Mathematics guide, Beyond AP, step 32 of 33Previous step Next step

Step 32: Vectors and Lines in 3D

By the EconTutorials team. 11-minute read with 6 practice questions.

Syllabus9709 P3 3.7 · IB HL 3.12–3.18

In 30 seconds

A vector has a size and a direction, written as a column or with $ \mathbf{i}, \mathbf{j}, \mathbf{k} $. Its length is $ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $, and $ \overrightarrow{AB} = \mathbf{b}-\mathbf{a} $: end minus start.
The scalar product $ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta $ gives angles, and $ \mathbf{a} \cdot \mathbf{b} = 0 $ means perpendicular.
A line is $ \mathbf{r} = \mathbf{a} + t\mathbf{b} $: start at a point, then move any multiple of a direction. In 3D two lines can be parallel, intersecting or skew (not parallel, never meeting).
Exam point: for the foot of the perpendicular from $ P $ to a line, write the general point $ N $, then solve $ \overrightarrow{PN} \cdot \mathbf{b} = 0 $. The vector product and planes are IB HL only.

Video lesson coming soonA short animated lesson for this step will appear here.

Vectors, notation and magnitude

Vectors are topic 3.7 of Cambridge 9709 Paper 3, usually one multi-part question on lines. In IB Analysis and Approaches all vectors are HL only (3.12 to 3.18), and HL adds the vector product and planes, covered in the last section. AP Calculus BC uses 2D vectors for motion (step 28).

A scalar answers “how much?”; a vector answers “how much, and which way?”. In three dimensions, $ \mathbf{i} $, $ \mathbf{j} $ and $ \mathbf{k} $ are vectors of length 1 along the $ x $-, $ y $- and $ z $-axes, and every vector is a combination of them. The same vector can be written two ways:

\[ 4\mathbf{i}-4\mathbf{j} + 7\mathbf{k} = \begin{pmatrix} 4 \\ -4 \\ 7 \end{pmatrix}. \]

In print vectors are bold; by hand, underline them. The numbers are the components, and two vectors are equal when every component matches.

The magnitude $ |\mathbf{a}| $ is the length of the arrow, by Pythagoras used twice (across the floor, then up):

\[ \left| \begin{pmatrix} x \\ y \\ z \end{pmatrix} \right| = \sqrt{x^2 + y^2 + z^2}, \qquad \hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\,\mathbf{a}. \]

Key termUnit vector: a vector of magnitude 1. Dividing a non-zero vector by its length gives the unit vector $ \hat{\mathbf{a}} $ (“a hat”) in the same direction.

Worked example

Let $ \mathbf{v} = 4\mathbf{i}-4\mathbf{j} + 7\mathbf{k} $. Find $ |\mathbf{v}| $, the unit vector in the direction of $ \mathbf{v} $, and the vectors of magnitude 27 that are parallel to $ \mathbf{v} $.

Step 1$ |\mathbf{v}| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9 $.

Step 2$ \hat{\mathbf{v}} = \tfrac{1}{9}(4\mathbf{i}-4\mathbf{j} + 7\mathbf{k}) $.

Step 3Length 27 is 27 unit vectors, so the vector is $ \pm 27\hat{\mathbf{v}} = \pm 3\mathbf{v} $.

The vectors are $ \pm(12\mathbf{i}-12\mathbf{j} + 21\mathbf{k}) $. The minus sign gives the one pointing the opposite way.

Vector arithmetic and position vectors

Adding, subtracting and scaling all work one component at a time. Adding means “do one journey, then the other”. A scalar multiple $ k\mathbf{a} $ stretches $ \mathbf{a} $ by the factor $ |k| $, and reverses it if $ k < 0 $. So non-zero vectors are parallel exactly when one is a multiple of the other: $ \mathbf{b} = k\mathbf{a} $. To test, divide matching components; every ratio must give the same $ k $.

