as this session's reading of the facts bank and the lesson file established, written down anywhere in this course's existing material; it is flagged there as reasoning derived for this document, not as a fact drawn from a primary Pearson source.

#1. The spec points themselves, and where this topic actually sits

Verbatim from the spec (Issue 3, April 2019, Unit P4, p.29, quoted in WMA14-verified-facts.md §1):

• 7.4 Position vectors. Guidance: $\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}=\mathbf b-\mathbf a$.
• 7.5 The distance between two points. Guidance: the distance $d$ between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$.

Read on their own, these are two short, self-contained facts — nothing in either guidance clause mentions a line, a foot of a perpendicular, or a reflection. Those harder ideas only enter once 7.4 and 7.5 are combined with 7.6 (vector equations of lines, forms $\mathbf r=\mathbf a+t\mathbf b$ and $\mathbf r=\mathbf c+t(\mathbf d-\mathbf c)$, and the conditions for two lines to be parallel, intersecting or skew) and 7.7 (the scalar product, and the fact that $\mathbf a\cdot\mathbf b=0$ for nonzero $\mathbf a,\mathbf b$ means $\mathbf a\perp\mathbf b$). The spec's own numbering places 7.4–7.5 immediately before 7.6–7.7, which is a genuinely useful ordering fact, not just trivia: position vectors and the distance formula are the vocabulary, and the foot-of-perpendicular problem is the first place in the unit that vocabulary gets used for something harder than direct computation. This is also exactly the ordering the lesson file's own header states it chose to follow (line 31 of that file), so nothing about the sequencing below is new — it is restated here only because it is the reason this deep-dive's own case taxonomy in §3 starts with the plain two-point case before the harder line-based case.

#2. The theory, stated more formally than a teaching lesson needs to

#2.1 A position vector has a fixed home; a free vector does not — and this is the single fact 7.4 exists to formalise

A point is a location. A vector is a length and a direction, with no location of its own — the same vector $(3,4,12)$ can be drawn starting anywhere, and every one of those arrows is the same vector. A position vector is the special case of tying a vector's tail to one fixed point, the origin $O$: the position vector of a point $A$ is $\overrightarrow{OA}$, conventionally written $\mathbf a$. Every point has exactly one position vector, and unlike a general (free) vector, a position vector's starting point is not optional — it is always $O$. This is the entire content of spec 7.4's identity $\overrightarrow{AB}=\mathbf b-\mathbf a$: it is a statement about how to convert between the "fixed-to-the-origin" bookkeeping of position vectors and the "free, tail-to-tip" bookkeeping of an ordinary displacement vector, via one subtraction. Nothing about this is specific to three dimensions, to this exam board, or to this course — it is the defining relationship between an affine point space and the vector space of displacements between its points, and 7.4's guidance clause states the $n$-dimensional-agnostic version of it directly.

The reason this distinction is worth restating formally, rather than assuming the lesson's own repeated warnings about it have done all the necessary work, is that §2.4 below turns entirely on keeping it straight: the single documented trap for this whole topic is, in the most literal possible sense, exactly the error of treating a position vector as if it already were the free vector a later step needs.

#2.2 The distance formula is $|\overrightarrow{AB}|$ written out in coordinates — not a second, independent fact to learn

Spec 7.5's guidance clause states the distance formula as a standalone result, $d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$, and it is entirely legitimate to use it exactly that way on paper. But it is worth being explicit that this is not a second piece of theory sitting alongside 7.4 — it is 7.4's own identity, composed with the ordinary 3D-Pythagoras magnitude formula: the distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is, by definition, the length of the displacement between them, $|\overrightarrow{AB}|=|\mathbf b-\mathbf a|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$, and squaring both sides recovers 7.5's formula exactly (the sign inside each square is irrelevant since every term is squared). Spec 7.5 is, in other words, spec 7.4 plus the magnitude formula already established for 7.2 — not an independent third fact. This is stated here as mathematical reasoning about how the spec's own numbered points relate to one another, not as a claim drawn from any primary Pearson document; no examiner report or mark scheme reviewed for this facts bank discusses the relationship between 7.4 and 7.5 explicitly.

