Quick questions on Vectors in two dimensions explained: O-Level E-Maths

Multiplying a vector by a number (scalar) multiplies each component:

The magnitude (length) of a vector is found by Pythagoras from its components:

A vector linking two points is found by subtracting their position vectors, "destination minus origin". The vector from $A(x_1, y_1)$ to $B(x_2, y_2)$ is $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$, and its magnitude is the distance $AB$, which is exactly the distance formula in disguise. So from $A(1, 2)$ to $B(4, 6)$, the vector is $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ with magnitude $5$. Remembering "tip minus tail" both gives the correct direction and ties vectors directly to the coordinate-geometry distance you already know.

Two vectors are parallel when one is a scalar multiple of the other, $\mathbf{b} = k\mathbf{a}$ for some number $k$. So $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$ are parallel because the second is twice the first, and a negative $k$ means they point in opposite directions. This test is the vector equivalent of equal gradients for parallel lines, and it lets you prove points are collinear: if $\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{AC}$, then $A$, $B$ and $C$ lie on one straight line. Checking for a common scalar factor is the standard way to detect parallel or collinear vectors.

Subtracting a negative component adds, as in $3 - (-5) = 8$.

Find $\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix}$. [1 mark]

Find $3\begin{pmatrix} -2 \\ 4 \end{pmatrix}$. [1 mark]

Find the magnitude of $\begin{pmatrix} 8 \\ 6 \end{pmatrix}$. [2 marks]