Fix an origin $ O $. Each point $ A $ then has a position vector $ \mathbf{a} = \overrightarrow{OA} $, whose components are just its coordinates. To go from $ A $ to $ B $, go back to $ O $ and out again: $ \overrightarrow{AB} = -\mathbf{a} + \mathbf{b} $.

\[ \overrightarrow{AB} = \mathbf{b}-\mathbf{a} \quad(\text{end minus start}), \qquad AB = |\mathbf{b}-\mathbf{a}|. \]

Worked example

$ A $ is $ (1, 0, 2) $ and $ B $ is $ (3, 6, 5) $. Find $ \overrightarrow{AB} $, the distance $ AB $, the point $ P $ on $ AB $ with $ AP : PB = 1 : 2 $, and show that $ C(7, 18, 11) $ lies on the line through $ A $ and $ B $.

Step 1$ \overrightarrow{AB} = \begin{pmatrix} 3 \\ 6 \\ 5 \end{pmatrix}-\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 3 \end{pmatrix} $, so $ AB = \sqrt{4 + 36 + 9} = 7 $.

Step 2$ P $ is a third of the way from $ A $: $ \overrightarrow{OP} = \mathbf{a} + \tfrac{1}{3}\overrightarrow{AB} $, so $ P = \left(\tfrac{5}{3}, 2, 3\right) $.

Step 3$ \overrightarrow{AC} = (6, 18, 9) = 3\overrightarrow{AB} $. So $ \overrightarrow{AC} $ is parallel to $ \overrightarrow{AB} $, and the two share the point $ A $.

Hence $ A $, $ B $ and $ C $ are collinear.

Common mistakeStopping at “parallel”. Two parallel vectors can sit on two different parallel lines. To prove three points collinear you must also say that the vectors share a point, as in “$ \overrightarrow{AC} = 3\overrightarrow{AB} $ and $ A $ is common”.

The scalar product and angles

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta \]

Here $ 0^\circ \le \theta \le 180^\circ $. The answer is a number, hence scalar product (or dot product); the two forms agree by the cosine rule. Three facts follow:

For non-zero vectors, $ \mathbf{a} \cdot \mathbf{b} = 0 $ exactly when they are perpendicular.
A positive dot product means an acute angle; a negative one means obtuse.
$ \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 $.

Worked example

The points $ P(2, 0, 1) $, $ Q(3, 2, 3) $ and $ R(4, 0, 4) $ form a triangle. Find angle $ PQR $.

Step 1The angle is at $ Q $, so use the two vectors that start at $ Q $: $ \overrightarrow{QP} = (-1, -2, -2) $ and $ \overrightarrow{QR} = (1, -2, 1) $.

Step 2$ \overrightarrow{QP} \cdot \overrightarrow{QR} = -1 + 4-2 = 1 $, $ |\overrightarrow{QP}| = 3 $, $ |\overrightarrow{QR}| = \sqrt{6} $.

Step 3$ \cos Q = \dfrac{1}{3\sqrt{6}} $, so $ Q = 82.2^\circ $ to 1 d.p.

Angle $ PQR = 82.2^\circ $. Using $ \overrightarrow{PQ} $ instead of $ \overrightarrow{QP} $ flips the sign of the dot product and gives the supplement, $ 97.8^\circ $: both vectors must point away from the vertex.

The economics of the dot product. A shopping basket of three goods is a vector $ \mathbf{x} $. With a price vector $ \mathbf{p} $, total spending is $ p_1x_1 + p_2x_2 + p_3x_3 = \mathbf{p} \cdot \mathbf{x} $. With prices $ \mathbf{p} = (3, 2, 5) $ dollars, the basket $ (4, 10, 2) $ costs $ 12 + 20 + 10 = \$42 $. A consumer with income $ M $ faces the budget constraint $ \mathbf{p} \cdot \mathbf{x} = M $. A move $ \mathbf{d} $ along the budget line costs nothing extra, so $ \mathbf{p} \cdot \mathbf{d} = 0 $: the price vector is perpendicular to the budget line.

