Paper 1 · Further Pure 1 ✦

Vectors in Three Dimensions

One new tool — the vector product — turns every 3-D geometry question into arithmetic. Lines, planes, angles, distances, and the two lines that never meet.

Syllabus 1.6Assumes: scalar product & line equations (9709 P3)MF19 gives a·b only

What the examiner expects you to do

Use the equation of a plane in any of the forms $a x + b y + c z = d$ , $𝐫 \cdot 𝐧 = p$ , $𝐫 = 𝐚 + λ 𝐛 + μ 𝐜$ , and convert between them.
Recall the vector product both as $| 𝐚 | | 𝐛 | \sin θ \hat{𝐧}$ and in component form.
Use lines, planes, and both products to solve problems on distances, angles and intersections — specifically:
- whether a line lies in a plane, is parallel to it, or meets it — and where;
- the foot of the perpendicular from a point to a plane;
- the angle between a line and a plane, and between two planes;
- an equation for the line of intersection of two planes;
- the shortest distance between two skew lines;
- an equation for the common perpendicular to two skew lines.

§1 The vector product

◆ Two descriptions of the same thing

Geometric: $𝐚 \times 𝐛 = | 𝐚 | | 𝐛 | \sin θ \hat{𝐧}$ , where $\hat{𝐧}$ is the unit vector perpendicular to both, in the right-hand-rule sense.

Component: for $𝐚 = a_{1} 𝐢 + a_{2} 𝐣 + a_{3} 𝐤$ and $𝐛 = b_{1} 𝐢 + b_{2} 𝐣 + b_{3} 𝐤$ ,

𝐚 \times 𝐛 = (a_{2} b_{3} - a_{3} b_{2}) 𝐢 + (a_{3} b_{1} - a_{1} b_{3}) 𝐣 + (a_{1} b_{2} - a_{2} b_{1}) 𝐤 .

The reliable way to write that down is the determinant layout:

𝐚 \times 𝐛 = | \begin{array}{ccc} 𝐢 & 𝐣 & 𝐤 \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array} |,

expanded along the top row with the $+ - +$ signs — so the $𝐣$ component carries a minus, which is where the second component's reversed order comes from.

◇ Properties you will actually use

$𝐚 \times 𝐛$ is perpendicular to both $𝐚$ and $𝐛$ — this is why it produces normals to planes.
$𝐛 \times 𝐚 = - (𝐚 \times 𝐛)$ : the order matters (unlike the scalar product).
$𝐚 \times 𝐛 = 𝟎$ exactly when $𝐚$ and $𝐛$ are parallel (including when one is zero).
$| 𝐚 \times 𝐛 |$ is the area of the parallelogram spanned by $𝐚$ and $𝐛$ — a useful picture of what the magnitude means.

✎ Worked example — and how to check it in one line

With $𝐚 = 2 𝐢 + 𝐣 - 𝐤$ and $𝐛 = 𝐢 + 3 𝐣 + 2 𝐤$ :

𝐚 \times 𝐛 = ((1) (2) - (- 1) (3)) 𝐢 + ((- 1) (1) - (2) (2)) 𝐣 + ((2) (3) - (1) (1)) 𝐤 = 5 𝐢 - 5 𝐣 + 5 𝐤 .

Always check with two dot products: $𝐚 \cdot (5, - 5,5) = 10 - 5 - 5 = 0$ ✓ and $𝐛 \cdot (5, - 5,5) = 5 - 15 + 10 = 0$ ✓. If either is non-zero you have made an arithmetic slip — and you have found it in five seconds.