#2.3 The foot-of-perpendicular condition, in closed form

The lesson file solves $\overrightarrow{AX}\cdot\mathbf d=0$ afresh for each individual scenario it presents, which is the right pedagogical choice for a teaching lesson — a student needs to see the substitution happen with real numbers, not memorise a formula that hides the mechanism. It is nonetheless worth writing the general solution down once, because doing so makes two things visible that case-by-case substitution does not.

For a line $l:\mathbf r=\mathbf c+t\mathbf d$ and an external point with position vector $\mathbf a$, the general point on $l$ is $\mathbf c+t\mathbf d$, so $\overrightarrow{AX}=(\mathbf c+t\mathbf d)-\mathbf a=(\mathbf c-\mathbf a)+t\mathbf d$. Setting $\overrightarrow{AX}\cdot\mathbf d=0$:

$$(\mathbf c-\mathbf a)\cdot\mathbf d+t(\mathbf d\cdot\mathbf d)=0 \quad\Longrightarrow\quad t^\ast=\frac{(\mathbf a-\mathbf c)\cdot\mathbf d}{\mathbf d\cdot\mathbf d}=\frac{(\mathbf a-\mathbf c)\cdot\mathbf d}{|\mathbf d|^2}.$$

This is the standard vector scalar projection formula — it says the parameter value at the foot of the perpendicular is the component of $\mathbf a-\mathbf c$ along $\mathbf d$, scaled by $1/|\mathbf d|^2$ to land in the line's own parametrisation. Two things follow immediately from having this closed form, neither of which is stated in the spec, the formula booklet, or anywhere in the facts bank's own §3 (the booklet-contents inventory) — flagged here in the same way the sibling parametric-curve deep-dive flags its own non-booklet identities, as correct standard mathematics rather than exam-board-sourced fact:

• The equation is only solvable because $\mathbf d\cdot\mathbf d\neq0$, which holds for any genuine direction vector (a line's direction vector is nonzero by definition — a "line" with $\mathbf d=\mathbf 0$ is not a line). This is the precise reason the other equation sometimes discussed as a hypothetical alternative slip, $\mathbf d\cdot\mathbf d=0$, is structurally different from the documented trap in §2.4 below: $\mathbf d\cdot\mathbf d$ is a fixed positive number with no $t$ in it at all, so that equation has no solution for any line, whereas $t^\ast$ above always exists and is always unique.
• The shortest distance itself has a closed form too, reached by substituting $t^\ast$ back into $|\overrightarrow{AX}|^2=|(\mathbf c-\mathbf a)+t\mathbf d|^2$ and simplifying (a short algebraic exercise, not reproduced step-by-step here since it is standard completing-the-square manipulation of the kind the lesson's own method-comparison block already demonstrates for specific numbers):

$$|\overrightarrow{AX}|^2_{\min}=|\mathbf a-\mathbf c|^2-\frac{\big[(\mathbf a-\mathbf c)\cdot\mathbf d\big]^2}{|\mathbf d|^2}.$$

This is a Pythagorean decomposition in disguise: $|\mathbf a-\mathbf c|^2$ is the squared distance from $A$ to the known point $C$ on the line, and the term subtracted off is the squared length of $C$'s own projection onto $\mathbf d$ — the decomposition is exactly the right-angled triangle $A,X,C$ that the lesson file's own derivation (its Pythagoras argument for why $X$ is the unique closest point) already establishes geometrically, just written as one formula instead of a proof by contradiction on an arbitrary other point $P$.

Both identities are checked against a fresh worked case in §3 below, and both are genuinely useful as an independent check on an exam answer reached by the lesson's own step-by-step method (form $\overrightarrow{AX}$, dot with $\mathbf d$, solve) — but neither is offered as an exam-room shortcut in place of showing the working the mark scheme actually rewards. §4 below states explicitly why: the real mark scheme's M1/dM1 structure credits the intermediate steps by name, and a candidate who jumps straight to a memorised closed-form formula without showing $\overrightarrow{AX}=\ldots$ and $\overrightarrow{AX}\cdot\mathbf d=0$ risks forfeiting exactly the marks this section's own derivation shows those steps are worth.