The price vector is perpendicular to the budget line

budget line 3x₁ + 2x₂ = 24price vector pmove d along the line

Figure 1. Prices p = (3, 2) dollars and income $24. A move d = (2, −3) along the budget line changes spending by p · d = 6 − 6 = 0, so p is at right angles to the line. Both axes use the same scale, so the right angle looks right.

Lines in 3D: parallel, intersecting or skew

In 3D there is no single gradient, so a line is given like directions: start here, then keep going that way.

\[ \mathbf{r} = \mathbf{a} + t\mathbf{b} \]

Here $ \mathbf{a} $ is the position vector of a known point, $ \mathbf{b} $ is a direction vector, and each real value of the parameter $ t $ gives one point $ \mathbf{r} $ of the line. For the line through $ A $ and $ B $, use $ \mathbf{r} = \overrightarrow{OA} + t\,\overrightarrow{AB} $. The equation is not unique: any point can be the start and any non-zero multiple of $ \mathbf{b} $ the direction.

The line r = a + tb through A(1, 0, 2) and B(3, 6, 5)

position vector adirection b = AB = (2, 6, 3)the line l

Figure 2. Start at A, then travel any multiple t of the direction b. Each value of t gives one point of the line: t = 1 gives B(3, 6, 5) and t = 2 gives D(5, 12, 8).

To test a point, find $ t $ from one component and check the other two. On $ \mathbf{r} = (1, 0, 2) + t(2, 6, 3) $, the point $ D(5, 12, 8) $ gives $ t = 2 $ in every component, so it lies on the line; $ E(-1, -6, 0) $ gives $ t = -1 $ from $ x $ and $ y $ but needs $ 2 + 3t = 0 $, so it does not. The angle between two lines is the angle between their direction vectors. Exams want the acute one, so take the modulus: $ \cos\theta = \dfrac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|} $.

In the plane, two different lines are parallel or they cross. In space there is a third case, like a road under a flyover.

Skew lines along the edges of a box

l₁ along the floorl₂ across the topcommon perpendicular

Figure 3. l₁ and l₂ point in different directions, so they are not parallel, and they are at different heights, so they never meet. Lines like this are skew. They can only occur in three dimensions.

Step	For $ \mathbf{r} = \mathbf{a}_1 + \lambda\mathbf{b}_1 $ and $ \mathbf{r} = \mathbf{a}_2 + \mu\mathbf{b}_2 $
1	Is $ \mathbf{b}_2 $ a multiple of $ \mathbf{b}_1 $? If so the lines are parallel (or the same line, if a point of one lies on the other).
2	If not, set the general points equal: three equations in two unknowns.
3	Solve two of the equations, then test the third. If it holds, the lines intersect; if not, they are skew.

Worked example

Show that $ l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} $ and $ l_2: \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} $ intersect. Find the point of intersection and the acute angle between the lines.

Step 1Not parallel: $ \tfrac{2}{1} \neq \tfrac{-3}{1} $. Equate: $ 1 + \lambda = 4 + 2\mu $, $ 2 + \lambda = -3\mu $, $ 2\lambda = 3 + \mu $.

Step 2Subtract the second from the first: $ -1 = 4 + 5\mu $, so $ \mu = -1 $ and $ \lambda = 1 $.

Step 3Third equation: $ 2\lambda = 2 $ and $ 3 + \mu = 2 $. It holds, so the lines meet, at $ \lambda = 1 $: the point $ (2, 3, 2) $. Check with $ \mu = -1 $ in $ l_2 $: $ (4-2, 0 + 3, 3-1) = (2, 3, 2) $.

Step 4$ \mathbf{b}_1 \cdot \mathbf{b}_2 = 2-3 + 2 = 1 $, $ |\mathbf{b}_1| = \sqrt{6} $, $ |\mathbf{b}_2| = \sqrt{14} $, so $ \cos\theta = \dfrac{1}{\sqrt{84}} $.

The lines meet at $ (2, 3, 2) $ at an acute angle of $ 83.7^\circ $.