§2 Planes — three forms, one object

◆ The three forms and how to move between them

Form	Looks like	Read off
Parametric	$𝐫 = 𝐚 + λ 𝐛 + μ 𝐜$	a point $𝐚$ and two directions in the plane
Scalar product	$𝐫 \cdot 𝐧 = p$	the normal $𝐧$
Cartesian	$a x + b y + c z = d$	the normal is $𝐧 = (a, b, c)$ , and $d = p$

Parametric ⟶ the other two: take $𝐧 = 𝐛 \times 𝐜$ , then $p = 𝐚 \cdot 𝐧$ .
Scalar ⟷ Cartesian: they are the same statement — writing $𝐫 = (x, y, z)$ turns $𝐫 \cdot (a, b, c) = d$ into $a x + b y + c z = d$ .
Cartesian ⟶ parametric: solve for one variable and let the other two be parameters.

✎ Worked example — the plane through three points

Find the Cartesian equation of the plane through $A (1,0,2)$ , $B (2,1,0)$ , $C (0,3,1)$ .

Two directions in the plane: $\vec{A B} = (1,1, - 2)$ and $\vec{A C} = (- 1,3, - 1)$ . A normal is their vector product:

𝐧 = \vec{A B} \times \vec{A C} = ((1) (- 1) - (- 2) (3), (- 2) (- 1) - (1) (- 1), (1) (3) - (1) (- 1)) = (5,3,4) .

Then $d = 𝐧 \cdot 𝐚 = 5 (1) + 3 (0) + 4 (2) = 13$ , so the plane is

5 x + 3 y + 4 z = 13 .

Check all three points: $A$ : $5 + 0 + 8 = 13$ ✓ $B$ : $10 + 3 + 0 = 13$ ✓ $C$ : $0 + 9 + 4 = 13$ ✓ — that is three free marks of insurance.

§3 Does the line meet the plane?

◆ Two dot products settle it completely

For the line $𝐫 = 𝐚 + λ 𝐝$ and the plane $𝐫 \cdot 𝐧 = p$ :

$𝐝 \cdot 𝐧$	$𝐚 \cdot 𝐧$	Conclusion
$\neq 0$	anything	the line crosses the plane at exactly one point
$= 0$	$= p$	the line lies in the plane
$= 0$	$\neq p$	the line is parallel to the plane and never meets it

The logic is simply that $𝐝 \cdot 𝐧 = 0$ means the line runs perpendicular to the normal, i.e. along the plane's direction; then one point decides whether it is in the plane or beside it.

✎ Worked example — finding the point of intersection

Where does $𝐫 = (1,0,2) + λ (1,2,1)$ meet $2 x - y + 3 z = 5$ ?

Here $𝐧 = (2, - 1,3)$ and $𝐝 \cdot 𝐧 = 2 - 2 + 3 = 3 \neq 0$ , so there is exactly one intersection. Substitute the parametric coordinates $(1 + λ, 2 λ, 2 + λ)$ into the plane:

2 (1 + λ) - 2 λ + 3 (2 + λ) = 5 ⟹ 8 + 3 λ = 5 ⟹ λ = - 1 .

So the point is $(0, - 2,1)$ . Check: $2 (0) - (- 2) + 3 (1) = 5$ ✓

✎ Worked example — the two degenerate cases

Same plane $2 x - y + 3 z = 5$ , so $𝐧 = (2, - 1,3)$ .

$𝐫 = (1,1,1) + μ (1,2,0)$ : $𝐝 \cdot 𝐧 = 2 - 2 + 0 = 0$ and $𝐚 \cdot 𝐧 = 2 - 1 + 3 = 4 \neq 5$ . Parallel, never meets.
$𝐫 = (1,0,1) + μ (1,2,0)$ : $𝐝 \cdot 𝐧 = 0$ and $𝐚 \cdot 𝐧 = 2 - 0 + 3 = 5 = p$ . The line lies in the plane.

Identical directions; only the starting point differs. Always test both dot products.

§4 Angles — and the one that uses sine

◆ Three formulas, one of them different

two lines: \cos θ = \frac{| 𝐝_{1} \cdot 𝐝_{2} |}{| 𝐝_{1} | | 𝐝_{2} |}, two planes: \cos θ = \frac{| 𝐧_{1} \cdot 𝐧_{2} |}{| 𝐧_{1} | | 𝐧_{2} |},

line and plane: \sin θ = \frac{| 𝐝 \cdot 𝐧 |}{| 𝐝 | | 𝐧 |}

The modulus signs give the acute angle, which is what is wanted unless the question says otherwise.

