Remember to register **here** for FREE to ask any questions you may come across in your QCE studies!Vectors go on to playing a huge role in Unit 3 in your Year 12 course. This follows from being introduced quite extensively in Unit 1 as well. It's

**assumed** that you know all of the Unit 1 material on vectors in order to continue onto further work later in Unit 3.

Here we summarise pretty much everything you need to know from Unit 1. At any point if you bump into issues (as a consequence of gaps in learning, forgetfulness etc.) with vectors, you can come back to this to get back on track.

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*Using directed line segments*- A

**vector** refers to a quantity assumed to have a magnitude, but also a

*direction* (e.g. up/down, left/right, compass directions).

- A

**scalar** refers to a quantity that only has magnitude. Scalars are usually just numbers.

Generally vectors are represented as directed line segments, i.e. a part of a line with a direction assigned to it. Colloquially we usually call this object an 'arrow'.

- As implied by its name, the

**direction** of a vector is where it is heading towards.

- The

**magnitude** of a vector reflects its

*length*Two vectors are

**equal** if they have the

*same* magnitude and direction. (Note that unlike points in the Cartesian plane, vectors do not

*need* to start from the origin.)

A

**scalar multiple** of a vector is the result of multiplying a number (scalar) to the vector. It has the effect of

- stretching its length (by the factor of the number) if it has absolute value greater than 1.

- shrinking its length (by the factor of the number) if it has absolute value less than 1.

- leaving its length unchanged if the number is 1 or -1.

- reversing the vector's direction if the number is negative (else the direction is unchanged).

To compute the

**sum** of two vectors, we use the "triangle rule".

- Join the start of your second vector to the end of the first vector.

- The vector from the start of your first vector, to the end of your second vector, represents the sum.

Subtraction of vectors can be computed by finding the "negative" of the second vector, that is, obtain the vector with the same length but

*opposite direction* to the second. Add that one to the first instead.

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*Vectors in the plane*In the Cartesian plane, a vector can always be split into its

**components**. The vector from a point \(A\) to another point \(B\) is found by computing the difference between their \(x\) and \(y\) coordinates. The differences are found from the final point to the initial point, and the vector is denoted \( \overrightarrow{AB} \).

The vector can be short-hand labelled as \( \mathbf{v} \) or \(\underset{\sim}{a}\). (You may, of course, be given different letters to use.)

Further, we can then represent the vector using

**column vector notation**.

- The first entry reflects how much we go in the \(x\) direction. If the vector points to the right, this entry will be positive. If it points to the left, it will be negative. If it does not go left or right, it will be 0.

- The second entry reflects how much we go in the \(y\) direction. If the vector points up, this entry will be positive. If it points down, it will be negative. If it does not go up or down, it will be 0.

Once again, note that the start (and end) points of a vector need not be the same, for us to obtain the same vector.

Note that the

**zero vector** \( \begin{pmatrix}0\\0\end{pmatrix} \) is essentially a vector that does not go anywhere - it starts and finishes at the same pooint.

The

**standard unit vectors** \( \hat{\mathbf{i}} \) and \(\hat{\mathbf{j}}\) are used to denote the following special vectors:

\begin{align*}

\hat{\mathbf{i}}&=\begin{pmatrix}1\\0\end{pmatrix}\\

\hat{\mathbf{j}}&=\begin{pmatrix}0\\1\end{pmatrix}

\end{align*}

There are multiple valid representations for the same vector. The column vector notation is one standard way of representing the vector.

