Vectors in 2 Dimensions (AQA A-Level Mathematics): Revision Notes

The magnitude of a vector is its length and is calculated as follows: Given a vector $\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, its magnitude $|\mathbf{a}|$ is given by:

📑Example 1: Finding the Distance Between Two Position Vectors

Find the distance between position vectors $\mathbf{A} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \ and \ \mathbf{B} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}$.

1. Calculate the vector $\overrightarrow{AB}$ by subtracting $\mathbf{A}$ from $\mathbf{B}$:

2. Find the magnitude $|\overrightarrow{AB}|$:

📑Example 2: Finding a Unit Vector Parallel to a Given Vector

Find a unit vector that runs parallel to $\mathbf{a} = \begin{pmatrix} 6 \\ 1 \\ 4 \end{pmatrix}$.

• Note: A unit vector is a vector of length 1.

3. Calculate the magnitude of $\mathbf{a}$:

4. To find the unit vector $\hat{\mathbf{a}}, :highlight[divide] \ \mathbf{a}$ by its magnitude:

📑Example 3:

Given that $\mathbf{\xi} = \begin{pmatrix} \lambda \\ -2\lambda \\ \mu \end{pmatrix} is :highlight[parallel] to \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}$:

a) Show that $3\lambda + 2\mu = 0$.

b) Given also that $|\mathbf{\xi}| = 2\sqrt{29}$, find the nonzero values of $\lambda$ and $\mu$.

Solution:

Part a:

• Note: In 3D, we discuss directions and not gradients. If two lines are parallel, they have the same direction, i.e., one direction is a multiple of the other. Thus, we can say:

$\mathbf{\xi}$ is equal to some multiple of $\begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}$.

• From this, we can form three simultaneous equations:

• Eliminating $k$ from the $x$ and $z$ equations:

• Since $\mu = -\frac{3\lambda}{2}$, then $2\mu = -3\lambda$.
• Therefore, $3\lambda + 2\mu = 0$.

Part b:

• Given:

• Squaring both sides:

• Case 1: If $\lambda = 4$, then:

• Case 2: If $\lambda = -4$, then: