The Scenario
Two delivery drones leave the same depot (the origin $O$). Drone A flies to position $A(3, 1, 2)$ km. Drone B flies to position $B(1, 4, -1)$ km.
Looking from above, their paths clearly diverge at some angle $\theta$. But what exactly is that angle? In 2D, you could use the tangent formula or draw a triangle. In 3D, the two vectors live in different planes โ there is no obvious right-angled triangle to exploit.
We need a new tool: an operation that takes two vectors and returns a single number (a scalar) that somehow encodes the angle between them.
The Question
Given the components of two 3D vectors $\vec{a}$ and $\vec{b}$, how can we find the angle $\theta$ between them โ using only arithmetic on the components, without drawing a diagram?
What single operation on $\vec{a}$ and $\vec{b}$ could produce a scalar that encodes the angle?
๐ก The Bridge (click to reveal)
The scalar product (also called the dot product) is defined as:
$$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$
This scalar has a remarkable geometric meaning:
$$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\,\cos\theta$$
Rearranging gives the angle formula:
$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$
Today we learn how to compute $\vec{a}\cdot\vec{b}$ from components, use it to find angles, and test whether two vectors are perpendicular.
The scalar product (or dot product) is an operation that takes two vectors and returns a single number (a scalar). It has two equivalent forms โ one algebraic (from components) and one geometric (involving the angle). Together they let us compute angles and test perpendicularity in 3D.
For $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$:
$$\boxed{\;\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\;}$$Multiply corresponding components, then add. The result is a scalar (a number), not a vector.
The scalar product has a geometric interpretation. If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0 \leq \theta \leq 180^\circ$), then:
$$\boxed{\;\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\,\cos\theta \quad\Longleftrightarrow\quad \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}\;}$$This is the tool for finding the angle between any two vectors in 2D or 3D โ just from their components.
Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular ($\vec{a} \perp \vec{b}$) if and only if their scalar product is zero:
$$\boxed{\;\vec{a} \cdot \vec{b} = 0 \quad\Longleftrightarrow\quad \vec{a} \perp \vec{b}\;}$$Why: If $\vec{a} \perp \vec{b}$ then $\theta = 90^\circ$ and $\cos 90^\circ = 0$, so $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}| \cdot 0 = 0$. Conversely, if $\vec{a}\cdot\vec{b} = 0$ and neither vector is zero, then $\cos\theta = 0$ so $\theta = 90^\circ$.
- To compute $\vec{a}\cdot\vec{b}$: multiply matching components and add.
- To find the angle between $\vec{a}$ and $\vec{b}$: compute $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$, then $\theta = \cos^{-1}(\cdots)$.
- To test perpendicularity: check whether $\vec{a}\cdot\vec{b} = 0$.
- Sign check: if $\vec{a}\cdot\vec{b} > 0$ the angle is acute; if $< 0$ the angle is obtuse; if $= 0$ it is $90^\circ$.
Four rounds, alternating teams: Team A takes Rounds 1 and 3, Team B takes Rounds 2 and 4. Teams huddle for 30 seconds and score 10 points for a correct answer. Misconceptions are exposed instantly by the scoreboard.
Five questions, 90 seconds, one attempt each. Immediate colour feedback shows the correct answer; the run advances automatically. 10 points per correct answer.
Relative to an origin O, the position vectors of the points A, B and C are given by $\overrightarrow{OA} = \begin{pmatrix}2 \\1 \\3\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}-2 \\4 \\3\end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix}-1 \\2 \\1\end{pmatrix}$. Find the perimeter of triangle ABC.
(Review from W8-L1: this uses vector magnitudes. We include it here as a warm-up before the scalar product examples.)
๐ Reveal Solution
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}-2 \\4 \\3\end{pmatrix} - \begin{pmatrix}2 \\1 \\3\end{pmatrix} = \begin{pmatrix}-4 \\3 \\0\end{pmatrix}$
$|\overrightarrow{AB}| = \sqrt{(-4)^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}-1 \\2 \\1\end{pmatrix} - \begin{pmatrix}-2 \\4 \\3\end{pmatrix} = \begin{pmatrix}1 \\-2 \\-2\end{pmatrix}$
$|\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = \begin{pmatrix}2 \\1 \\3\end{pmatrix} - \begin{pmatrix}-1 \\2 \\1\end{pmatrix} = \begin{pmatrix}3 \\-1 \\2\end{pmatrix}$
$|\overrightarrow{CA}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$
$\text{Perimeter} = |\overrightarrow{AB}| + |\overrightarrow{BC}| + |\overrightarrow{CA}| = 5 + 3 + \sqrt{14}$
$\boxed{\text{Perimeter} = 8 + \sqrt{14} \approx 11.74}$
Find the angles between these pairs of vectors.
(i) $\begin{pmatrix}2 \\1 \\3\end{pmatrix}$ and $\begin{pmatrix}2 \\-1 \\4\end{pmatrix}$ โ (ii) $\begin{pmatrix}1 \\-1 \\0\end{pmatrix}$ and $\begin{pmatrix}3 \\1 \\5\end{pmatrix}$ โ (iii) $3\hat{i} + 2\hat{j} - 2\hat{k}$ and $-4\hat{i} - \hat{j} + 3\hat{k}$
๐ Reveal Solution
$\vec{a}\cdot\vec{b} = 2(2) + 1(-1) + 3(4) = 4 - 1 + 12 = 15$
$|\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}$, โ $|\vec{b}| = \sqrt{4 + 1 + 16} = \sqrt{21}$
$\cos\theta = \dfrac{15}{\sqrt{14}\,\sqrt{21}} = \dfrac{15}{\sqrt{294}} = \dfrac{15}{7\sqrt{6}} \approx 0.8748$
$\theta = \cos^{-1}(0.8748) \approx 28.97^\circ$
Answer: $\theta \approx 29^\circ$ (to the nearest degree)
$\vec{a}\cdot\vec{b} = 1(3) + (-1)(1) + 0(5) = 3 - 1 + 0 = 2$
$|\vec{a}| = \sqrt{1 + 1 + 0} = \sqrt{2}$, โ $|\vec{b}| = \sqrt{9 + 1 + 25} = \sqrt{35}$
$\cos\theta = \dfrac{2}{\sqrt{2}\,\sqrt{35}} = \dfrac{2}{\sqrt{70}} \approx 0.2390$
$\theta = \cos^{-1}(0.2390) \approx 76.2^\circ$
Answer: $\theta \approx 76^\circ$ (to the nearest degree)
$\vec{a}\cdot\vec{b} = 3(-4) + 2(-1) + (-2)(3) = -12 - 2 - 6 = -20$
$|\vec{a}| = \sqrt{9 + 4 + 4} = \sqrt{17}$, โ $|\vec{b}| = \sqrt{16 + 1 + 9} = \sqrt{26}$
$\cos\theta = \dfrac{-20}{\sqrt{17}\,\sqrt{26}} = \dfrac{-20}{\sqrt{442}} \approx -0.9513$
$\theta = \cos^{-1}(-0.9513) \approx 162.0^\circ$
Answer: $\theta \approx 162^\circ$ (to the nearest degree) โ the angle is obtuse because $\vec{a}\cdot\vec{b} < 0$.
Find the angles between these vectors.
(i) $2\hat{i} + 3\hat{j}$ and $4\hat{i} + \hat{j}$ โ (ii) $2\hat{i} - \hat{j}$ and $\hat{i} + 2\hat{j}$ โ (iii) $\begin{pmatrix}-1 \\ -1\end{pmatrix}$ and $\begin{pmatrix}-1 \\ -2\end{pmatrix}$
(iv) $4\hat{i} + \hat{j}$ and $\hat{i} + \hat{j}$ โ (v) $\begin{pmatrix}2 \\ 3\end{pmatrix}$ and $\begin{pmatrix}-6 \\ 4\end{pmatrix}$ โ (vi) $\begin{pmatrix}3 \\ -1\end{pmatrix}$ and $\begin{pmatrix}-6 \\ 2\end{pmatrix}$
๐ก Hint
For each pair: compute $\vec{a}\cdot\vec{b}$, then $|\vec{a}|$ and $|\vec{b}|$, then $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$. If $\vec{a}\cdot\vec{b} = 0$, the angle is $90^\circ$ โ no need to compute further!
๐ Reveal Solution
$\vec{a}\cdot\vec{b} = 8 + 3 = 11$, โ $|\vec{a}| = \sqrt{13}$, โ $|\vec{b}| = \sqrt{17}$
$\cos\theta = \dfrac{11}{\sqrt{221}} \approx 0.7399$ โ $\theta \approx 42.3^\circ$
$\vec{a}\cdot\vec{b} = 2 - 2 = 0$ โ $\Rightarrow$ โ $\theta = 90^\circ$ (perpendicular!)