⚠ Why the line-and-plane one is a sine

The formula measures the angle between the line and the normal, not the plane. The angle to the plane is $90^{\circ}$ minus that, and $\cos (90^{\circ} - θ) = \sin θ$ . Writing $\cos$ here is the single most common error in the whole topic — and it is worth several marks a year.

✎ Worked examples

Two planes $2 x - y + 3 z = 5$ and $x + y + z = 1$ : $𝐧_{1} = (2, - 1,3), 𝐧_{2} = (1,1,1)$ , so

\cos θ = \frac{| 2 - 1 + 3 |}{\sqrt{14} \sqrt{3}} = \frac{4}{\sqrt{42}} ⟹ θ = 51 . 9^{\circ} (1 d.p.) .

Line and plane, $𝐝 = (1,2,1)$ and $2 x - y + 3 z = 5$ :

\sin θ = \frac{| 2 - 2 + 3 |}{\sqrt{6} \sqrt{14}} = \frac{3}{\sqrt{84}} ⟹ θ = 19 . 1^{\circ} (1 d.p.) .

§5 Distances, feet of perpendiculars, and the line where two planes meet

◆ Distance from a point to a plane

For the point $(x_{0}, y_{0}, z_{0})$ and the plane $a x + b y + c z = d$ :

distance = \frac{| a x_{0} + b y_{0} + c z_{0} - d |}{\sqrt{a^{2} + b^{2} + c^{2}}} .

Substitute the point into the plane, see how far the answer misses $d$ , and divide by $| 𝐧 |$ .

✎ Worked example — foot of the perpendicular (and the distance for free)

Find the foot of the perpendicular from $P (4,1, - 2)$ to $2 x - y + 3 z = 5$ , and the distance.

Travel from $P$ along the normal: the line $𝐫 = (4,1, - 2) + t (2, - 1,3)$ . Substitute into the plane:

2 (4 + 2 t) - (1 - t) + 3 (- 2 + 3 t) = 5 ⟹ 1 + 14 t = 5 ⟹ t = \frac{2}{7} .

So the foot is $(\frac{32}{7}, \frac{5}{7}, - \frac{8}{7})$ . Check it is on the plane: $\frac{64}{7} - \frac{5}{7} - \frac{24}{7} = \frac{35}{7} = 5$ ✓

The distance is the length of that step, $| t | | 𝐧 | = \frac{2}{7} \sqrt{14} = \frac{2 \sqrt{14}}{7}$ . The formula agrees: $\frac{| 8 - 1 - 6 - 5 |}{\sqrt{14}} = \frac{4}{\sqrt{14}} = \frac{2 \sqrt{14}}{7} \approx 1.069$ . ✓

◆ The line of intersection of two planes

The line lies in both planes, so its direction is perpendicular to both normals:

𝐝 = 𝐧_{1} \times 𝐧_{2} .

For a point on it, set one variable to a convenient value (usually $z = 0$ ) and solve the two Cartesian equations simultaneously.

✎ Worked example

Find the line of intersection of $2 x - y + 3 z = 5$ and $x + y + z = 1$ .

𝐝 = (2, - 1,3) \times (1,1,1) = ((- 1) (1) - (3) (1), (3) (1) - (2) (1), (2) (1) - (- 1) (1)) = (- 4,1,3) .

Put $z = 0$ : $2 x - y = 5$ and $x + y = 1$ . Adding gives $3 x = 6$ , so $x = 2, y = - 1$ . Hence

𝐫 = (2, - 1,0) + t (- 4,1,3) .

Check: $(2, - 1,0)$ satisfies both planes ✓, and $𝐝 \cdot 𝐧_{1} = - 8 - 1 + 9 = 0$ , $𝐝 \cdot 𝐧_{2} = - 4 + 1 + 3 = 0$ ✓

If setting $z = 0$ gives an inconsistent or degenerate pair, the line is parallel to the plane $z = 0$ — just set a different variable to zero instead.

