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Cambridge International AS & A Level · Topic 3.7
A-Level Maths 9709 / Topic 3.7 / AS & A Level

Add a third direction.

Paper 3 only. Everything you know about column vectors in the plane carries straight into space: add a \(z\)-component, a unit vector \(\mathbf{k}\), and one more square under the square root.

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The key idea

A vector in space is \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = \begin{pmatrix}x\\y\\z\end{pmatrix}\), with magnitude \(\sqrt{x^{2}+y^{2}+z^{2}}\). The displacement from \(A\) to \(B\) is \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\): end minus start.

x y z O P(2, 3, 6) 2 3 6 OP = 2i + 3j + 6k |OP| = √(4 + 9 + 36) = √49 = 7
Fig. 1: P(2, 3, 6) sits at a corner of a 2 by 3 by 6 box; its position vector is 2i + 3j + 6k and its distance from O is the length of the box diagonal, 7
Section 01

Notation and magnitude.

In the lab, build a vector from its components and watch its length update: the length is Pythagoras on the components. In three dimensions the rule simply gains a third square. Learn every notation Cambridge uses, because questions switch between them freely.

Worked example Magnitude and unit vector
Given \(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\), find \(|\mathbf{a}|\) and the unit vector in the direction of \(\mathbf{a}\).
  1. Square each component: \(2^{2} = 4\), \((-3)^{2} = 9\), \(6^{2} = 36\). A negative component still gives a positive square.
  2. \(|\mathbf{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\).
  3. Divide the vector by its magnitude: \(\hat{\mathbf{a}} = \dfrac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\).
  4. Check: \(\left(\tfrac{2}{7}\right)^{2} + \left(\tfrac{3}{7}\right)^{2} + \left(\tfrac{6}{7}\right)^{2} = \tfrac{49}{49} = 1\).
Answer: \(|\mathbf{a}| = 7\); unit vector \(\tfrac{2}{7}\mathbf{i} - \tfrac{3}{7}\mathbf{j} + \tfrac{6}{7}\mathbf{k}\).
Worked example Displacement and distance
The points \(A\) and \(B\) have coordinates \((1, 2, -1)\) and \((4, -2, 11)\). Find \(\overrightarrow{AB}\) and the distance \(AB\).
  1. End minus start: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}4-1\\-2-2\\11-(-1)\end{pmatrix} = \begin{pmatrix}3\\-4\\12\end{pmatrix}\).
  2. \(AB = \sqrt{3^{2} + (-4)^{2} + 12^{2}} = \sqrt{9 + 16 + 144} = \sqrt{169}\).
Answer: \(\overrightarrow{AB} = 3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}\), \(AB = 13\).
Your turn
Find the possible values of \(k\) for which \(|2k\mathbf{i} + k\mathbf{j} + 2\mathbf{k}| = 7\).
Examiner trap

\(\overrightarrow{AB}\) is \(\mathbf{b} - \mathbf{a}\), not \(\mathbf{a} - \mathbf{b}\). Getting it backwards gives \(\overrightarrow{BA}\): the same length but the opposite direction, which ruins every later step that uses the direction.

Section 02

Adding, subtracting and scaling.

Vectors add component by component, and a scalar multiplies every component. Geometrically, adding is following one arrow then the next; multiplying by a number stretches the arrow (and reverses it if the number is negative).

Worked example Completing a parallelogram
\(OABC\) is a parallelogram, where \(O\) is the origin, \(A\) is \((2, 1, 3)\) and \(C\) is \((-1, 4, 2)\). Find the position vector of \(B\), and show that the diagonals bisect each other.
  1. Go \(O \to A\) then \(A \to B\), where \(\overrightarrow{AB} = \overrightarrow{OC}\) (opposite sides of a parallelogram).
  2. \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = \begin{pmatrix}2\\1\\3\end{pmatrix} + \begin{pmatrix}-1\\4\\2\end{pmatrix} = \begin{pmatrix}1\\5\\5\end{pmatrix}\).
  3. Midpoint of \(OB\): \(\tfrac{1}{2}(1, 5, 5) = (0.5, 2.5, 2.5)\).
  4. Midpoint of \(AC\): \(\tfrac{1}{2}(2 + (-1),\ 1 + 4,\ 3 + 2) = (0.5, 2.5, 2.5)\), the same point.
Answer: \(\overrightarrow{OB} = \mathbf{i} + 5\mathbf{j} + 5\mathbf{k}\); both diagonals have midpoint \((0.5, 2.5, 2.5)\).
Worked example Parallel vectors
Find \(p\) and \(q\) given that \(2\mathbf{i} + p\mathbf{j} - 4\mathbf{k}\) is parallel to \(3\mathbf{i} + 6\mathbf{j} + q\mathbf{k}\).
  1. Parallel means \(\begin{pmatrix}2\\p\\-4\end{pmatrix} = \lambda\begin{pmatrix}3\\6\\q\end{pmatrix}\).
  2. The \(\mathbf{i}\)-components are both known: \(2 = 3\lambda\), so \(\lambda = \tfrac{2}{3}\).
  3. \(\mathbf{j}\): \(p = 6 \times \tfrac{2}{3} = 4\).
  4. \(\mathbf{k}\): \(-4 = \tfrac{2}{3}q\), so \(q = -6\).
Answer: \(p = 4\), \(q = -6\).
Worked example Collinear points
Show that \(A(1, 2, 3)\), \(B(3, 3, 1)\) and \(C(7, 5, -3)\) are collinear, and find the ratio \(AB : BC\).
  1. \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (2, 1, -2)\).
  2. \(\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = (4, 2, -4) = 2\overrightarrow{AB}\), so the two vectors are parallel.
  3. They share the point \(B\), so \(A\), \(B\), \(C\) lie on one straight line.
  4. Since \(\overrightarrow{BC} = 2\overrightarrow{AB}\), \(BC\) is twice as long as \(AB\).
Answer: Collinear, with \(AB : BC = 1 : 2\).
Examiner trap

