8.3	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

\(|r|=\sqrt{x^2+y^2}\).

\(\blacksquare\) 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

\((a\utilde{i}+b\utilde{j})+(c\utilde{i}+d\utilde{j})=(a+c)\utilde{i}+(b+d)\utilde{j}\)

\(\blacksquare\) Subtraction

\((a\utilde{i}+b\utilde{j})-(c\utilde{i}+d\utilde{j})=(a-c)\utilde{i}+(b-d)\utilde{j}\)

\(\blacksquare\) Multiplication by scalar

\(k(a\utilde{i}+b\utilde{j})=ka\utilde{i}+kb\utilde{j}\)

Example 1:

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

Solution:

Based on the diagram,

\(\utilde{q}=2\utilde{i}+\utilde{j}\),

hence,

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

Example 2:

Determine unit vector for the following:

\(\overrightarrow{AD}=3 \utilde{i}+4 \utilde{j}\)

Solution:

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

Solution:

Given that

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

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

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

Hence,

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

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

Hence,

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

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 \(200 \text{ m}\).

Solution:

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.

Hence, \(v=1.25 \utilde{i} + 2.5 \utilde{j}\). 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 \(200 \text{ m}= 0.2 \text{ km}\), hence,

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