$\vec{a}\cdot\vec{b} = 1 + 2 = 3$, โ $|\vec{a}| = \sqrt{2}$, โ $|\vec{b}| = \sqrt{5}$
$\cos\theta = \dfrac{3}{\sqrt{10}} \approx 0.9487$ โ $\theta \approx 18.4^\circ$
$\vec{a}\cdot\vec{b} = 4 + 1 = 5$, โ $|\vec{a}| = \sqrt{17}$, โ $|\vec{b}| = \sqrt{2}$
$\cos\theta = \dfrac{5}{\sqrt{34}} \approx 0.8575$ โ $\theta \approx 31.0^\circ$
$\vec{a}\cdot\vec{b} = -12 + 12 = 0$ โ $\Rightarrow$ โ $\theta = 90^\circ$ (perpendicular!)
$\vec{a}\cdot\vec{b} = -18 - 2 = -20$, โ $|\vec{a}| = \sqrt{10}$, โ $|\vec{b}| = \sqrt{40} = 2\sqrt{10}$
$\cos\theta = \dfrac{-20}{\sqrt{10} \cdot 2\sqrt{10}} = \dfrac{-20}{20} = -1$ โ $\theta = 180^\circ$ (anti-parallel)
Note: $\vec{b} = -2\vec{a}$, so the vectors point in opposite directions.
Relative to an origin O, the position vectors of the points P and Q are given by $\overrightarrow{OP} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\overrightarrow{OQ} = \hat{i} + x\hat{j} - 2\hat{k}$. Find the values of $x$ for which the magnitude of PQ is 7.
๐ก Hint
Find $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ in terms of $x$. Then set $|\overrightarrow{PQ}|^2 = 7^2 = 49$ and solve for $x$.
๐ Reveal Solution
$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = (\hat{i} + x\hat{j} - 2\hat{k}) - (3\hat{i} + \hat{j} + 4\hat{k}) = -2\hat{i} + (x-1)\hat{j} - 6\hat{k}$
$\overrightarrow{PQ} = \begin{pmatrix}-2 \\ x-1 \\ -6\end{pmatrix}$
$|\overrightarrow{PQ}|^2 = (-2)^2 + (x-1)^2 + (-6)^2 = 4 + (x-1)^2 + 36 = (x-1)^2 + 40$
Since $|\overrightarrow{PQ}| = 7$: โ $(x-1)^2 + 40 = 49$
$(x-1)^2 = 9$
$x - 1 = \pm 3$
$x = 4$ โ or โ $x = -2$
$\boxed{x = 4 \text{ or } x = -2}$
Relative to an origin O, the position vectors of points A and B are $2\hat{i} + \hat{j} + 2\hat{k}$ and $3\hat{i} - 2\hat{j} + p\hat{k}$ respectively.
(i) Find the value of $p$ for which OA and OB are perpendicular.
(ii) In the case where $p = 6$, use a scalar product to find angle AOB, correct to the nearest degree.
(iii) Express the vector $\overrightarrow{AB}$ in terms of $p$ and hence find the values of $p$ for which the length of AB is 3.5 units.
Cambridge International AS & A Level Mathematics 9709 Paper 1 Q10 June 2008
๐ก Hint 1 (for part i)
OA and OB are perpendicular when $\overrightarrow{OA} \cdot \overrightarrow{OB} = 0$. Compute the dot product in terms of $p$ and set it to zero.
๐ก Hint 2 (for part iii)
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$. Express it in terms of $p$, then set $|\overrightarrow{AB}|^2 = 3.5^2 = 12.25$ and solve.
