Vectors — position vectors, distance, and the foot of a perpendicular

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Vectors in two and three dimensions

Spec 7.1 states the content in five words: "Vectors in two and three dimensions." A vector is written in component form, $(x,y)$ or $(x,y,z)$ , or equivalently using the unit vectors $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ along the three axes: $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ . Both notations describe exactly the same object; which one a question uses is a presentation choice, not a difference in the underlying mathematics.

Two of a 3D vector's three components are already familiar from two-dimensional work; the third, $z$ (or $\mathbf{k}$ ), is the genuinely new ingredient this unit adds. Every algebraic rule that already works in two dimensions — adding component-wise, scaling by a constant, finding a magnitude — extends to three dimensions with nothing conceptually new to learn, only one more coordinate to carry through the working every time.

One distinction is worth stating before anything else in this lesson, because the rest of it is built on keeping it straight: a vector has a magnitude and a direction, but no fixed location — the SAME vector $(2,1,-3)$ can be drawn starting from anywhere, and it is still the same vector. A POINT, by contrast, is a fixed location with no length or direction of its own. Confusing the two — treating a point like a vector, or a vector like it has a fixed 'home' — is the single most common root cause of the errors this lesson documents, in both its foundational and its harder content.

Magnitude, and the unit vector in the direction of a

Spec 7.2's own guidance states the two skills required in one sentence: "Students should be able to find a unit vector in the direction of a, and be familiar with |a|." $|\mathbf{a}|$ is the MAGNITUDE of $\mathbf{a}$ — its length, a single non-negative number, computed as $|\mathbf{a}| = \sqrt{a_1^2+a_2^2+a_3^2}$ for $\mathbf{a}=(a_1,a_2,a_3)$ .

A UNIT vector is a vector with magnitude exactly 1 — and "the unit vector in the direction of $\mathbf{a}$ " means: keep pointing the same way $\mathbf{a}$ points, but rescale the length to exactly 1. That rescaling is $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ — dividing every component of $\mathbf{a}$ by the single number $|\mathbf{a}|$ . This is not a separate fact to memorise: scaling a vector by a positive constant $k$ scales its magnitude by that same $k$ (spec 7.3's own rule, met properly two blocks below), so scaling $\mathbf{a}$ by $\dfrac{1}{|\mathbf{a}|}$ scales its magnitude by $\dfrac{1}{|\mathbf{a}|}$ too — giving a new magnitude of $|\mathbf{a}| \times \dfrac{1}{|\mathbf{a}|} = 1$ , exactly as required, for any nonzero $\mathbf{a}$ .

A genuinely fast way to catch a unit-vector slip before it costs a mark: square every component of the claimed answer and add them up. If the total isn't 1, the division was never actually carried out on every component, or was carried out by the wrong number entirely — this check appears explicitly in the worked example below, and is worth running as a habit on every unit-vector answer, not just when something already looks wrong.

Mechanism

Where |v| = √(x² + y² + z²) actually comes from — Pythagoras, run twice

This formula is not a new rule to memorise; it is ordinary 2D Pythagoras, applied twice. Take $\mathbf{v}=(x,y,z)$ . First, project $\mathbf{v}$ straight down onto the $xy$ -plane — this projection is the vector $(x,y,0)$ , and by the familiar 2D case, its length is $\sqrt{x^2+y^2}$ . Second, notice that $\mathbf{v}$ itself, its projection $(x,y,0)$ , and a vertical segment of length $|z|$ (running straight up or down from the projection's endpoint to $\mathbf{v}$ 's own endpoint) form a second right-angled triangle — right-angled specifically because the $z$ -axis is perpendicular to the entire $xy$ -plane, so it is perpendicular to the projection no matter which direction the projection itself points in. Applying Pythagoras to THIS triangle: $|\mathbf{v}|^2 = \left(\sqrt{x^2+y^2}\right)^2 + z^2 = x^2+y^2+z^2$ , so $|\mathbf{v}| = \sqrt{x^2+y^2+z^2}$ — the 3D magnitude formula, reached by two applications of a 2D fact already known, rather than handed down as something new about three dimensions specifically.

Vector addition, scalar multiplication, and their geometric interpretation

Spec 7.3 covers two algebraic operations, each with a matching geometric picture, since the spec's own wording asks for both: "Algebraic operations of vector addition and multiplication by scalars, and their geometrical interpretations." Algebraically, both are component-wise: $(a_1,a_2,a_3)+(b_1,b_2,b_3) = (a_1+b_1,\,a_2+b_2,\,a_3+b_3)$ , and $k(a_1,a_2,a_3) = (ka_1,ka_2,ka_3)$ for a scalar $k$ . Nothing about either rule changes between two and three dimensions beyond carrying one more coordinate through.

Geometrically, scaling $\mathbf{a}$ by $k$ scales its LENGTH by $|k|$ and, if $k$ is negative, reverses its direction — $2\mathbf{a}$ points the same way as $\mathbf{a}$ but twice as far; $-\mathbf{a}$ points exactly the opposite way, at the same length. Addition follows the triangle law: to draw $\mathbf{a}+\mathbf{b}$ , draw $\mathbf{a}$ from any starting point, then draw $\mathbf{b}$ starting from WHERE $\mathbf{a}$ ENDED — the sum $\mathbf{a}+\mathbf{b}$ is the single vector from $\mathbf{a}$ 's start to $\mathbf{b}$ 's end. Subtraction is addition of a negative: $\mathbf{a}-\mathbf{b} = \mathbf{a}+(-\mathbf{b})$ , so drawing $\mathbf{a}-\mathbf{b}$ means reversing $\mathbf{b}$ first and then applying the same tip-to-tail rule — not simply flipping the arrow drawn for $\mathbf{a}$ , which is a different vector entirely.

Diagram — Vector addition and subtraction: the triangle law

x-axis: horizontal displacement · y-axis: vertical displacement

a: Drawn from a chosen starting point to wherever it ends — its own length and direction, nothing else.
b, drawn starting from where a ends: The tip-to-tail placement that makes the triangle law work: b's tail sits exactly at a's tip.
a + b (the resultant): The single vector from a's starting point straight to b's ending point — the third side of the triangle formed by a and b placed tip-to-tail.

