The Scalar Product (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Using unit vectors

Calculate the value of:

• a) $\mathbf{i} \cdot \mathbf{i}$
• b) $\mathbf{k} \cdot \mathbf{j}$
• c) $\mathbf{i} \cdot (\mathbf{j} + \mathbf{k})$

Solution:

a) $\mathbf{i} \cdot \mathbf{i} = 1 \times 1 \times \cos 0° = 1$

The unit vector $\mathbf{i}$ has magnitude 1, and the angle between $\mathbf{i}$ and itself is 0°.

b) $\mathbf{k} \cdot \mathbf{j} = 1 \times 1 \times \cos 90° = 0$

The unit vectors $\mathbf{k}$ and $\mathbf{j}$ are perpendicular (in the positive $z$- and $y$-directions respectively), so the angle between them is 90°.

c) $\mathbf{i} \cdot (\mathbf{j} + \mathbf{k}) = \mathbf{i} \cdot \mathbf{j} + \mathbf{i} \cdot \mathbf{k} = 0 + 0 = 0$

Using the distributive property, we can expand the brackets and calculate each term separately.

Worked Example 2: Calculating scalar product and finding angles

Given $\mathbf{a} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -4 \\ -1 \\ 5 \end{pmatrix}$, calculate:

• a) The scalar product $\mathbf{a} \cdot \mathbf{b}$
• b) The angle between vectors $\mathbf{a}$ and $\mathbf{b}$

Solution:

a) Using the component formula:

$\mathbf{a} \cdot \mathbf{b} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -4 \\ -1 \\ 5 \end{pmatrix}$

$= (1 \times -4) + (-3 \times -1) + (2 \times 5)$

$= -4 + 3 + 10 = 9$

b) To find the angle, we first need the magnitudes of both vectors.

For vector $\mathbf{a}$: $|\mathbf{a}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

For vector $\mathbf{b}$: $|\mathbf{b}| = \sqrt{(-4)^2 + (-1)^2 + 5^2} = \sqrt{16 + 1 + 25} = \sqrt{42}$

Now we can rearrange the scalar product formula to find $\theta$:

$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$\cos\theta = \frac{9}{\sqrt{14}\sqrt{42}} = \frac{9}{\sqrt{588}} = \frac{9}{14\sqrt{3}} = \frac{3\sqrt{3}}{14}$

Therefore: $\theta = \cos^{-1}\left(\frac{3\sqrt{3}}{14}\right) = 68.2°$ (to 1 decimal place)

Worked Example 3: Finding the obtuse angle between two lines

Given that $L_1$ and $L_2$ have equations $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ and $L_2: \frac{x+4}{2} = \frac{y+3}{-1} = \frac{z-1}{1}$, and they intersect at point $A$, find the obtuse angle between $L_1$ and $L_2$.

Solution:

Step 1: Identify the direction vectors

From the vector equation of $L_1$, the direction vector is $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.

From the Cartesian form of $L_2$, the direction vector is $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

Step 2: Calculate the scalar product

$\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = (1 \times 2) + (-2 \times -1) + (1 \times 1) = 2 + 2 + 1 = 5$

Step 3: Find the magnitudes

$\left|\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\right| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$

$\left|\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\right| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$

Step 4: Calculate the acute angle

$\cos\theta = \frac{5}{\sqrt{6}\sqrt{6}} = \frac{5}{6}$

$\theta = \cos^{-1}\left(\frac{5}{6}\right) = 33.6°$

Step 5: Find the obtuse angle

Obtuse angle $= 180° - 33.6° = 146.4°$ (to 1 decimal place)

Worked Example 4: Proving commutativity

Show that $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ for any 3D vectors $\mathbf{a}$ and $\mathbf{b}$.

Solution:

Let $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$

Then: $\mathbf{a} \cdot \mathbf{b} = (a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}) \cdot (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k})$

Expanding using the distributive property and normal algebraic techniques:

Since $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are perpendicular unit vectors, we know that products of different unit vectors equal zero, and products of identical unit vectors equal 1. Therefore:

$= a_1b_1 + a_2b_2 + a_3b_3$

By expanding $\mathbf{b} \cdot \mathbf{a}$ using the same method, we get:

$\mathbf{b} \cdot \mathbf{a} = b_1a_1 + b_2a_2 + b_3a_3$

Since multiplication of real numbers is commutative ($a_1b_1 = b_1a_1$, etc.), we have:

$\mathbf{b} \cdot \mathbf{a} = a_1b_1 + a_2b_2 + a_3b_3 = \mathbf{a} \cdot \mathbf{b}$

Therefore, $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ as required.