§6 Skew lines: shortest distance and the common perpendicular

◇ What "skew" means

Two lines in 3-D are skew if they are neither parallel nor intersecting — they pass by each other at different heights. Test: $𝐝_{1} \times 𝐝_{2} \neq 𝟎$ (not parallel) and the lines have no common point.

◆ Shortest distance

For $𝐫 = 𝐚_{1} + λ 𝐝_{1}$ and $𝐫 = 𝐚_{2} + μ 𝐝_{2}$ :

shortest distance = \frac{| (𝐚_{2} - 𝐚_{1}) \cdot (𝐝_{1} \times 𝐝_{2}) |}{| 𝐝_{1} \times 𝐝_{2} |} .

In words: project the gap between the two starting points onto the common normal direction. A bonus test comes free — if that numerator is $0$ (and the lines are not parallel) the distance is zero, so the lines intersect.

Why that is the shortest distance

$𝐧 = 𝐝_{1} \times 𝐝_{2}$ is perpendicular to both lines, so it is the direction of the shortest crossing. Take any point on each line; the vector between them is $𝐚_{2} - 𝐚_{1}$ plus something along $𝐝_{1}$ and $𝐝_{2}$ . Those extra parts contribute nothing in the $𝐧$ direction, because $𝐧 \cdot 𝐝_{1} = 𝐧 \cdot 𝐝_{2} = 0$ . So the separation measured along $\hat{𝐧}$ is the same wherever you start:

d = | (𝐚_{2} - 𝐚_{1}) \cdot \hat{𝐧} | = \frac{| (𝐚_{2} - 𝐚_{1}) \cdot (𝐝_{1} \times 𝐝_{2}) |}{| 𝐝_{1} \times 𝐝_{2} |} . ■

◆ The common perpendicular — find the two feet

Let $P = 𝐚_{1} + λ 𝐝_{1}$ and $Q = 𝐚_{2} + μ 𝐝_{2}$ be the feet. $\vec{P Q}$ must be perpendicular to both directions, giving two linear equations in $λ$ and $μ$ :

\vec{P Q} \cdot 𝐝_{1} = 0, \vec{P Q} \cdot 𝐝_{2} = 0 .

Solve them, get $P$ and $Q$ , and the common perpendicular is the line through $P$ with direction $𝐝_{1} \times 𝐝_{2}$ — equivalently, through $P$ and $Q$ .

✎ Worked example — the full skew-lines question

$l_{1} : 𝐫 = (0,3,2) + λ (1,0,1)$ and $l_{2} : 𝐫 = (1, - 1,2) + μ (0,1,1)$ .

(a) The common normal direction.

𝐧 = 𝐝_{1} \times 𝐝_{2} = (1,0,1) \times (0,1,1) = ((0) (1) - (1) (1), (1) (0) - (1) (1), (1) (1) - (0) (0)) = (- 1, - 1,1) .

It is not $𝟎$ , so the lines are not parallel.

(b) Shortest distance. With $𝐚_{2} - 𝐚_{1} = (1, - 4,0)$ :

d = \frac{| (1) (- 1) + (- 4) (- 1) + (0) (1) |}{\sqrt{1 + 1 + 1}} = \frac{| 3 |}{\sqrt{3}} = \sqrt{3} .

Non-zero, so the lines really are skew.

(c) The feet. $P = (λ, 3, 2 + λ)$ and $Q = (1, - 1 + μ, 2 + μ)$ , so $\vec{P Q} = (1 - λ, μ - 4, μ - λ)$ . The two perpendicularity conditions are

\vec{P Q} \cdot (1,0,1) = 1 + μ - 2 λ = 0, \vec{P Q} \cdot (0,1,1) = 2 μ - λ - 4 = 0 .