The parallelogram law: An equivalent picture: draw a and b from the SAME starting point instead of tip-to-tail, complete the parallelogram they form, and a+b is the diagonal from that shared start to the opposite corner. Same resultant, same length, same direction — just a different construction reaching it.
a − b, drawn from a common origin: If a and b are drawn from the same starting point (as position vectors often are), a − b is the vector from B's endpoint TO A's endpoint — pointing from wherever b ends back to wherever a ends, not the other way round. This is exactly the geometric picture spec 7.4's own identity AB = b − a is built on, met properly in the next block.
Order does not matter for addition, but does for subtraction: a + b and b + a reach the identical resultant either way. a − b and b − a do not — they point in exactly opposite directions along the same line, one the negative of the other.

Common error: Drawing a − b by simply reversing the arrow already drawn for a, rather than reversing b first and then applying the ordinary tip-to-tail rule.

Correct: a − b = a + (−b): reverse b's direction (keep its length exactly the same), THEN add tip-to-tail as for ordinary addition. Reversing a instead produces b − a, not a − b — the correct-looking picture for the wrong calculation.

Position vectors, and the identity the rest of this unit is built on

A POSITION vector is the vector from the origin $O$ to a specific point: the position vector of $A$ is $\overrightarrow{OA}$ , almost always written $\mathbf{a}$ (lower-case, matching the point's own capital letter). Every point in this unit has exactly one position vector, and — unlike a general vector — a position vector's own starting point is fixed: always the origin.

Spec 7.4's , quoted verbatim from its own guidance: $\overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ . This is not a separate fact to memorise on top of the triangle law just derived — it IS the triangle law, applied to the specific triangle $O$ - $A$ - $B$ . Travelling from $O$ to $B$ can be done two ways: directly (the vector $\mathbf{b}$ ), or via $A$ — first to $A$ (the vector $\mathbf{a}$ ), then from $A$ to $B$ (the vector $\overrightarrow{AB}$ , whatever it turns out to be). Both routes reach the same place, so $\mathbf{a} + \overrightarrow{AB} = \mathbf{b}$ , and rearranging gives $\overrightarrow{AB} = \mathbf{b}-\mathbf{a}$ directly — reached from the addition rule already known, not introduced as a new one.

The one sign to get right, every time: AB (FROM A TO B) is $\mathbf{b}-\mathbf{a}$ — the DESTINATION's position vector minus the STARTING point's. Writing it backwards, $\mathbf{a}-\mathbf{b}$ , gives $\overrightarrow{BA}$ instead — the reverse vector, with every component's sign flipped. This is worth deliberately slowing down for on a first read of any question, because both directions are genuinely used throughout this unit (this lesson's own foot-of-perpendicular section later needs the vector from a fixed external point A to a moving point X, specifically in that order), and the two are easy to swap under time pressure.

Mechanism

The distance formula is |AB| written out in coordinates — not a separate rule

The DISTANCE between two points $A=(x_1,y_1,z_1)$ and $B=(x_2,y_2,z_2)$ is defined as the length of the vector between them: $d = |\overrightarrow{AB}|$ . Two facts already established supply everything needed to expand this into spec 7.5's own formula. First, by the 7.4 identity just derived, $\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = (x_2-x_1,\ y_2-y_1,\ z_2-z_1)$ . Second, by the magnitude formula from this lesson's own earlier mechanism block, the length of THAT vector is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ . So $d^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$ — spec 7.5's guidance, quoted verbatim as $d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$ , reached as a direct consequence of the position-vector identity and the magnitude formula rather than a third new thing to learn. One genuinely useful consequence of the SQUARING in this formula, worth stating explicitly: $(x_1-x_2)^2 = (x_2-x_1)^2$ regardless of which point is labelled which, so the DISTANCE between two points doesn't depend on which one is called A and which is called B — unlike the vector $\overrightarrow{AB}$ itself, which very much does (that's exactly what the previous block's sign warning was about). Distance is symmetric; the vector between two points is not — two related facts that are easy to blur into one if the derivation above is skipped and both are simply asserted as rules.

Worked, in full

Forming AB, finding the distance AB, and the unit vector in its direction (VERIDIAN-original)

01
$A=(1,2,2)$ , $B=(4,6,14)$ , so $\mathbf{a}=(1,2,2)$ and $\mathbf{b}=(4,6,14)$ . By spec 7.4's identity, $\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = (4-1,\ 6-2,\ 14-2) = (3,4,12)$ .
Earns: B1 — independent mark for forming AB in the correct order (destination minus start), not the reverse.
02
Find $|\overrightarrow{AB}|^2$ first, by summing the squares of its three components: $3^2+4^2+12^2 = 9+16+144 = 169$ .
Earns: M1 — method mark for attempting the sum-of-squares (magnitude²) calculation on the correct vector.
03
$|\overrightarrow{AB}| = \sqrt{169} = 13$ . This IS the distance $AB$ (spec 7.5) — no separate formula is needed, since the distance between two points is defined as exactly the magnitude of the vector between them, already computed in stage 1.
Earns: A1 — the correct magnitude and, in the same step, the correct distance. A script that computes AB and its magnitude but never states this number IS the distance has done all the arithmetic and still risks the mark reserved for the concluding statement.
04
The unit vector in the direction of $\overrightarrow{AB}$ : divide every component of $\overrightarrow{AB}$ by its own magnitude, 13: $\hat{\mathbf{v}} = \dfrac{1}{13}(3,4,12) = \left(\dfrac{3}{13},\ \dfrac{4}{13},\ \dfrac{12}{13}\right)$ .
Earns: M1 — method mark for dividing every component by the magnitude found in stage 3, not just some of them.
05
Check: $\left(\dfrac{3}{13}\right)^2 + \left(\dfrac{4}{13}\right)^2 + \left(\dfrac{12}{13}\right)^2 = \dfrac{9+16+144}{169} = \dfrac{169}{169} = 1$ . The result genuinely has magnitude 1, confirming it is a real unit vector rather than merely a scaled-down version of AB.
Earns: A1 — the correct unit vector, with the magnitude-1 check carried out rather than assumed. This is the fast self-check named in this lesson's own teach block above, run for real here rather than only described.