๐ Reveal Solution
$\overrightarrow{OA} = (2, 1, 2)$, โ $\overrightarrow{OB} = (3, -2, p)$
$\overrightarrow{OA} \cdot \overrightarrow{OB} = 2(3) + 1(-2) + 2(p) = 6 - 2 + 2p = 4 + 2p$
Perpendicular $\Rightarrow 4 + 2p = 0 \Rightarrow p = -2$
Answer: $p = -2$
$\overrightarrow{OA} \cdot \overrightarrow{OB} = 4 + 2(6) = 16$
$|\overrightarrow{OA}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
$|\overrightarrow{OB}| = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$
$\cos\theta = \dfrac{16}{3 \times 7} = \dfrac{16}{21} \approx 0.7619$
$\theta = \cos^{-1}(0.7619) \approx 40.4^\circ$
Answer: $\theta \approx 40^\circ$ (to the nearest degree)
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3-2,\; -2-1,\; p-2) = (1, -3, p-2)$
$|\overrightarrow{AB}|^2 = 1^2 + (-3)^2 + (p-2)^2 = 1 + 9 + (p-2)^2 = 10 + (p-2)^2$
$|\overrightarrow{AB}| = 3.5 \Rightarrow |\overrightarrow{AB}|^2 = 12.25$
$10 + (p-2)^2 = 12.25$
$(p-2)^2 = 2.25$
$p - 2 = \pm 1.5$
$p = 3.5$ โ or โ $p = 0.5$
Answer: $p = 3.5$ or $p = 0.5$
Relative to an origin O, the position vectors of the points A and B are given by $\overrightarrow{OA} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\overrightarrow{OB} = 4\hat{i} - 3\hat{j} + 2\hat{k}$.
(i) Use a scalar product to find angle AOB, correct to the nearest degree.
(ii) Find the unit vector in the direction of $\overrightarrow{AB}$.
(iii) The point C is such that $\overrightarrow{OC} = 6\hat{j} + p\hat{k}$, where p is a constant. Given that the lengths of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are equal, find the possible values of p.
๐ก Hint 1 (for part i)
Compute $\overrightarrow{OA}\cdot\overrightarrow{OB}$, $|\overrightarrow{OA}|$, and $|\overrightarrow{OB}|$. Then use $\cos\theta = \dfrac{\overrightarrow{OA}\cdot\overrightarrow{OB}}{|\overrightarrow{OA}||\overrightarrow{OB}|}$.
๐ก Hint 2 (for part iii)
Find $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$ in terms of $p$. Set $|\overrightarrow{AC}|^2 = |\overrightarrow{AB}|^2$ and solve for $p$. You will need $|\overrightarrow{AB}|$ from part (ii).
๐ Reveal Solution
$\overrightarrow{OA} = (2, 3, -1)$, โ $\overrightarrow{OB} = (4, -3, 2)$
$\overrightarrow{OA}\cdot\overrightarrow{OB} = 2(4) + 3(-3) + (-1)(2) = 8 - 9 - 2 = -3$
$|\overrightarrow{OA}| = \sqrt{4 + 9 + 1} = \sqrt{14}$
$|\overrightarrow{OB}| = \sqrt{16 + 9 + 4} = \sqrt{29}$
$\cos\theta = \dfrac{-3}{\sqrt{14}\,\sqrt{29}} = \dfrac{-3}{\sqrt{406}} \approx -0.1489$
$\theta = \cos^{-1}(-0.1489) \approx 98.6^\circ$
Answer: $\theta \approx 99^\circ$ (to the nearest degree) โ the angle is obtuse since $\overrightarrow{OA}\cdot\overrightarrow{OB} < 0$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4-2,\; -3-3,\; 2-(-1)) = (2, -6, 3)$
$|\overrightarrow{AB}| = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$
Unit vector $= \dfrac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \dfrac{1}{7}(2, -6, 3) = \dfrac{2}{7}\hat{i} - \dfrac{6}{7}\hat{j} + \dfrac{3}{7}\hat{k}$
Answer: $\dfrac{2}{7}\hat{i} - \dfrac{6}{7}\hat{j} + \dfrac{3}{7}\hat{k}$
$\overrightarrow{OC} = (0, 6, p)$, โ $\overrightarrow{OA} = (2, 3, -1)$
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (0-2,\; 6-3,\; p-(-1)) = (-2, 3, p+1)$
$|\overrightarrow{AC}|^2 = (-2)^2 + 3^2 + (p+1)^2 = 4 + 9 + (p+1)^2 = 13 + (p+1)^2$
From part (ii): $|\overrightarrow{AB}| = 7$, so $|\overrightarrow{AB}|^2 = 49$.
$13 + (p+1)^2 = 49$
$(p+1)^2 = 36$
$p + 1 = \pm 6$
$p = 5$ โ or โ $p = -7$
Answer: $p = 5$ or $p = -7$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides next lesson's review.