#2.4 What the documented trap equation actually solves — a structural observation, not a Pearson-sourced fact

WMA14-verified-facts.md §5.7 and the lesson file's own header both establish, with a direct primary-source quote, that the single most consequential documented error on this question type is dotting the line's raw general point directly against $\mathbf d$ — $(\mathbf c+t\mathbf d)\cdot\mathbf d=0$ — instead of first subtracting $\mathbf a$ to form $\overrightarrow{AX}$. Both existing course documents correctly state that this equation is solvable (it genuinely contains $t$) and explain why it is wrong (the vector tested for perpendicularity was never $\overrightarrow{AX}$). Neither source states what that equation actually computes, beyond "the wrong thing." It is worth being precise about this, because the answer is a specific, well-posed problem, not an arbitrary meaningless expression: setting the line's own general point $X$ perpendicular to $\mathbf d$, i.e. $X\cdot\mathbf d=0$, is exactly the condition $\S2.3$ derives for the foot of the perpendicular from the origin to $l$ — because $\overrightarrow{OX}=X-\mathbf 0=X$, so "$\overrightarrow{OX}\cdot\mathbf d=0$" and "$X\cdot\mathbf d=0$" are the same equation.

In other words: a candidate who makes this documented substitution has not written down nonsense. They have correctly solved the foot-of-perpendicular problem for a different external point — the origin — while believing they were solving it for $A$. This reframes the trap from "a broken equation that happens to still have a solution" into "the right method, applied to the wrong point, because the point it was actually being applied to (a position vector's own origin-anchored bookkeeping) got substituted in place of the point the question asked about" — which is the same point §2.1 makes about position vectors having a fixed home, now made completely concrete via the equation the examiner's report itself documents.

A short illustration, using fresh numbers not drawn from any real question or from this course's existing lesson. Take $l:\mathbf r=(1,-2,3)+t(2,1,-2)$. Solving $X\cdot\mathbf d=0$ (the documented wrong-equation shape) for this line: $(1+2t,\,-2+t,\,3-2t)\cdot(2,1,-2)=2(1+2t)+(-2+t)-2(3-2t)=9t-6=0\Rightarrow t=\tfrac23$, giving the point $\left(\tfrac73,-\tfrac43,\tfrac53\right)$ — which is, independently checked, exactly the foot of the perpendicular from $O=(0,0,0)$ to $l$, and nothing to do with whatever external point $A$ a real question might have specified. This is offered as a worked illustration of the structural point above, not as a claim about any specific past-paper question — no primary source reviewed for this facts bank states this origin-equivalence explicitly, and it is not attributed to one.

#3. The real distinct cases, each with a short, fresh micro-example

Spec 7.4–7.5's own content is thin enough (two short guidance clauses, §1 above) that it does not itself generate multiple "sub-cases" the way, say, binomial expansion or integration by parts do. The genuine structural variation in this topic comes from how 7.4–7.5 combine with 7.6–7.7, and the real anchor question (§4) exercises exactly two of the four cases below in one nine-mark item. All four numeric examples were computed and independently cross-checked with sympy; none reuses a scenario, a direction vector, or a numeric answer from the lesson file, the sibling skew-lines lesson, or either other technique-hub document.

Case A — forming $\overrightarrow{AB}$ and $|\overrightarrow{AB}|$ directly from two given position vectors (spec 7.4+7.5, no line involved). This is the case the spec's own guidance clauses describe literally, with no harder machinery layered on. Micro-example: $A$ and $B$ have position vectors $\mathbf a=(2,0,-3)$ and $\mathbf b=(3,4,5)$. Then $\overrightarrow{AB}=\mathbf b-\mathbf a=(1,4,8)$, and $|\overrightarrow{AB}|=\sqrt{1^2+4^2+8^2}=\sqrt{81}=9$ — confirmed independently, $1+16+64=81$.

Case B — the foot of the perpendicular from an external point to a line (7.4+7.5 combined with 7.6+7.7; the hard case, and the one the real anchor question tests). Line $l:\mathbf r=(1,-2,3)+t(2,1,-2)$, external point $A=(5,2,0)$ (not on $l$: substituting each coordinate into $l$'s equation gives three inconsistent values of $t$, confirming this directly). Forming the general point $X=(1+2t,\,-2+t,\,3-2t)$ and $\overrightarrow{AX}=X-\mathbf a=(2t-4,\,t-4,\,3-2t)$, the perpendicularity condition $\overrightarrow{AX}\cdot\mathbf d=0$ expands to $2(2t-4)+1(t-4)-2(3-2t)=9t-18=0$, giving $t=2$. Substituting back, $X=(5,0,-1)$, and the shortest distance is $|\overrightarrow{AX}|=|(0,-2,-1)|=\sqrt{0+4+1}=\sqrt5$. Independently cross-checked against §2.3's closed form: $t^\ast=\dfrac{(\mathbf a-\mathbf c)\cdot\mathbf d}{|\mathbf d|^2}=\dfrac{(4,4,-3)\cdot(2,1,-2)}{9}=\dfrac{18}{9}=2$, and the distance-squared shortcut gives $|\mathbf a-\mathbf c|^2-\dfrac{[(\mathbf a-\mathbf c)\cdot\mathbf d]^2}{|\mathbf d|^2}=41-\dfrac{324}{9}=41-36=5$, matching $(\sqrt5)^2$ exactly.