Complete it yourself

Complete the chain — CD, the distance CD, and the unit vector in its direction (VERIDIAN-original)

01
$C=(-1,5,2)$ , $D=(1,8,8)$ , so $\mathbf{c}=(-1,5,2)$ and $\mathbf{d}=(1,8,8)$ . By spec 7.4's identity, $\overrightarrow{CD} = \mathbf{d}-\mathbf{c} = (1-(-1),\ 8-5,\ 8-2) = (2,3,6)$ .
02
Find $|\overrightarrow{CD}|^2$ : $2^2+3^2+6^2 = 4+9+36 = 49$ .

In your own words

In one sentence: why does computing AB specifically require b − a, rather than a − b — and what does a − b actually represent instead?

A genuinely hard application: the foot of the perpendicular from a point to a line

The rest of this lesson combines everything above with two facts already established elsewhere in this unit's own "Vectors — scalar product, angle-finding, and skew/parallel/intersecting lines" lesson, recalled here rather than re-derived: (1) a line's vector equation, $\mathbf{r} = \mathbf{c} + t\mathbf{d}$ , where $\mathbf{c}$ is the position vector of one known point on the line, $\mathbf{d}$ is the line's direction, and $t$ ranges over every real number as it sweeps out the whole line (spec 7.6); and (2) the scalar-product perpendicularity test — for nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ , $\mathbf{u}\cdot\mathbf{v}=0$ exactly when $\mathbf{u}$ and $\mathbf{v}$ are perpendicular (spec 7.7).

The question type: given a point $A$ that is NOT on a line $l$ , find the point $X$ on $l$ that is closest to $A$ — called the foot of the perpendicular from $A$ to $l$ — and, usually as a following part, the shortest distance from $A$ to $l$ itself. This is a genuinely real, examined combination of spec 7.4 and 7.5 (position vectors; distance) with the line-equation and scalar-product tools above, and the research bank behind this lesson records one specific, well-documented way real scripts get the very first line of it wrong — covered in full below.

Mechanism

Setting up the right equation: AX · d = 0, not the general point dotted with d

Let $X$ be the general point on line $l: \mathbf{r}=\mathbf{c}+t\mathbf{d}$ , so $X$ has position vector $\mathbf{c}+t\mathbf{d}$ for whichever value of $t$ is eventually found. The vector from the fixed external point $A$ (position vector $\mathbf{a}$ ) to this general point is $\overrightarrow{AX} = (\mathbf{c}+t\mathbf{d})-\mathbf{a}$ — by spec 7.4's own identity, exactly as used throughout this lesson so far, just with $X$ 's position vector written in terms of the unknown $t$ instead of fixed numbers. Forming this subtraction is its own genuine step, not a formality to skip: $\overrightarrow{AX}$ and $X$ 's own position vector $\mathbf{c}+t\mathbf{d}$ are two different vectors, and only the first of them is the one this method actually needs. $X$ is specifically the FOOT of the perpendicular from $A$ — the closest point on the whole line to $A$ — exactly when $\overrightarrow{AX}$ is perpendicular to the line's own direction $\mathbf{d}$ . This is not merely asserted: for any OTHER point $P$ on $l$ with $P\neq X$ , the vector $\overrightarrow{XP}$ lies along $\mathbf{d}$ (both $X$ and $P$ sit on the same line), so if $\overrightarrow{AX}\perp\mathbf{d}$ then $\overrightarrow{AX}\perp\overrightarrow{XP}$ too, and the triangle $AXP$ has a genuine right angle at $X$ . Pythagoras applied to that triangle gives $|AP|^2 = |AX|^2+|XP|^2$ , and since $P\neq X$ forces $|XP|^2>0$ , this means $|AP|^2>|AX|^2$ , so $|AP|>|AX|$ — every OTHER point on $l$ really is strictly further from $A$ than $X$ is, a consequence of the right angle rather than a fact taken on faith. (The diagram below plots exactly this: $|\overrightarrow{AX}|$ as $t$ varies traces a single curve with one minimum, occurring precisely at the value of $t$ where $\overrightarrow{AX}\perp\mathbf{d}$ — the same dip a differentiation or completed-square approach to minimising $|\overrightarrow{AX}|^2$ would find directly, met properly in the method-comparison block below.) By the perpendicularity test recalled above, that condition is $\overrightarrow{AX}\cdot\mathbf{d}=0$ . Substituting the expression for $\overrightarrow{AX}$ in terms of $t$ turns this into ONE linear equation in the single unknown $t$ — solvable directly, since $\mathbf{d}$ is a fixed, known vector and every other quantity in the dot product is either a fixed number or a multiple of $t$ . Once $t$ is found, substituting it back into $\mathbf{c}+t\mathbf{d}$ gives $X$ itself, and the shortest distance from $A$ to $l$ is simply $|\overrightarrow{AX}|$ at that value of $t$ — the SAME vector already used to set up the equation, not a separate calculation. The real, examiner-report-documented error on exactly this question type is not d·d = 0 — it is subtler, and more dangerous precisely because it does not fail so visibly. It comes from skipping the subtraction of $\mathbf{a}$ altogether: writing the general point $\mathbf{c}+t\mathbf{d}$ itself where $\overrightarrow{AX}$ should be, and dotting THAT against $\mathbf{d}$ — $(\mathbf{c}+t\mathbf{d})\cdot\mathbf{d}=0$ . This equation still contains $t$ , in the $t\,\mathbf{d}\cdot\mathbf{d}$ term, and is genuinely solvable — on the real exam question this trap is documented against, it gives a specific, plausible-looking wrong value ( $t=\tfrac{37}{50}$ , against a correct value of $\tfrac{2}{5}$ ), not an equation that visibly breaks. It scores no marks not because it can't be solved, but because the vector tested for perpendicularity was never actually $\overrightarrow{AX}$ — it was the general point's own position vector from the origin, a different (and, for this purpose, meaningless) quantity that just happens to still depend on $t$ . Genuinely dotting the direction vector with itself, $\mathbf{d}\cdot\mathbf{d}=0$ , is a separate, much rarer, purely hypothetical slip — that one really IS structurally unsolvable, since $\mathbf{d}\cdot\mathbf{d}=|\mathbf{d}|^2$ is a fixed positive number containing no $t$ at all — but it is not the error the examiner report below documents, and should not be confused with it.