Compare $ l_3: \mathbf{r} = (0, 1, 5) + \mu(1, 0, -1) $ with $ l_1 $. The directions are not multiples. Equating, the $ y $-components give $ \lambda = -1 $, then $ x $ gives $ \mu = 0 $, but the $ z $-components become $ -2 = 5 $. Not parallel and not meeting, so skew.

Exam tipTo “show that the lines are skew” you must prove two things and say both: the directions are not parallel, and the three equations have no common solution. End with a sentence such as “not parallel and do not intersect, so skew”. Always give the two lines different parameters, $ \lambda $ and $ \mu $.

The foot of the perpendicular

The shortest distance from a point $ P $ to a line is at the point $ N $ where $ PN $ meets the line at right angles. $ N $ is the foot of the perpendicular, and $ |\overrightarrow{PN}| $ is the perpendicular distance. Method: write the general point $ N $ in terms of $ t $, solve $ \overrightarrow{PN} \cdot \mathbf{b} = 0 $, then substitute.

Worked example

Find the foot of the perpendicular from $ P(7, 6, 1) $ to the line $ \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} $, and the perpendicular distance from $ P $ to the line.

Step 1General point $ N = (1 + 2t,\ t,\ 1 + 2t) $, so $ \overrightarrow{PN} = (2t-6, t-6, 2t) $.

Step 2$ \overrightarrow{PN} \cdot (2, 1, 2) = (4t-12) + (t-6) + 4t = 9t-18 = 0 $, so $ t = 2 $.

Step 3$ N = (5, 2, 5) $ and $ \overrightarrow{PN} = (-2, -4, 4) $, so $ PN = \sqrt{4 + 16 + 16} = 6 $.

The foot is $ N(5, 2, 5) $ and the perpendicular distance is 6. Check: $ (-2)(2) + (-4)(1) + (4)(2) = 0 $.

Foot of the perpendicular from P(7, 6, 1)

l: r = (1, 0, 1) + t(2, 1, 2)PN, perpendicular to l

Figure 4. The view in the flat plane that contains l and P, with true lengths. Of all the points on l, N is the closest to P: there PN meets the direction b at a right angle, so PN · b = 0. The slanted distance AP is longer.

IB HL only: the vector product and planes

IB Analysis and Approaches HL only (3.16 to 3.18). The vector product and planes are not in Cambridge 9709.

The vector product (cross product) of two vectors is a vector perpendicular to both:

\[ \mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1 \end{pmatrix}, \qquad |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta. \]

It follows the right-hand rule, $ \mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b} $, and its length is the area of the parallelogram on $ \mathbf{a} $ and $ \mathbf{b} $, so a triangle has area $ \tfrac{1}{2}|\mathbf{a} \times \mathbf{b}| $. A plane is fixed by one point $ A $ and a normal $ \mathbf{n} $ perpendicular to it: $ \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} $, or in Cartesian form $ n_1x + n_2y + n_3z = d $. The coefficients of $ x, y, z $ are a normal vector.

Worked example

(IB HL) For $ A(1, 1, 0) $, $ B(3, 2, 2) $, $ C(2, 3, 1) $, find the area of triangle $ ABC $ and a Cartesian equation of the plane $ ABC $. Then find where the line $ \mathbf{r} = (0, 1, 2) + t(2, 1, 1) $ meets the plane, and the acute angle between them.

Step 1$ \overrightarrow{AB} = (2, 1, 2) $, $ \overrightarrow{AC} = (1, 2, 1) $, so $ \overrightarrow{AB} \times \overrightarrow{AC} = (1-4,\ 2-2,\ 4-1) = (-3, 0, 3) $.

Step 2Area $ = \tfrac{1}{2}\sqrt{9 + 0 + 9} = \tfrac{3\sqrt{2}}{2} \approx 2.12 $.

Step 3Use $ \mathbf{n} = (1, 0, -1) $, a multiple of the cross product: $ x-z = d $, and $ A $ gives $ d = 1 $. Plane: $ x-z = 1 $. Check $ B $: $ 3-2 = 1 $; $ C $: $ 2-1 = 1 $.