Case C — the boundary case: $A$ already lies on $l$. Not documented in any primary source this facts bank has reviewed for this topic — no examiner report or mark scheme discusses it — but worth stating explicitly as the genuine edge of the method's domain, because a student meeting it without warning could reasonably wonder whether "the closest point on the line to a point already on the line" is a well-formed question at all. It is, and the same method answers it without needing a special case: using the same line as Case B, take $A$ to be the point on $l$ itself at $t=1$, i.e. $A=(3,-1,1)$. Solving $\overrightarrow{AX}\cdot\mathbf d=0$ exactly as before gives $9t-9=0\Rightarrow t=1$ — the unique solution is $t=1$, so $X=A$ and the shortest distance is $0$, exactly as geometric intuition demands. The method does not need to detect this case separately; the algebra collapses to the right answer on its own, because $\overrightarrow{AX}=\mathbf 0$ trivially satisfies $\overrightarrow{AX}\cdot\mathbf d=0$ regardless of $\mathbf d$, and the equation's solution is unique since $\mathbf d\cdot\mathbf d\neq0$ (§2.3).

Case D — the reflection of $A$ in $l$, via the closed form. Reusing Case B's own $X=(5,0,-1)$ and $A=(5,2,0)$: the real anchor question's own method (§4 below), $\overrightarrow{AX}=\overrightarrow{XB}$, gives $B=2X-\mathbf a=(10,0,-2)-(5,2,0)=(5,-2,-2)$ — independently checked, the midpoint of $A=(5,2,0)$ and $B=(5,-2,-2)$ is $(5,0,-1)=X$, confirming the reflection is correct. This reuses spec 7.4's own identity a second time (a midpoint is itself a position-vector average, the same structure as $\overrightarrow{AB}=\mathbf b-\mathbf a$ rearranged), exactly as the real mark scheme's own method credit does — not a new formula, as both the lesson file and codex-formula-and-method-bank.md §2.10 already state.

#4. The exam-angle synthesis — what one real, richly-detailed question actually shows

The honest denominator, stated first. signal-frequency-and-boundaries.md §1 counts this spec area (7.4–7.5, foot-of-perpendicular) as evidenced in 1 of the 7 series this facts bank mined question-by-question, and states explicitly that a 1-of-7 count here is a sourcing artefact rather than proof of rarity — 7.4 and 7.5 are compulsory, explicitly spec-listed skills, and this course's own review has simply not yet turned up a second series that examines the harder foot-of-perpendicular application specifically. Everything below rests on that single real anchor: October 2021, WMA14/01, Q7(a)+(b), 9 marks total (7+2), Publications Code WMA14_01_2110_MS, Question Paper Log Number P69198A — independently re-verified against both the primary-source mark scheme and the primary-source examiner's report this session's predecessor audit fetched directly from qualifications.pearson.com (WMA14-verified-facts.md §5.7).

The question's own structure, read as a dependency graph rather than narrated part by part. Part (a) asks for the foot of the perpendicular $X$ and the shortest distance from $A(9,-3,2)$ to $l:\mathbf r=(4,2,-3)+\lambda(-4,-3,5)$; part (b) asks for the reflection $B$ of $A$ in $l$. The real mark scheme's own dependency chain for part (a) is M1 (form $\overrightarrow{AX}$ from the general point, either order) → M1 (the full method: dot with $\mathbf d$, solve for $\lambda$) → A1 ($\lambda=2/5$) → dM1, explicitly marked dependent on the preceding M1, for substituting $\lambda$ back in to get $X$'s coordinates. Part (b) is a separate two-mark item, credited as: "Uses AX = XB or similar correct method" (M1), correct position vector of $B$ (A1) — eight words of direct quotation, the mark scheme's own phrasing, cited to Oct 2021 Q7(b). This dependency structure is exactly what §2.10 of codex-formula-and-method-bank.md and the FORGE guide's own §1 already state in general terms ("write every intermediate stage as its own visible line"); what this section adds is naming the specific mark-type sequence for this exact question — M1 → M1 → A1 → dM1, then a separate M1 → A1 — rather than repeating the general principle.