Diagram — The foot of the perpendicular from a point to a line — what AX · d = 0 actually tests

x-axis: position along the line, parameterised by t · y-axis: perpendicular distance from A

The line l: r = c + t·d: Every point on l has the form c + t·d for some real t — c is one known fixed point on the line, d is its fixed direction. As t ranges over every real number, this expression sweeps out every point on l, one t per point.
AX, for the general point X on l: AX = X − a changes as X moves along the line — unlike d, which never changes, AX is genuinely a function of t. This is the vector the whole method is built around, and confusing it with d (a fixed vector) is exactly the trap this diagram's own commonError documents.

X, the foot of the perpendicular: The one specific point on l where AX is perpendicular to d — where AX · d = 0. At every other point on l, AX and d meet at some other angle, and |AX| is strictly larger than it is at X.
A, the external point: Fixed, and not on the line l. (If A were on l, the foot of the perpendicular would simply be A itself, and the shortest distance would be zero — a degenerate case the same method still handles correctly, since AX would then be the zero vector, trivially satisfying AX · d = 0.)
The right angle at X: The defining geometric fact: of every point on l, X is the one closest to A, and the segment AX meets the line at exactly 90°. AX · d = 0 is exactly this fact written algebraically — the same perpendicularity test used for angle-finding elsewhere in this unit, applied here to AX and d rather than to two lines' own direction vectors.

Common error: Skipping the subtraction of a: dotting the line's raw general point c + t·d directly against d, as if the general point itself already were AX, rather than first forming AX = (c + t·d) − a and dotting THAT against d.

Correct: AX = X − a is a genuinely independent step, not something to fold into the perpendicularity test. (c + t·d)·d = 0 still contains t and is solvable — it just tests the wrong vector, so the value of t it produces is a specific, plausible-looking wrong answer, not evidence the working has broken down. (This is a different error from d·d = |d|² = 0, which really is a fixed positive number with no t in it at all and cannot be solved for anything — a separate, much rarer, purely hypothetical slip, not the one this examiner report documents.)

examiner-report · Oct 2021 · Q7(a)

Worked, in full

Finding the foot of the perpendicular from a point to a line, and the shortest distance (VERIDIAN-original)

01
Line $l: \mathbf{r}=(1,0,2)+t(1,2,2)$ ; point $A=(4,3,2)$ , so $\mathbf{a}=(4,3,2)$ . The general point on $l$ is $X=(1+t,\ 2t,\ 2+2t)$ . Form $\overrightarrow{AX} = X - \mathbf{a} = (1+t-4,\ 2t-3,\ 2+2t-2) = (t-3,\ 2t-3,\ 2t)$ — the vector from A to the GENERAL point on the line, not the general point $X$ itself left unsubtracted.
Earns: B1 — independent mark for forming AX correctly, in terms of t, as (position vector of the general point on the line) minus (position vector of A).
02
Set $\overrightarrow{AX}\cdot\mathbf{d} = 0$ , using $\mathbf{d}=(1,2,2)$ : $(t-3)(1)+(2t-3)(2)+(2t)(2) = 0$ .
Earns: M1 — method mark for the correct perpendicularity condition, AX · d = 0, quoting it before substituting — WMA14's own general marking principles record that quoting a formula first earns this mark even if a small slip follows, while the mark can otherwise only be inferred from correct working with values and lost to any mistake in it (facts bank §4).
03
Expand and collect: $(t-3) + (4t-6) + (4t) = 9t - 9 = 0 \Rightarrow t = 1$ .
Earns: A1 — the correct value of t.
04
Substitute $t=1$ back into the line's equation: $X = (1+1,\ 2(1),\ 2+2(1)) = (2,2,4)$ .
Earns: dM1 — method mark, dependent on the previous mark (solving AX·d=0 for t), for substituting the found value of t into the line's equation to obtain the coordinates of the foot of the perpendicular.
05
$\overrightarrow{AX}$ at $t=1$ : $(1-3,\ 2-3,\ 2) = (-2,-1,2)$ . Shortest distance $= |\overrightarrow{AX}| = \sqrt{4+1+4} = \sqrt{9} = 3$ .
Earns: A1 — the correct distance, reusing the vector already formed in stage 1 rather than starting a fresh calculation.
06
Extension (Q7(b)-style): let $B$ be the image of $A$ after reflection in $l$ . $X$ is then the midpoint of $A$ and $B$ — the same relationship AX = XB — so $\mathbf{b} = 2\mathbf{x}-\mathbf{a}$ , reusing spec 7.4's own position-vector identity rather than a new formula: $\mathbf{b} = 2(2,2,4)-(4,3,2) = (4,4,8)-(4,3,2) = (0,1,6)$ .
Earns: M1 A1 — method mark for using AX = XB (X as the midpoint of A and B) to write b = 2x − a, and the correct position vector of B. Mirrors the real anchor question's own part (b), 2 further marks on top of the 7 for part (a): "Uses AX = XB or similar correct method M1 / point B has position vector ... A1" (Pearson Edexcel WMA14/01 Oct 2021 Mark Scheme, Q7(b)).