From the first, $μ = 2 λ - 1$ ; substituting gives $3 λ - 6 = 0$ , so $λ = 2$ and $μ = 3$ . Hence

P = (2,3,4), Q = (1,2,5) .

(d) The common perpendicular. $\vec{P Q} = (- 1, - 1,1)$ , which is exactly $𝐧$ ✓, and $| \vec{P Q} | = \sqrt{3}$ , agreeing with part (b) ✓. An equation is

𝐫 = (2,3,4) + t (- 1, - 1,1) .

Those two checks — that $\vec{P Q}$ is parallel to $𝐧$ , and that its length matches the formula — catch essentially every arithmetic slip.

§7 Interactive: rotate the 3-D scene

Drag the view round to see what "skew" really looks like, and slide the lines apart. The panel computes the shortest distance from the formula and, separately, by directly minimising $| P Q |$ over all $λ$ and $μ$ — if the two agree, the formula is doing what it claims.

Turn the view 40°

Tilt the view 25°

Slide the second line 0.0

line 1 / the line line 2 / the plane common perpendicular / foot

turn it round ✎

§8 Examiner traps & A* checklist

⚠ The eight most common mark-losers

Using $\cos$ for the angle between a line and a plane. It is $\sin θ = | 𝐝 \cdot 𝐧 | / (| 𝐝 | | 𝐧 |)$ .
Dropping the minus on the $𝐣$ component of a vector product.
Computing $𝐛 \times 𝐚$ when the question wanted $𝐚 \times 𝐛$ — the sign flips.
Forgetting the modulus in a distance, and reporting a negative one.
Using the same parameter $λ$ for both lines. They are independent — use $λ$ and $μ$ .
Concluding "skew" without checking the lines are not parallel first — $𝐝_{1} \times 𝐝_{2} = 𝟎$ means parallel, and then the skew formula is meaningless.
Answering "the shortest distance" when the question asked for an equation of the common perpendicular — you need the feet, not just the length.
Not dividing by $| 𝐧 |$ : $(𝐚_{2} - 𝐚_{1}) \cdot (𝐝_{1} \times 𝐝_{2})$ on its own is a volume, not a distance.

★ Before you turn the page — the free checks

Dot your vector product with both original vectors: two zeros, or you have an error.
Substitute every given point back into any plane equation you produce.
Check $\vec{P Q}$ is parallel to $𝐝_{1} \times 𝐝_{2}$ and that $| \vec{P Q} |$ equals your distance.

§9 Video explainers (curated)

▶The vector product
Curated search — component form, determinant layout, properties.↗ ▶Planes, lines & intersections
Curated search — the three forms, and where a line meets a plane.↗ ▶Skew lines & the common perpendicular
Curated search — the formula, and finding the two feet.↗

§10 Check yourself

Score 0 / 5

Q1 · vector product

With $𝐚 = (2,1, - 1)$ and $𝐛 = (1,3,2)$ , the vector product $𝐚 \times 𝐛$ is:

Solution

$((1) (2) - (- 1) (3), (- 1) (1) - (2) (2), (2) (3) - (1) (1)) = (5, - 5,5)$ . The distractor $(- 5,5, - 5)$ is $𝐛 \times 𝐚$ ; $(2,3, - 2)$ is the meaningless component-by-component product. Check: $𝐚 \cdot (5, - 5,5) = 0$ ✓

Q2 · angle between a line and a plane

The acute angle $θ$ between the line of direction $(1,2,1)$ and the plane $2 x - y + 3 z = 5$ satisfies:

Solution

$𝐝 \cdot 𝐧 = 2 - 2 + 3 = 3$ , $| 𝐝 | = \sqrt{6}$ , $| 𝐧 | = \sqrt{14}$ . Because the formula compares the line with the normal, the result is $\sin θ = \frac{| 𝐝 \cdot 𝐧 |}{| 𝐝 | | 𝐧 |} = \frac{3}{\sqrt{84}}$ , giving $θ = 19 . 1^{\circ}$ . Choosing $\cos$ here would give $70 . 9^{\circ}$ — the angle to the normal, not to the plane.