The examiner's report credits two named alternative methods, not just the AX·d=0 route. Beyond the dot-product method this whole document has otherwise used throughout, the Oct 2021 report states plainly: "Other methods were seen in part (a)" — including minimising $|\overrightarrow{AX}|$ directly, via either differentiation or completing the square. This is a genuinely useful, exam-actionable fact distinct from the trap material: a candidate who is unsure of the vector method under time pressure has a real, examiner-documented fallback, and the general P4 marking principle that a correctly-attempted three-term-quadratic solution earns credit "by factorisation, by the quadratic formula... or by completing the square" (WMA14-verified-facts.md §4) applies here too, once $|\overrightarrow{AX}|^2$ is expressed as a quadratic in the line's parameter. The lesson file's own method-comparison block already demonstrates both alternative routes on its own worked numbers; this section states only that the examiner's report itself is the source naming them as credited, not merely as mathematically valid.

The single documented trap, restated precisely rather than narrated again, and now explained structurally (§2.4). The report's own account is that most incorrect responses set the line's raw general point dotted with $\mathbf d$ equal to zero, rather than $\overrightarrow{AX}$ dotted with $\mathbf d$ — "This immediately resulted in 0 marks for (a)(i)," a nine-word direct quotation. Independently recomputed this session (matching the facts bank's own prior recomputation): that wrong equation gives $\lambda=37/50$, a specific, plausible-looking wrong value, against the correct $\lambda=2/5$ — not an equation that visibly breaks. §2.4 above adds the structural reason this is true: the wrong equation is not gibberish, it is the correct equation for a different problem (the foot of the perpendicular from the origin, not from $A$), which is precisely why it produces a plausible number rather than an obviously-impossible one.

What this synthesis will not do, on the strength of one question. Exactly as the sibling area-under-a-parametric-curve deep-dive states for its own single-citation topic: one real, richly-detailed, independently-verified data point exists for this spec area, and it is treated here as exactly that — a detailed anatomy of one question, not a frequency claim, a likelihood, or a grade-boundary-relevant statement. signal-frequency-and-boundaries.md §2 already establishes that no grade-boundary primary source has been reviewed for this paper at all; this document introduces none either.

#Sources used

research/veridian/WMA14-verified-facts.md §1 (spec text for 7.1–7.7, verbatim), §4 (mark-scheme conventions, verbatim, including the general three-term-quadratic and formula-first principles applied to this topic in §4 of this document), §5.7 (the full Oct 2021 Q7(a)+(b) entry — mark-scheme dependency chain, both quoted examiner's-report passages, and the independently-recomputed $\lambda=37/50$ vs $2/5$ arithmetic), §7 (confirms this spec area is not among the topics judged to have enough verified material for a standalone lesson on its own evidence). content/veridian/curriculum/wma14/position-vectors-distance-and-foot-of-perpendicular.ts (full file, including its own header/flag disclosure of exactly which content is VERIDIAN-original scaffolding versus the one real citation, checked to avoid duplicating any worked exemplar, MCQ, or trap-taxonomy entry). codex-formula-and-method-bank.md §2.10 and §3 Part 1 (Vectors), forge-exam-technique-guide.md §1 and its foot-of-perpendicular checklist item, and signal-frequency-and-boundaries.md §1–§3, all cross-checked to avoid restating their own content. The scalar-projection closed form and Pythagorean distance-squared identity in §2.3, and the origin-equivalence observation in §2.4, are standard vector algebra derived directly from the same perpendicularity condition ($\overrightarrow{AX}\cdot\mathbf d=0$) this facts bank's own citations already establish as the real, examined method — presented here as mathematical reasoning, not attributed to any Pearson document, since no primary source reviewed states either explicitly. Every fresh numeric example in §2.4 and §3 was computed and independently checked with sympy before being written in.

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