Source — Examiner report, Oct 2021

"most incorrect responses attempted to set [the general point on the line] . [the direction vector] = 0 rather than AX . [the direction vector] = 0... This immediately resulted in 0 marks for (a)(i)"

Complete it yourself

Complete the chain — a different point and line, same method (VERIDIAN-original)

01
Line $p: \mathbf{r}=(3,5,-1)+u(2,-2,1)$ ; point $A=(8,3,3)$ , so $\mathbf{a}=(8,3,3)$ . General point on $p$ : $X=(3+2u,\ 5-2u,\ -1+u)$ . Form $\overrightarrow{AX}=X-\mathbf{a} = (3+2u-8,\ 5-2u-3,\ -1+u-3) = (2u-5,\ -2u+2,\ u-4)$ .
02
Set $\overrightarrow{AX}\cdot\mathbf{d}=0$ , using $\mathbf{d}=(2,-2,1)$ : $(2u-5)(2)+(-2u+2)(-2)+(u-4)(1) = 0$ .

Same question, every valid method

Point $A=(4,3,2)$ . Line $l: \mathbf{r}=(1,0,2)+t(1,2,2)$ . Find the shortest distance from A to l. (4) — the same scenario as the worked-chain above, VERIDIAN-original, shown here through two further legitimate routes to the identical answer.

3 valid methods · every one reaches 3 · 4 marks available

01
Form AX = (t-3, 2t-3, 2t); set AX·d = 0: (t-3)+2(2t-3)+2(2t) = 9t-9 = 0.
Method mark for the correct perpendicularity condition, correctly expanded.
M1
02
t = 1.
Correct value of the parameter.
A1
03
X = (2,2,4); AX = (-2,-1,2).
Method mark for substituting back to find the foot and its vector from A.
M1
04
Distance = |AX| = √9 = 3.
Correct answer only.
A1

The default, most direct route on this paper — one perpendicularity condition, one linear equation in t, done. Prefer this whenever the question asks only for the distance or the foot itself, which is most of the time this question type is set.

In your own words

In one sentence: why does AX · d = 0 correctly find the closest point on a line to an external point A, while d · d = 0 cannot — even before any numbers are substituted in?

Marked, line by line

(a) Find the coordinates of the point Y on the line $m: \mathbf{r} = (1,2,-3) + s(0,3,4)$ that is closest to the point $B(1,9,-2)$ . (4) (b) Hence find the shortest distance from B to m. (2) (c) Hence find the position vector of B′, the image of B after reflection in m. (2) — VERIDIAN-original question, in the style of the real question this centrepiece is modelled on (Oct 2021, Q7(a)+(b): finding the foot of the perpendicular from a point to a line, the resulting distance, and the position vector of the reflected point). Not a reproduction of any past-paper question or its actual numbers; the per-line mark allocations for (a) and (b) are modelled on WMA14's own verified mark-scheme conventions (facts bank §4) rather than transcribed from a real scheme, which for an original question does not exist — part (c)'s method mirrors the real Q7(b) mark scheme's own method credit ('Uses AX = XB or similar correct method'), reusing the position-vector identity (spec 7.4) already taught rather than inventing a different one.

8 marks available

(a) — 4 marks

01
General point on m: $M(s) = (1,\ 2+3s,\ -3+4s)$ . Position vector of B: $\mathbf{b}=(1,9,-2)$ . Form $\overrightarrow{BM} = M(s) - \mathbf{b} = (0,\ 3s-7,\ 4s-1)$ — the vector from B to the GENERAL point on the line, formed as its own independent step rather than folded into the next line.
Method mark for forming BM as (general point on the line) minus (the fixed external point) — the vector that must be tested for perpendicularity to the line's direction, not the general point left unsubtracted (see the trap taxonomy below) and not the line's own direction vector dotted with itself.
M1
02
Set $\overrightarrow{BM}\cdot\mathbf{d}=0$ , using $\mathbf{d}=(0,3,4)$ : $(0)(0)+(3s-7)(3)+(4s-1)(4) = 9s-21+16s-4 = 25s-25 = 0$ .
Method mark, dependent on the previous line, for the correct perpendicularity condition, expanded correctly.
dM1
03
s = 1.
Correct value of the parameter, correct answer only.
A1
04
Y = M(1) = (1, 2+3, -3+4) = (1,5,1).
Method mark, dependent on the previous mark (solving BM·d=0 for s), for substituting the found value of s into the line's equation to obtain the coordinates of the foot of the perpendicular — not an independent accuracy mark, since it cannot be earned without a correct value of s to substitute.
dM1

(b) — 2 marks

101
$\overrightarrow{BY} = Y - \mathbf{b} = (1-1,\ 5-9,\ 1-(-2)) = (0,-4,3)$ .
Method mark for forming BY (or reusing BM at s=1 already found in part (a), oe) and attempting its magnitude.
M1
102
Distance $= |\overrightarrow{BY}| = \sqrt{0+16+9} = \sqrt{25} = 5$ .
Correct answer only — an exact integer here, so there is no decimal-rounding decision to make.
A1

(c) — 2 marks

201
Let B′ be the image of B after reflection in m. Y is the closest point on m to B, so Y is also the midpoint of B and B′ — the same relationship as AX = XB in the real anchor question — giving $\mathbf{b'} = 2\mathbf{y}-\mathbf{b}$ .
Method mark for using BY = YB′ (Y as the midpoint of B and its reflection) to write b′ = 2y − b — reusing spec 7.4's own position-vector identity rather than a new formula. Mirrors the real anchor question's own part (b): "Uses AX = XB or similar correct method" (Pearson Edexcel WMA14/01 Oct 2021 Mark Scheme, Q7(b)) — mark-scheme evidence only; the examiner's report records no part-(b)-specific wrong path beyond a general note that many candidates did not attempt it.
M1
202
$\mathbf{b'} = 2(1,5,1) - (1,9,-2) = (2,10,2)-(1,9,-2) = (1,1,4)$ .
Correct position vector of B′ — correct answer only.
A1