Step 4Substitute $ (2t, 1 + t, 2 + t) $: $ 2t-(2 + t) = 1 $, so $ t = 3 $ and the point is $ (6, 4, 5) $.

Step 5The dot product with the normal gives the angle to the normal, so use sine for the angle to the plane: $ \sin\phi = \dfrac{|(2, 1, 1) \cdot (1, 0, -1)|}{\sqrt{6}\sqrt{2}} = \dfrac{1}{\sqrt{12}} $, so $ \phi = 16.8^\circ $.

Area $ \tfrac{3\sqrt{2}}{2} $, plane $ x-z = 1 $, meeting point $ (6, 4, 5) $, angle $ 16.8^\circ $.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Let $ \mathbf{a} = 3\mathbf{i}-\mathbf{j} + 2\mathbf{k} $ and $ \mathbf{b} = \mathbf{i} + 4\mathbf{j}-2\mathbf{k} $. Find $ 2\mathbf{a}-\mathbf{b} $ and its magnitude. Show worked answer

$ 2\mathbf{a} = 6\mathbf{i}-2\mathbf{j} + 4\mathbf{k} $.
$ 2\mathbf{a}-\mathbf{b} = 5\mathbf{i}-6\mathbf{j} + 6\mathbf{k} $.
$ |2\mathbf{a}-\mathbf{b}| = \sqrt{25 + 36 + 36} = \sqrt{97} \approx 9.85 $.

EasyQ2. Find the unit vector in the direction of $ 2\mathbf{i}-6\mathbf{j} + 3\mathbf{k} $. Show worked answer

Magnitude: $ \sqrt{4 + 36 + 9} = \sqrt{49} = 7 $.
Unit vector: $ \tfrac{1}{7}(2\mathbf{i}-6\mathbf{j} + 3\mathbf{k}) $.
Check: $ \tfrac{4 + 36 + 9}{49} = 1 $.

MediumQ3. The points $ A(1, 2, 3) $, $ B(2, 0, 1) $ and $ C(3, 2, 1) $ form a triangle. Find angle $ ABC $. Show worked answer

Vectors out of $ B $: $ \overrightarrow{BA} = (-1, 2, 2) $, $ \overrightarrow{BC} = (1, 2, 0) $.
$ \overrightarrow{BA} \cdot \overrightarrow{BC} = -1 + 4 + 0 = 3 $; $ |\overrightarrow{BA}| = 3 $, $ |\overrightarrow{BC}| = \sqrt{5} $.
$ \cos B = \dfrac{3}{3\sqrt{5}} = \dfrac{1}{\sqrt{5}} $.
Angle $ ABC = 63.4^\circ $.

MediumQ4. Decide whether the lines $ \mathbf{r} = (2, -1, 0) + s(1, 2, -2) $ and $ \mathbf{r} = (5, 3, -1) + t(-2, -4, 4) $ are the same line, parallel, intersecting or skew. Show worked answer

$ (-2, -4, 4) = -2(1, 2, -2) $, so the directions are parallel.
Is $ (5, 3, -1) $ on the first line? $ (5, 3, -1)-(2, -1, 0) = (3, 4, -1) $.
Ratios with $ (1, 2, -2) $: $ 3, 2, \tfrac{1}{2} $. Not equal, so the point is not on the first line.
The lines are parallel and distinct.

MediumQ5. Find the foot of the perpendicular from $ P(3, 4, -1) $ to the line $ \mathbf{r} = (0, 0, 1) + t(1, 1, 0) $, and the perpendicular distance. Show worked answer

General point $ N = (t, t, 1) $, so $ \overrightarrow{PN} = (t-3, t-4, 2) $.
$ \overrightarrow{PN} \cdot (1, 1, 0) = 2t-7 = 0 $, so $ t = 3.5 $.
$ N = (3.5, 3.5, 1) $, $ \overrightarrow{PN} = (0.5, -0.5, 2) $.
Distance $ = \sqrt{0.25 + 0.25 + 4} = \sqrt{4.5} = \tfrac{3}{\sqrt{2}} \approx 2.12 $.

