Vectors in a Cartesian Plane


 Vectors in a Cartesian Plane

\(\blacksquare\) Vectors in a Cartesian plane can be written as
(a) \(x\utilde{i}+y\utilde{j}\)
(b) \(\begin{pmatrix} x \\ y \end{pmatrix}\), where

\(\utilde{i}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} ,\utilde{j}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}\)

\(\blacksquare\) \(\utilde{i}\) and \(\utilde{j}\) are vectors of magnitude \(1\) unit that parallel to axis and axis respectively.
\(\blacksquare\) If \(A \begin{pmatrix} x_1, y_1 \end{pmatrix}\) is a point on a Cartesian plane, the vector formed from the origin \(O\) to point \(A\) is
\(\begin{aligned} \overrightarrow{OA}&=x_1\utilde{i}+y_1\utilde{j} \end{aligned}\).
\(\begin{aligned} \overrightarrow{OA} \end{aligned}\) is known as a position vector.

\(\blacksquare\) The magnitude for a vector

\(\utilde{r}=x\utilde{i}+y\utilde{j}\) is


\(\blacksquare\) The unit vector in the direction of \(\utilde{r}\) is


\(\begin{aligned} \hat{\utilde{r}}&=\dfrac{\utilde{r}}{|\utilde{r}|} \\\\ &=\dfrac{x\utilde{i}+y\utilde{j}}{\sqrt{x^2+y^2}} \end{aligned}\)

\(\blacksquare\) The magnitude of the unit vector in the direction of a vector is \(1\) unit.
\(\blacksquare\) Addition


\(\blacksquare\) Subtraction


\(\blacksquare\) Multiplication by scalar


Example 1:


The diagram shows a vector \(\utilde{q}\) drawn on the Cartesian plane.

Find the magnitude of the vector \(\utilde{q}\).

Based on the diagram,

\(\begin{aligned} |q|&=\sqrt{2^2+1^2} \\\\ &=\sqrt{4+1} \\\\ &=\sqrt{5} \text{ units}. \end{aligned} \)

Example 2:
Determine unit vector for the following:
\(\overrightarrow{AD}=3 \utilde{i}+4 \utilde{j}\)
The unit vector in the direction of \(\utilde{r}\) is

\(\begin{aligned} \hat{\utilde{r}}&=\dfrac{\utilde{r}}{|\utilde{r}|} \\\\ &=\dfrac{x\utilde{i}+y\utilde{j}}{\sqrt{x^2+y^2}} \end{aligned}\)

Hence, unit vector in the direction of \(\overrightarrow{AD}\)

\(\begin{aligned} &=\dfrac{\overrightarrow{AD}}{\left|\overrightarrow{AD}\right|} \\\\&=\dfrac{3\utilde{i}+4\utilde{j}}{\sqrt{3^2+4^2}} \\\\ &=\dfrac{3\utilde{i}+4\utilde{j}}{\sqrt{25}} \\\\ &=\dfrac{3\utilde{i}+4\utilde{j}}{5} \\\\ &=\dfrac{3}{5}\utilde{i} +\dfrac{4}{5}\utilde{j}. \end{aligned}\)

Example 3:

Given the vectors

\(\textbf{p}=-5 \utilde{i}+2\utilde{j}\),

\(\textbf{q}= \utilde{i}+3\utilde{j}\),

\(\textbf{r}=2 \utilde{i}-7\utilde{j}\).


Find \(3\textbf{p}-\textbf{q}+2\textbf{r}\).

Given that

\(\textbf{p}=-5 \utilde{i}+2\utilde{j}\),

\(\textbf{q}= \utilde{i}+3\utilde{j}\),

\(\textbf{r}=2 \utilde{i}-7\utilde{j}\).


\(\begin{aligned} \textbf{p}&=-5 \utilde{i}+2\utilde{j} \\\\ 3\textbf{p}&=3(-5 \utilde{i}+2\utilde{j}) \\\\ &=-15 \utilde{i}+6\utilde{j}. \end{aligned}\)


\(\begin{aligned} \textbf{r}&=2 \utilde{i}-7\utilde{j} \\\\ 2\textbf{r}&=2(2 \utilde{i}-7\utilde{j}) \\\\ &=4\utilde{i}-14\utilde{j}. \end{aligned}\)


\(\begin{aligned} &3\textbf{p}-\textbf{q}+2\textbf{r} \\\\ &=&-15 \utilde{i}+6\utilde{j} \\ &(-)&\quad \utilde{i}+3\utilde{j} \\ \hline &&-15\utilde{i}+3\utilde{j} \\ &(+) &4 \utilde{i}-14\utilde{j}\\ \hline &&-11\utilde{i}-11\utilde{j} \end{aligned}\)

Example 4:

The current of a river is flowing parallel to its bank with a velocity of \(1.25 \text{ km h}^{-1}\).

A swimmer is swimming at \(2.5 \text{ km h}^{-1}\) perpendicularly to the river bank across the river.

Calculate the time taken, in hour, if the width of the river is \(200 \text{ m}\).


Assuming the direction of the water current is along the positive \(x-\)axis and the direction of the swimmer is along the positive \(y-\)axis.

The magnitude of the water current\(=1.25 \text{ km h}^{-1}\)
The magnitude of the swimmer\(=2.5 \text{ km h}^{-1}\)

Let \(v \text{ km h}^{-1}\) represent the resultant velocity of the swimmer.




\(v=1.25 \utilde{i} + 2.5 \utilde{j}\).

The magnitude of the resultant velocity of the swimmer is


\(\begin{aligned} |\utilde{v}|&= \sqrt{1.25^2+2.5^2} \\\\ &=2.795 \text{ km h}^{-1}. \end{aligned}\)


\(\text{Time taken}=\dfrac{\text{Displacement}}{\text{Velocity}}\)


Since the width of the river is \(200 \text{ m}= 0.2 \text{ km}\),



\(\begin{aligned} \text{Time taken}&=\dfrac{0.2}{2.795} \\\\ &=0.07156 \,\text{ hour}. \end{aligned}\)