Named traps

general-point-substituted-for-ax: Confirmed directly on exactly this question type, and more precisely than "dots the direction vector with itself" would suggest: the real, examiner-report-documented error is not dotting the direction vector with itself (that equation has no t in it and is trivially, visibly unsolvable) — it is skipping the step that forms AX = (general point) − a, and instead dotting the raw general point on the line straight against d: "most incorrect responses attempted to set [the general point on the line] . [the direction vector] = 0 rather than AX . [the direction vector] = 0... This immediately resulted in 0 marks for (a)(i)" (Oct 2021, Q7(a)). This equation genuinely DOES contain t and is solvable — on the real exam question it gives a specific, plausible-looking wrong value ( $\lambda=\tfrac{37}{50}$ , versus the correct $\lambda=\tfrac{2}{5}$ ) — which is exactly why it is dangerous: nothing about the arithmetic looks broken, only the choice of vector is wrong. Contrast this with genuinely dotting the direction vector with itself, d·d = 0: that IS structurally unsolvable (no t at all) — a different, much rarer, and merely hypothetical slip, not the one this examiner report documents. This is the single documented error this section of the facts bank contains for this question type; a real, single-source citation, not padded into a broader pattern.
ab-formed-in-the-wrong-order: A general, VERIDIAN-identified trap rather than a sourced one: forming AB as a − b instead of b − a gives BA (every component's sign reversed), the reverse of what was actually asked for. Unlike the distance formula, which is symmetric because it squares each difference (so it doesn't matter which point is subtracted from which), the vector AB is NOT symmetric — getting the order backwards produces a genuinely wrong vector, not merely a differently-labelled correct one.
magnitude-missing-the-final-square-root: A general, VERIDIAN-identified trap: stopping at the sum of squares (e.g. reporting 169 instead of 13) rather than taking the final square root. This is a genuinely easy line to skip, since every individual term along the way — each component squared, then summed — was computed correctly; the omission is the very last step, not an error anywhere inside the working.
unit-vector-not-actually-normalised: A general, VERIDIAN-identified trap: giving the original vector itself, or a partially-scaled version of it, as "the unit vector," rather than dividing every one of its components by its own magnitude. The fast check named earlier in this lesson — do the squares of the claimed answer's own components sum to exactly 1? — catches this immediately, and is worth running as a habit rather than only when something already looks suspicious.

Retrieval — with feedback on every choice

Question 1

2 marks

Find the unit vector in the direction of $\mathbf{a} = (6,-8,0)$ .

Question 2

3 marks

Find the distance between $P=(1,-2,4)$ and $Q=(5,1,-8)$ .

Question 3

3 marks

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $3\mathbf{a}-2\mathbf{b}$ .

Question 4

2 marks

$A$ and $B$ have position vectors $\mathbf{a}=(3,0,-2)$ and $\mathbf{b}=(-1,5,4)$ . Find $\overrightarrow{AB}$ .

Question 5

2 marks

Find $|\mathbf{v}|$ where $\mathbf{v}=(4,4,7)$ .

Question 6

3 marks

$A$ is a point not on line $l: \mathbf{r} = \mathbf{c}+t\mathbf{d}$ ; $X$ is the general point on $l$ . Which equation, once solved for $t$ , correctly gives the point on $l$ closest to $A$ ?

Question 7

3 marks

Point $A=(3,1,-1)$ . Line $n: \mathbf{r}=t(1,1,1)$ . Find the EXACT shortest distance from A to n.

Reference — not a study method, a lookup

|v| = √(x²+y²+z²) — 3D Pythagoras. Unit vector in the direction of v: v/|v| (check: its own components' squares sum to 1).
AB = OB − OA = b − a (destination minus start — NOT a − b, which gives BA, the reverse).
Addition/subtraction: component-wise. Scalar k·v: scales magnitude by |k|; reverses direction if k < 0.
Distance between (x₁,y₁,z₁) and (x₂,y₂,z₂): d² = (x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)² — literally |AB|² written out in coordinates.
Foot of perpendicular from A to line r = c + t·d: form AX = X − a first (X = c + t·d), then solve AX·d = 0 for t. NOT the raw general point X dotted with d instead of AX — (c+t·d)·d = 0 still solves, just for the wrong point; skipping the subtraction of a is the error, not unsolvability.
Shortest distance from A to the line = |AX| at that t. Same t also minimises |AX|² — check via differentiation or completing the square.
Reflection of A in the line: X is the midpoint of A and its image B, so B = 2X − A (spec 7.4's identity, reused rather than a new formula).

Not affiliated with or endorsed by Pearson Edexcel. This lesson's sourcing is deliberately uneven, and disclosed as such rather than smoothed over: the foundational content (spec 7.1–7.5 as ordinary vector arithmetic — every prequestion apart from prequestion 3's choice B, the "vectors in two/three dimensions" and "magnitude and unit vector" and "vector addition/scalar multiplication" and "position vectors" teach blocks, both Pythagoras-based mechanism derivations, the first worked-chain and chain-drill, three of the four trap-taxonomy items, and MCQs 1–5) is entirely VERIDIAN-original scaffolding. WMA14-verified-facts.md has no examiner-report or mark-scheme material specifically about basic vector arithmetic errors — its own §5.7 section is about the harder foot-of-perpendicular application, not this — so nothing above dresses this foundational content in a false citation to look better-sourced than it is. The foot-of-perpendicular centrepiece — prequestion 3's choice B, the transition teach block, the "Setting up the right equation" mechanism block, the diagram's commonError, the centrepiece worked-chain's embeddedEvidence, the method-comparison's credit for the differentiation/completing-the-square alternative, the marked-solution's commonWrongPath, the first trap-taxonomy item, MCQs 6–7, and the reference-card — draws on exactly ONE real, verified past-paper question, October 2021 Q7, the sole question the facts bank's own §5.7 subsection cites for this content — but on TWO distinct citations from it: the examiner's report on part (a) (the AX-vs-general-point setup trap, corrected in a later audit of this lesson from an earlier, less precise "d·d=0" description — see the Lesson-audit log in WMA14-verified-facts.md — to the report's own actual wrong left-hand side, and the differentiation/completing-the-square alternative credited for the same part), and the mark scheme's own part (b) (the reflection point, added in that same audit to the centrepiece worked-chain's final stage and the marked-solution's part (c): B = 2X − A, credited as "Uses AX = XB or similar correct method"). The examiner's report has no part-(b)-specific wrong-path detail, so nothing above invents a documented (b)-specific trap — the (b) addition is mark-scheme method only, labelled as such. That is still genuinely thinner evidence than this unit's sibling lesson's three-series-confirmed and two-series-confirmed traps — one question, two of its parts, not multiple series — and this lesson does not round it up, split it into separately-numbered findings, or imply broader confirmation than one series actually supports. The general mark-scheme conventions used throughout — M/A/B/dM mark types, "quote the formula first," and the exact-answer/no-rounded-decimals rule cited in MCQ 7 — are real and separately verified (facts bank §4, stable wording checked across Jan 2021 → Jun 2024 mark schemes), cited as general convention rather than as evidence specific to this topic. All numeric scenarios in this lesson (the two foundational worked items, the four foot-of-perpendicular scenarios and their reflection-point extension, and the five further MCQ calculations — MCQs 1–5, each with its own numeric vector arithmetic; MCQ 6 is conceptual, no numbers, and is not counted here) were computed and independently checked with sympy before being written in, including the deliberately wrong parameter values in the trap material — t = −5/9 in the centrepiece worked-chain, s = 6/25 in the marked-solution's commonWrongPath, and t = 0 in MCQ 7 (a degenerate case, since that scenario's line passes through the origin) — genuinely produced by applying the documented wrong method (dotting the line's raw general point against d, having skipped the subtraction of a) to each scenario's own numbers, not reverse-engineered from a target answer.