Q3 · line and plane

The line $𝐫 = (1,1,1) + μ (1,2,0)$ and the plane $2 x - y + 3 z = 5$ :

Solution

$𝐝 \cdot 𝐧 = (1) (2) + (2) (- 1) + (0) (3) = 0$ , so the line is not heading through the plane. Then $𝐚 \cdot 𝐧 = 2 - 1 + 3 = 4 \neq 5$ , so the starting point is not on the plane. Parallel, and never meets. (A line and a plane can never meet at exactly two points — if they share two points the whole line lies in the plane.)

Q4 · distance to a plane

The distance from $P (4,1, - 2)$ to the plane $2 x - y + 3 z = 5$ is:

Solution

$\frac{| 2 (4) - (1) + 3 (- 2) - 5 |}{\sqrt{2^{2} + (- 1)^{2} + 3^{2}}} = \frac{| - 4 |}{\sqrt{14}} = \frac{4}{\sqrt{14}} = \frac{2 \sqrt{14}}{7} \approx 1.07$ . Dividing by $14$ instead of $\sqrt{14}$ is the standard slip.

Q5 · full method — try it on paper first

$l_{1} : 𝐫 = (1,2,3) + λ (1,0,1)$ and $l_{2} : 𝐫 = (2,0,1) + μ (0,1,1)$ .

(i) Show that $l_{1}$ and $l_{2}$ are skew. (ii) Find the shortest distance between them. (iii) Find an equation for their common perpendicular.

Solution

(i) First, not parallel:

𝐧 = 𝐝_{1} \times 𝐝_{2} = (1,0,1) \times (0,1,1) = ((0) (1) - (1) (1), (1) (0) - (1) (1), (1) (1) - (0) (0)) = (- 1, - 1,1) \neq 𝟎 .

Next, no common point: equating the two position vectors gives $1 + λ = 2$ , so $λ = 1$ ; and $2 = μ$ from the $y$ -components. But then the $z$ -components need $3 + λ = 1 + μ$ , i.e. $4 = 3$ , which is false. No solution, so the lines do not meet. Not parallel and non-intersecting ⟹ skew.

(ii) With $𝐚_{2} - 𝐚_{1} = (1, - 2, - 2)$ :

d = \frac{| (1) (- 1) + (- 2) (- 1) + (- 2) (1) |}{\sqrt{1 + 1 + 1}} = \frac{| - 1 |}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577 .

(Non-zero, which confirms part (i) again.)

(iii) Let $P = (1 + λ, 2, 3 + λ)$ on $l_{1}$ and $Q = (2, μ, 1 + μ)$ on $l_{2}$ , so

\vec{P Q} = (1 - λ, μ - 2, μ - λ - 2) .

Perpendicular to both directions:

\vec{P Q} \cdot (1,0,1) = μ - 2 λ - 1 = 0, \vec{P Q} \cdot (0,1,1) = 2 μ - λ - 4 = 0 .

The first gives $μ = 2 λ + 1$ ; substituting into the second gives $3 λ - 2 = 0$ , so $λ = \frac{2}{3}$ and $μ = \frac{7}{3}$ . Hence

P = (\frac{5}{3}, 2, \frac{11}{3}), Q = (2, \frac{7}{3}, \frac{10}{3}), \vec{P Q} = (\frac{1}{3}, \frac{1}{3}, - \frac{1}{3}) .

Check: $\vec{P Q} = - \frac{1}{3} (- 1, - 1,1)$ is parallel to $𝐧$ ✓, and $| \vec{P Q} | = \sqrt{3 \cdot \frac{1}{9}} = \frac{1}{\sqrt{3}}$ , matching part (ii) ✓

An equation for the common perpendicular is therefore

𝐫 = (\frac{5}{3}, 2, \frac{11}{3}) + t (1,1, - 1) . ■

FP1 · Topic 6 · Vectors