# Vector Basics

## Vector Basics Revision

**Vector Basics**

This section covers what **vectors **are and how to **add **them together.

**What are Vectors?**

A vector contains both **size **and **direction**, for example a velocity of 5\text{ m/s} on a bearing of 120\degree

As shown on the right, vectors are drawn as lines with **arrowheads** on them.

The magnitude (size) of the vector is expressed by the length of the line, these are sometimes drawn to scale.

The direction of the vector is shown by the direction of the arrow.

**Note:** Vectors represented by a single letter are either written using bold font, \boldsymbol{a}, or with an underline, \underline{a}.

**Finding Resultant Vectors**

Vectors can be added together by drawing them, then the single vector that goes from the start of the first vector to the end of the final vector is known as the **resultant** vector.

We can see that to get to the end point of \boldsymbol{r}, we need to go along the vector \boldsymbol{a} then the vector \boldsymbol{b}.

Thus, \boldsymbol{r}=\boldsymbol{a}+\boldsymbol{b}

The **order** that you add the vectors in doesn’t matter, the resultant vector is always the **same**.

**Subtracting Vectors**

**Subtracting **a vector is the same as **adding a negative** vector.

If \boldsymbol{a} is a vector, then \boldsymbol{-a} is a vector of the same size, in the **opposite **direction.

Therefore, subtracting a vector is the same as adding the negative vector.

E.g. \boldsymbol{a-b}=\boldsymbol{a+(-b)}

**Multiplying Vectors by Scalars**

When you multiply a vector by a **scalar **(numeric value), this **changes the length** of the vector, but has no affect on the direction.

If a vector is multiplied by a non-zero scalar, the new vector is always **parallel **to the original vector, e.g. \boldsymbol{a} is parallel to 3\boldsymbol{a}

To show two vectors are parallel you need to show that they are **scalar multiples** of each other.

**Example:** \overrightarrow{AB}=\boldsymbol{x}, \overrightarrow{BC}=\boldsymbol{y}

The point M lies on the line AB and divides \overrightarrow{AB} in the ratio 3:2 and N lies on the line BC and divides \overrightarrow{BC} in the ratio 2:3

Show that \overrightarrow{MN} is parallel to \overrightarrow{AC}

\overrightarrow{AC}=\boldsymbol{x}+\boldsymbol{y}

M divides \overrightarrow{AB} in the ratio 3:2 so M is \dfrac{3}{5} of the way along \overrightarrow{AB}. Therefore, \overrightarrow{AM}=\dfrac{3}{5}\boldsymbol{x}, so \overrightarrow{MB}=\dfrac{2}{5}\boldsymbol{x}.

Similarly, N is \dfrac{2}{5} of the way along \overrightarrow{BC}, so \overrightarrow{BN}=\dfrac{2}{5}\boldsymbol{y}

Finally, \overrightarrow{MN}=\dfrac{2}{5}\boldsymbol{x}+\dfrac{2}{5}\boldsymbol{y}=\dfrac{2}{5}(\boldsymbol{x}+\boldsymbol{y})=\dfrac{2}{5}\overrightarrow{AC}, which shows \overrightarrow{MN} is parallel to \overrightarrow{AC}

## Vector Basics Example Questions

**Question 1:** Find the resultant vector \boldsymbol{r} of the two single vectors \boldsymbol{a} and 2\boldsymbol{b} using the diagram below.

**[1 mark]**

Going from the start of \boldsymbol{a} to the end of 2\boldsymbol{b}, \boldsymbol{r}=\boldsymbol{a}+2\boldsymbol{b}

**Question 2: **Using the diagram below, write down the following vectors in terms of \boldsymbol{a},\boldsymbol{b} and \boldsymbol{c}

a) \overrightarrow{XY}

b) \overrightarrow{XZ}

c) \overrightarrow{ZY}

**[3 marks]**

\overrightarrow{XY}=-\boldsymbol{c}+\boldsymbol{a}

\overrightarrow{XZ}=-\boldsymbol{c}+\boldsymbol{b}

\overrightarrow{ZY}=-\boldsymbol{b}+\boldsymbol{a}

**Question 3: **\overrightarrow{XY}=\boldsymbol{p}, \overrightarrow{XZ}=\boldsymbol{q}

A divides \overrightarrow{XY} in the ratio 5:1 and B divides \overrightarrow{XZ} in the ratio 5:1

Show that \overrightarrow{YZ} is parallel to \overrightarrow{AB}

**[3 marks]**

A divides \overrightarrow{XY} in the ratio 5:1 so A is \dfrac{5}{6} of the way along \overrightarrow{XY}. Therefore, \overrightarrow{AX}=-\dfrac{5}{6}\boldsymbol{p}. \\[1.2em]

Similarly, B is \dfrac{5}{6} of the way along \overrightarrow{XZ}, so \overrightarrow{XB}=\dfrac{5}{6}\boldsymbol{q}\\[1.2em]

Finally, \overrightarrow{AB}=-\dfrac{5}{6}\boldsymbol{p}+\dfrac{5}{6}\boldsymbol{q}=\dfrac{5}{6}(-\boldsymbol{p}+\boldsymbol{q})=\dfrac{5}{6}\overrightarrow{YZ}, which shows \overrightarrow{AB} is parallel to \overrightarrow{YZ}

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