Question 12 marks

Find the unit vector in the direction of $\mathbf{a} = (6,-8,0)$ .

A $(6,-8,0)$
This is $\mathbf{a}$ itself, not normalised at all — its own magnitude is 10, not 1, so it cannot be the unit vector in its own direction.
$\left(\dfrac{3}{5},\ -\dfrac{4}{5},\ 0\right)$
Correct. $|\mathbf{a}| = \sqrt{36+64+0} = \sqrt{100} = 10$ , so $\hat{\mathbf{a}} = \dfrac{1}{10}(6,-8,0) = \left(\dfrac{3}{5},-\dfrac{4}{5},0\right)$ . Check: $\left(\dfrac{3}{5}\right)^2+\left(\dfrac{4}{5}\right)^2+0 = \dfrac{9+16}{25}=1$ . ✓
C $\left(\dfrac{3}{5},\ \dfrac{4}{5},\ 0\right)$
The magnitude and division are both correct, but the sign on the second component has been dropped — $-8/10=-4/5$ , not $+4/5$ . A unit vector must point in exactly the SAME direction as the original, sign included.
D $(-3,4,0)$
This treats the "magnitude" as the sum of the raw components, $6+(-8)+0=-2$ , then divides by $-2$ — not the actual magnitude formula, and a magnitude can never legitimately be negative in the first place.

Traps tested: Unit vector not actually normalised · Sign dropped during normalisation · Magnitude confused with sum of components

Question 23 marks

Find the distance between $P=(1,-2,4)$ and $Q=(5,1,-8)$ .

A $5$
This drops the $z$ -coordinate entirely and finds only the 2D distance in the $x$ – $y$ plane: $\sqrt{4^2+3^2}=\sqrt{25}=5$ . All three dimensions must be included for a genuine 3D distance.
B $169$
This is $d^2$ , not $d$ — every squared term was computed correctly, but the final square root was never taken.
C $19$
This adds the ABSOLUTE VALUES of the coordinate differences directly — $|4|+|3|+|-12|=19$ — rather than squaring, summing, and rooting. The distance formula is never a straight sum of differences, however the differences are combined.
$13$
Correct. The difference in each coordinate is $(4,3,-12)$ , so $d^2 = 4^2+3^2+(-12)^2 = 16+9+144=169$ , and $d=\sqrt{169}=13$ .

Traps tested: Dimension dropped from distance formula · Magnitude missing the final square root · Manhattan distance used instead of euclidean

Question 33 marks

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $3\mathbf{a}-2\mathbf{b}$ .

$(8,-11,5)$
Correct. $3\mathbf{a}=(6,-3,9)$ , $2\mathbf{b}=(-2,8,4)$ , so $3\mathbf{a}-2\mathbf{b} = (6-(-2),\ -3-8,\ 9-4) = (8,-11,5)$ .
B $(4,5,13)$
This is $3\mathbf{a}+2\mathbf{b}$ — a sign error on the whole second term, adding instead of subtracting it.
C $(7,-7,7)$
This is $3\mathbf{a}-\mathbf{b}$ — $\mathbf{b}$ was subtracted without first being scaled by 2, so only one of the two scalar multiplications in the question was actually carried out.
D $(4,-9,-1)$
This is $\mathbf{a}-2\mathbf{b}$ — $\mathbf{a}$ was left unscaled while $\mathbf{b}$ was correctly scaled by 2, the reverse omission from option C.

Traps tested: Sign error on scaled vector · Scalar multiplication only partially applied

Question 42 marks

$A$ and $B$ have position vectors $\mathbf{a}=(3,0,-2)$ and $\mathbf{b}=(-1,5,4)$ . Find $\overrightarrow{AB}$ .

A $(2,5,2)$
This is $\mathbf{a}+\mathbf{b}$ — the two position vectors added rather than subtracted, which has no fixed geometric meaning as the vector between A and B.
B $(4,-5,-6)$
This is $\mathbf{a}-\mathbf{b} = \overrightarrow{BA}$ — the reverse vector, every component's sign flipped from the correct answer.
$(-4,5,6)$
Correct. $\overrightarrow{AB}=\mathbf{b}-\mathbf{a} = (-1-3,\ 5-0,\ 4-(-2)) = (-4,5,6)$ .
D $(-4,5,-6)$
The $x$ - and $y$ -components are correct, but the $z$ -component has been computed as $a_z-b_z=-2-4=-6$ instead of $b_z-a_z=4-(-2)=6$ — a sign slip on just one coordinate, from mixing up the subtraction order in only that component.

