Displacement is the change in position of an object from point A to point B. It is represented as $\vec{AB}$ — it tells you how far and in which direction the object moved.

A position vector is a vector that starts at the origin O(0,0) and ends at a given point. For example, the position vector of point A(3, 2) is $\vec{OA} = (3, 2)$. Position vectors are named using the coordinates of their endpoints.

A vector in the xy-plane can be written using components: $\vec{v} = (x, y)$

The unit vectors $\mathbf{i}$ and $\mathbf{j}$ point along the x-axis and y-axis respectively: $\mathbf{i} = (1, 0), \quad \mathbf{j} = (0, 1)$

Any vector can be written in terms of $\mathbf{i}$ and $\mathbf{j}$: $\vec{v} = x\mathbf{i} + y\mathbf{j}$

Example 1: Write $\mathbf{a} = (-3, 4)$ in terms of $\mathbf{i}$ and $\mathbf{j}$.

Solution: $\mathbf{a} = -3\mathbf{i} + 4\mathbf{j}$

The magnitude (or length) of a vector $\vec{v} = (x, y)$ is found using the Pythagorean theorem: $|\vec{v}| = \sqrt{x^2 + y^2}$

Example 2: Find the magnitude of $\mathbf{v} = (-3, 4)$.

Solution: $|\mathbf{v}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

A unit vector has magnitude equal to 1. The unit vector in the direction of $\vec{v}$ is: $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$

Example 3: Find a unit vector in the direction of $\mathbf{u} = (12, 5)$.

Solution: $|\mathbf{u}| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ $\hat{u} = \frac{(12, 5)}{13} = \left(\frac{12}{13}, \frac{5}{13}\right)$

The direction of a vector can be expressed as a bearing, measured clockwise from the North (000°).

• North: 000°
• East: 090°
• South: 180°
• West: 270°

Example 4: A town P is 140 km at a bearing of 070° from town Q. Find the displacement from Q to P.

Solution: The displacement vector points from Q to P at a bearing of 070°. Its magnitude is 140 km.

The resultant of two or more vectors is their sum. Vectors can be added using:

Triangle Law

Join the end of the first vector to the start of the second vector. The resultant goes from the start of the first to the end of the second.

Parallelogram Law

When two vectors share a common starting point, complete a parallelogram. The resultant is the diagonal through that point.

Polygon Law

For more than two vectors, join them head-to-tail. The resultant goes from the start of the first to the end of the last.

Example 5: If $\mathbf{a} = (3, 2)$ and $\mathbf{b} = (1, 4)$, find $\mathbf{a} + \mathbf{b}$.

Solution: $\mathbf{a} + \mathbf{b} = (3+1, 2+4) = (4, 6)$

Subtracting a vector is the same as adding its opposite: $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})$

Example 6: If $\mathbf{p} = (8, 6)$ and $\mathbf{q} = (3, 4)$, find $\mathbf{p} - \mathbf{q}$.

Solution: $\mathbf{p} - \mathbf{q} = (8-3, 6-4) = (5, 2)$

Multiplying a vector by a scalar (a number) scales its magnitude but keeps its direction (or reverses it if the scalar is negative).

If $\mathbf{v} = (x, y)$ and $k$ is a scalar, then: $k\mathbf{v} = (kx, ky)$

Example 7: If $\mathbf{a} = 3\mathbf{i} + 3\mathbf{j}$, find $2\mathbf{a}$.

Solution: $2\mathbf{a} = 2(3\mathbf{i} + 3\mathbf{j}) = 6\mathbf{i} + 6\mathbf{j}$

Example 8: If $\mathbf{a} = 3\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = 5\mathbf{i} + 4\mathbf{j}$, find $-5\mathbf{a} + 3\mathbf{b}$.

Solution: $-5\mathbf{a} + 3\mathbf{b} = -5(3\mathbf{i} + 3\mathbf{j}) + 3(5\mathbf{i} + 4\mathbf{j})$ $= (-15\mathbf{i} - 15\mathbf{j}) + (15\mathbf{i} + 12\mathbf{j})$ $= (-15 + 15)\mathbf{i} + (-15 + 12)\mathbf{j}$ $= 0\mathbf{i} - 3\mathbf{j} = -3\mathbf{j}$

Two or more vectors are equivalent if they have the same magnitude and direction. They can start at different points but still represent the same displacement.

Vectors are used in everyday life in Tanzania, such as when navigating between towns. For example, a bus driver travelling from Dar es Salaam to Dodoma needs to know the displacement (distance and direction) to plan the route efficiently. Engineers also use vectors to calculate forces on structures like bridges, ensuring they can withstand loads safely.