ExamQ6. The lines $ l $ and $ m $ have equations $ l: \mathbf{r} = \mathbf{i}-2\mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j}-\mathbf{k}) $ and $ m: \mathbf{r} = \mathbf{i}-5\mathbf{j} + 4\mathbf{k} + \mu(\mathbf{i} + 2\mathbf{j}-\mathbf{k}) $. (a) Show that $ l $ and $ m $ intersect, and find the position vector of the point of intersection. [4] (b) Find the acute angle between $ l $ and $ m $. [3] (c) Find the position vector of the foot of the perpendicular from the origin to $ l $, and hence the perpendicular distance from $ O $ to $ l $. [3] Show worked answer

(a) $ 1 + 2\lambda = 1 + \mu $, $ -2 + \lambda = -5 + 2\mu $, $ 3-\lambda = 4-\mu $. [M1]
First: $ \mu = 2\lambda $. Second: $ -2 + \lambda = -5 + 4\lambda $, so $ \lambda = 1 $, $ \mu = 2 $. [A1]
Third: $ 3-1 = 2 $ and $ 4-2 = 2 $, so it holds and the lines intersect. [A1]
Point: $ 3\mathbf{i}-\mathbf{j} + 2\mathbf{k} $. [A1]
(b) $ (2, 1, -1) \cdot (1, 2, -1) = 2 + 2 + 1 = 5 $; both lengths are $ \sqrt{6} $. [M1]
$ \cos\theta = \tfrac{5}{6} $ [M1], so $ \theta = 33.6^\circ $. [A1]
(c) $ \overrightarrow{ON} = (1 + 2\lambda, -2 + \lambda, 3-\lambda) $; $ \overrightarrow{ON} \cdot (2, 1, -1) = 6\lambda-3 = 0 $, so $ \lambda = \tfrac{1}{2} $. [M1]
$ \overrightarrow{ON} = 2\mathbf{i}-1.5\mathbf{j} + 2.5\mathbf{k} $. [A1]
$ ON = \sqrt{4 + 2.25 + 6.25} = \sqrt{12.5} \approx 3.54 $. [A1]
Total: 10 marks.

Check yourself

1. What is the magnitude of $ 2\mathbf{i}-3\mathbf{j} + 6\mathbf{k} $?

$ \sqrt{4 + 9 + 36} = \sqrt{49} = 7 $. Square every component, including the negative one, before adding.

2. For which value of $ k $ are $ (k, 2, -1) $ and $ (3, k, 4) $ perpendicular?

The dot product is $ 3k + 2k-4 = 5k-4 $. Setting it to 0 gives $ k = 0.8 $.

3. Two lines have non-parallel directions, and the three equations from equating their general points have no common solution. The lines are

Not parallel and not meeting is the definition of skew lines, which only happen in three dimensions.

Your score: 0 out of 3

The short version

$ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $; a unit vector is $ \mathbf{a}/|\mathbf{a}| $; $ \overrightarrow{AB} = \mathbf{b}-\mathbf{a} $. Parallel means one vector is a multiple of the other.
$ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta $; zero means perpendicular. For an angle in a triangle, both vectors point away from the vertex.
A line is $ \mathbf{r} = \mathbf{a} + t\mathbf{b} $. Check the directions first, then equate with different parameters and test the third equation: intersecting or skew.
Foot of the perpendicular: general point $ N $, solve $ \overrightarrow{PN} \cdot \mathbf{b} = 0 $, and the distance is $ |\overrightarrow{PN}| $.
IB HL only: $ \mathbf{a} \times \mathbf{b} $ is perpendicular to both, with length equal to the parallelogram area; a plane is $ \mathbf{r} \cdot \mathbf{n} = d $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 32: Vectors and Lines in 3D

Vectors, notation and magnitude

Vector arithmetic and position vectors

The scalar product and angles

Lines in 3D: parallel, intersecting or skew

The foot of the perpendicular

IB HL only: the vector product and planes

Practise: 6 questions

Check yourself

The short version

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