Parallel vectors need ONE common factor for every component. \((2, 4, 6)\) and \((1, 2, 4)\) look alike, but the factors are \(2, 2, 1.5\), so they are not parallel.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Find \(|\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}|\).

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\(A\) is \((2, 5, 1)\) and \(B\) is \((3, 1, 4)\). Find \(\overrightarrow{AB}\).

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Find the unit vector in the direction of \(3\mathbf{i} - 4\mathbf{k}\).

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Which vector is parallel to \(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\)?

Section 03

Position vectors and midpoints.

Every point is reached from the origin by its position vector, so to find a new point, start at a known point and add the right multiple of a direction. The only ratio result Cambridge expects you to quote is the midpoint; anything else, build from a route.

Worked example Midpoint
Find the position vector of the midpoint \(M\) of \(AB\), where \(A\) is \((3, -1, 4)\) and \(B\) is \((5, 7, -2)\).
  1. \(\overrightarrow{OM} = \tfrac{1}{2}\left[\begin{pmatrix}3\\-1\\4\end{pmatrix} + \begin{pmatrix}5\\7\\-2\end{pmatrix}\right] = \tfrac{1}{2}\begin{pmatrix}8\\6\\2\end{pmatrix}\).
Answer: \(\overrightarrow{OM} = 4\mathbf{i} + 3\mathbf{j} + \mathbf{k}\).
Worked example A point a set distance along a line
\(A\) is \((1, 0, 2)\) and \(B\) is \((3, -4, 6)\). The point \(P\) lies on the ray from \(A\) through \(B\), with \(AP = 9\). Find the position vector of \(P\).
  1. \(\overrightarrow{AB} = (2, -4, 4)\) and \(|\overrightarrow{AB}| = \sqrt{4 + 16 + 16} = 6\).
  2. Unit vector from \(A\) towards \(B\): \(\tfrac{1}{6}(2, -4, 4) = \tfrac{1}{3}(1, -2, 2)\).
  3. Nine units in that direction: \(9 \times \tfrac{1}{3}(1, -2, 2) = (3, -6, 6)\).
  4. \(\overrightarrow{OP} = \overrightarrow{OA} + (3, -6, 6) = (1 + 3,\ 0 - 6,\ 2 + 6)\).
Answer: \(\overrightarrow{OP} = 4\mathbf{i} - 6\mathbf{j} + 8\mathbf{k}\) (beyond \(B\), since \(AP = 9 > AB = 6\)).
Your turn
\(A\) is \((1, 2, 3)\) and \(B\) is \((7, 5, -3)\). Find the position vector of the point \(P\) on \(AB\) with \(AP : PB = 1 : 2\).
Examiner trap

The midpoint is half the SUM of the position vectors, \(\tfrac{1}{2}(\mathbf{a} + \mathbf{b})\). Half the difference, \(\tfrac{1}{2}(\mathbf{b} - \mathbf{a})\), is the step from \(A\) to the midpoint, not the midpoint itself.

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\(OABC\) is a parallelogram with \(O\) the origin, \(\overrightarrow{OA} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\overrightarrow{OB} = 5\mathbf{i} + 2\mathbf{j} + \mathbf{k}\). Find \(\overrightarrow{OC}\).

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\(A\) is \((1, -2, 4)\) and \(B\) is \((5, 2, 2)\). \(M\) is the midpoint of \(AB\). Find the distance \(OM\).

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The vector \(\overrightarrow{OP}\) has magnitude 15 and the same direction as \(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\). Find \(\overrightarrow{OP}\).

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Given that \(|p\mathbf{i} + (p - 2)\mathbf{j} + 4\mathbf{k}| = 6\), find the possible values of \(p\).

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Vectors in space, sorted.

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Six original Cambridge-style questions
Displacements and distances, unit vectors, parallel and collinear points, a parallelogram, and a drone flying a set distance in a given direction.