## Vectors in a Cartesian Plane

 8.3 Vectors in a Cartesian Plane

 Vectors in a Cartesian Plane Vectors in a Cartesian plane can be written as $$x\utilde{i}+y\utilde{j}$$. $$\begin{pmatrix} x \\ y \end{pmatrix}$$, where $$\utilde{i}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} ,\utilde{j}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$.  $$\utilde{i}$$ and $$\utilde{j}$$ are vectors of magnitude $$1$$ unit that parallel to $$x$$-axis and $$y$$-axis respectively.  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: $$\overrightarrow{OA}=x_1\utilde{i}+y_1\utilde{j}$$, $$\overrightarrow{OA}$$ is known as a position vector. The magnitude for a vector $$\utilde{r}=x\utilde{i}+y\utilde{j}$$ is : $$|\utilde{r}|=\sqrt{x^2+y^2}$$.  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}  The magnitude of the unit vector in the direction of a vector is $$1$$ unit. Addition: $$(a\utilde{i}+b\utilde{j})+(c\utilde{i}+d\utilde{j})=(a+c)\utilde{i}+(b+d)\utilde{j}$$ Subtraction:  $$(a\utilde{i}+b\utilde{j})-(c\utilde{i}+d\utilde{j})=(a-c)\utilde{i}+(b-d)\utilde{j}$$ Multiplication by scalar: $$k(a\utilde{i}+b\utilde{j})=ka\utilde{i}+kb\utilde{j}$$

Example $$1$$
 Question

The diagram shows a vector $$\utilde{b}$$ drawn on the Cartesian plane.

Find the magnitude of the vector $$\utilde{b}$$.

 Solution

Based on the diagram,

$$\utilde{b}=2\utilde{i}+2\utilde{j}$$,

Hence,

\begin{aligned} |\utilde{b}|&=\sqrt{2^2+2^2} \\\\ &=\sqrt{4+4} \\\\ &=\sqrt{8} \text{ units}. \end{aligned}

Example $$2$$
 Question

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 $$3$$
 Question

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 $$4$$
 Question

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

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

Calculate the time taken, in hour, if the width of the river is $$200$$ 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 $$=1.25$$ km h$$^{-1}$$
The magnitude of the swimmer $$=2.5$$ km h$$^{-1}$$

Let $$v$$ 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}

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}

## Vectors in a Cartesian Plane

 8.3 Vectors in a Cartesian Plane

 Vectors in a Cartesian Plane Vectors in a Cartesian plane can be written as $$x\utilde{i}+y\utilde{j}$$. $$\begin{pmatrix} x \\ y \end{pmatrix}$$, where $$\utilde{i}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} ,\utilde{j}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$.  $$\utilde{i}$$ and $$\utilde{j}$$ are vectors of magnitude $$1$$ unit that parallel to $$x$$-axis and $$y$$-axis respectively.  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: $$\overrightarrow{OA}=x_1\utilde{i}+y_1\utilde{j}$$, $$\overrightarrow{OA}$$ is known as a position vector. The magnitude for a vector $$\utilde{r}=x\utilde{i}+y\utilde{j}$$ is : $$|\utilde{r}|=\sqrt{x^2+y^2}$$.  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}  The magnitude of the unit vector in the direction of a vector is $$1$$ unit. Addition: $$(a\utilde{i}+b\utilde{j})+(c\utilde{i}+d\utilde{j})=(a+c)\utilde{i}+(b+d)\utilde{j}$$ Subtraction:  $$(a\utilde{i}+b\utilde{j})-(c\utilde{i}+d\utilde{j})=(a-c)\utilde{i}+(b-d)\utilde{j}$$ Multiplication by scalar: $$k(a\utilde{i}+b\utilde{j})=ka\utilde{i}+kb\utilde{j}$$

Example $$1$$
 Question

The diagram shows a vector $$\utilde{b}$$ drawn on the Cartesian plane.

Find the magnitude of the vector $$\utilde{b}$$.

 Solution

Based on the diagram,

$$\utilde{b}=2\utilde{i}+2\utilde{j}$$,

Hence,

\begin{aligned} |\utilde{b}|&=\sqrt{2^2+2^2} \\\\ &=\sqrt{4+4} \\\\ &=\sqrt{8} \text{ units}. \end{aligned}

Example $$2$$
 Question

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 $$3$$
 Question

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 $$4$$
 Question

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

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

Calculate the time taken, in hour, if the width of the river is $$200$$ 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 $$=1.25$$ km h$$^{-1}$$
The magnitude of the swimmer $$=2.5$$ km h$$^{-1}$$

Let $$v$$ 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}

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}