- A

**component form** can also be used to represent a vector. In general, every vector can be written as a sum of a scalar multiple of \( \hat{\mathbf{i}} \) and \(\hat{\mathbf{j}}\). In particular,

\[ \begin{pmatrix}a\\b\end{pmatrix} = a \hat{\mathbf{i}} + b\hat{\mathbf{j}} .\]

Addition and subtraction of vectors becomes easy when we have component or column-vector form. Just add/subtract the components individually!\[ (2\mathbf{i} - \mathbf{j}) + (3\mathbf{i} + 4\mathbf{j}) = 5\mathbf{i} + 3\mathbf{j} \]

Scalar multiplication also becomes easy - simply multiply each component by the number given!

\[ -2\begin{pmatrix}4\\ -1 \end{pmatrix} = \begin{pmatrix} -8\\2 \end{pmatrix} \]

For vectors in a plane, the

**magnitude** of the vector can be computed using Pythagoras' theorem.

\[ | a\mathbf{i} + b\mathbf{j}| = \left|\begin{pmatrix} a\\b \end{pmatrix}\right| = \sqrt{a^2+b^2} \]

The "direction" is found depending on which quadrant the point is in, much like how you handle it with complex numbers. Depending on which quadrant you're in, you can assign an angle \(\theta\) such that \( -180^\circ < \theta \leq 180^\circ\) in degrees, or \( -\pi < \theta \leq \pi\) in radians.

In each case, it is true that for the vector \( a\mathbf{i}+b\mathbf{j} \), \( \boxed{\tan \theta = \frac{b}{a}} \).

Note: Usually nobody remembers these problems. People just remember the method - figure out where the angle is, and adjust it however appropriate.Every vector can also be represented in a

**polar form**, in which it is expressed in terms of its magnitude and its direction angle, We write

\[ a\mathbf{i} + b\mathbf{j} = [r, \theta] \]

where \(r\) is its magnitude \( \sqrt{a^2+b^2} \) and \(\theta\) is its direction angle. In general, the conversions are

\begin{align*}

x&=r\cos\theta\\

y&=r\sin\theta\\

r&=\sqrt{x^2+y^2}\\

\tan\theta &= \frac{y}{x}

\end{align*}

(A possibly useful property: if \(\lambda\) is a scalar, then \( |\lambda \mathbf{v}| = |\lambda| |\mathbf{v}| \).)

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*Further work on vectors*A

**unit vector** in the direction of another vector \(\mathbf{v}\) is the vector pointing in the same direction, but with length 1. It is found by essentially dividing the original vector by its length, and is denoted \( \hat{\mathbf{v}} \). That is,

\[ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \]

(The standard unit vectors are special cases of these vectors - they just happen to point in the directions of the \(x\) and \(y\) axes.)

Vectors that are scalar multiples of each other will always be

**parallel**, since they either point in the same direction or in opposite directions.

The vector \( \overrightarrow{OM}\) corresponding to the

**midpoint** of an interval (line segment) \(AB\) can be computed by considering

\[ \overrightarrow{OM} = \frac12 \left( \overrightarrow{OA} + \overrightarrow{OB} \right). \]

Given two vectors \( \mathbf{u} = \begin{pmatrix}u_1\\ u_2\end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix}v_1\\v_2\end{pmatrix} \), the scalar product is defined as

\[ \mathbf{u}\cdot \mathbf{v} = \begin{pmatrix}u_1\\u_2\end{pmatrix} \begin{pmatrix}v_1\\v_2\end{pmatrix} = u_1v_1+u_2v_2. \]

(That is, multiply component wise, and then add those products up.)

It may be worth noting that \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\).

The dot product has a useful geometric interpretation in that

\[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta \]

where \(\cos \theta\) reflects the

**acute/obtuse angle between** the vectors \(\mathbf{u}\) and \(\mathbf{v}\).

Vectors that are

**perpendicular** have dot product

**equal to 0**.

The

**projection** of a vector \( \mathbf{a} \) in the direction of another vector \(\mathbf{b}\) is the resulting vector from when we "push" \(\mathbf{a}\) onto \(\mathbf{b}\). It is given by

\[ \mathbf{a}\text{ on }\mathbf{b} = \frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} = (\mathbf{a}\cdot \hat{\mathbf{b}}) \mathbf{b}. \]

*Note: Diagram looks a bit unconventional because it appears that \(\mathbf{v}\) and \(\mathbf{w}\) point in opposite ways. The point of this diagram is to show that this doesn't really matter!*It can be worth noting that \( \mathbf{a} - (\mathbf{a}\text{ on }\mathbf{b})\) is also perpendicular to \(\mathbf{b}\).