Traps tested: Addition used in place of subtraction for ab · Ab formed in the wrong order

Question 52 marks

Find $|\mathbf{v}|$ where $\mathbf{v}=(4,4,7)$ .

A $\sqrt{32}$
This drops the $z$ -component before squaring anything: $\sqrt{4^2+4^2}=\sqrt{32}$ . All three components of a 3D vector must be included in its magnitude.
B $81$
This is $|\mathbf{v}|^2$ , not $|\mathbf{v}|$ — the sum of squares is correct, but the final square root is missing.
C $15$
This adds the raw, unsquared components: $4+4+7=15$ — not related to the vector's actual length.
$9$
Correct. $|\mathbf{v}|^2 = 4^2+4^2+7^2 = 16+16+49 = 81$ , so $|\mathbf{v}|=\sqrt{81}=9$ .

Traps tested: Dimension dropped from magnitude formula · Magnitude missing the final square root · Magnitude confused with sum of components

Question 63 marks

$A$ is a point not on line $l: \mathbf{r} = \mathbf{c}+t\mathbf{d}$ ; $X$ is the general point on $l$ . Which equation, once solved for $t$ , correctly gives the point on $l$ closest to $A$ ?

$(X-A)\cdot\mathbf{d} = 0$
Correct. The vector from A to the general point on the line, dotted with the line's own direction, set to zero — exactly the perpendicularity condition that picks out the one point on l where AX meets the line at 90°.
B $X\cdot\mathbf{d} = 0$
The single most-documented real error on this exact question type: using the general point $X=\mathbf{c}+t\mathbf{d}$ directly, as if it already were the vector from A to X, rather than first forming $X-A$ . Unlike dotting d with itself, this equation genuinely contains t and IS solvable — it just solves for where the general point itself (a position vector from the origin) is perpendicular to d, not where AX is. A real examiner report records this specific substitution scoring "0 marks for (a)(i)" despite being solvable (Oct 2021, Q7(a)) — the failure is the missing subtraction of A, not an unsolvable equation.
C $(X-A)\cdot(X-A) = 0$
This sets $|AX|^2$ to zero, which only holds if X and A are the same point — solving it (where a real solution exists) would find where the line passes through A, not the closest point to A on a line A isn't on.
D $\mathbf{c}\cdot\mathbf{d} = 0$
This tests whether the line's own fixed known point is perpendicular to its direction — a fact about the line's own position relative to the origin, with no connection to where the closest point to A actually is.

Traps tested: General point substituted for ax · Squared distance set to zero instead of perpendicularity · Wrong fixed vectors tested for perpendicularity

Question 73 marks

Point $A=(3,1,-1)$ . Line $n: \mathbf{r}=t(1,1,1)$ . Find the EXACT shortest distance from A to n.

A $2.83$ (3 s.f.)
This is a correctly-rounded DECIMAL approximation to the right answer — but the question asks for the exact value, and a real mark-scheme convention states plainly that resorting to a rounded decimal when an exact answer is required normally costs the mark, whatever the decimal itself rounds to.
$2\sqrt{2}$
Correct. General point $X=(t,t,t)$ ; $\overrightarrow{AX}=(t-3,\ t-1,\ t+1)$ ; setting $\overrightarrow{AX}\cdot(1,1,1)=0$ gives $(t-3)+(t-1)+(t+1)=3t-3=0 \Rightarrow t=1$ . $\overrightarrow{AX}=(-2,0,2)$ , so distance $=\sqrt{4+0+4}=\sqrt{8}=2\sqrt{2}$ . "Exact" is doing real work in this prompt: a real mark-scheme convention states directly that "where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals" (facts bank §4, Jun 2024 MS, cross-checked against Jan 2021 MS).
C $2$
This drops the $z$ -component of AX before taking the square root: $\sqrt{(-2)^2+0^2}=\sqrt{4}=2$ , omitting the third dimension entirely from an otherwise correct method.
D $\sqrt{11}$
This is the real, examiner-report-documented error, not an unsolvable dead end: skip forming AX = X − A, and instead dot the line's raw general point X = t(1,1,1) directly against d. Because this particular line happens to pass through the origin, that equation degenerates to $(t\mathbf{d})\cdot\mathbf{d}=0$ , i.e. $3t=0$ , giving $t=0$ — genuinely solvable, just for the wrong condition (most lines don't pass through the origin, so this equation usually solves to a less obviously-wrong value than t=0). Substituting back gives $X=(0,0,0)$ , and $|X-A|=\sqrt{9+1+1}=\sqrt{11}$ — a specific, plausible-looking wrong answer, not evidence the calculation "can't be done." A real examiner report on this question type: "most incorrect responses attempted to set [the general point on the line] . [the direction vector] = 0 rather than AX . [the direction vector] = 0... This immediately resulted in 0 marks" (Oct 2021, Q7(a)).

Traps tested: Rounded decimal given when exact surd required · Dimension dropped from magnitude formula · General point substituted for ax

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Oct 2021 · Q7(a) — cited directly in this lesson
Examiner report: Oct 2021 · Q7(a) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

Vectors — position vectors, distance, and the foot of a perpendicular

Vectors in two and three dimensions

Magnitude, and the unit vector in the direction of a

Where |v| = √(x² + y² + z²) actually comes from — Pythagoras, run twice

Vector addition, scalar multiplication, and their geometric interpretation

Position vectors, and the identity the rest of this unit is built on

The distance formula is |AB| written out in coordinates — not a separate rule

Forming AB, finding the distance AB, and the unit vector in its direction (VERIDIAN-original)

Complete the chain — CD, the distance CD, and the unit vector in its direction (VERIDIAN-original)

A genuinely hard application: the foot of the perpendicular from a point to a line

Setting up the right equation: AX · d = 0, not the general point dotted with d

Finding the foot of the perpendicular from a point to a line, and the shortest distance (VERIDIAN-original)

Complete the chain — a different point and line, same method (VERIDIAN-original)

Deep dive

Practice this for real