\(\utilde{i}\) 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: \(k(a\utilde{i}+b\utilde{j})=ka\utilde{i}+kb\utilde{j}\)
The diagram shows a vector \(\utilde{b}\) drawn on the Cartesian plane.
Find the magnitude of the vector \(\utilde{b}\).
Based on the diagram,
\(\utilde{b}=2\utilde{i}+2\utilde{j}\),
Hence,
\(\begin{aligned} |\utilde{b}|&=\sqrt{2^2+2^2} \\\\ &=\sqrt{4+4} \\\\ &=\sqrt{8} \text{ units}. \end{aligned}\)
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 \(\overrightarrow{AD}\):
\(\begin{aligned} &=\dfrac{\overrightarrow{AD}}{\left|\overrightarrow{AD}\right|} \\\\&=\dfrac{3\utilde{i}+4\utilde{j}}{\sqrt{3^2+4^2}} \\\\ &=\dfrac{3\utilde{i}+4\utilde{j}}{\sqrt{25}} \\\\ &=\dfrac{3\utilde{i}+4\utilde{j}}{5} \\\\ &=\dfrac{3}{5}\utilde{i} +\dfrac{4}{5}\utilde{j}. \end{aligned}\)
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,
\(\begin{aligned} \textbf{p}&=-5 \utilde{i}+2\utilde{j} \\\\ 3\textbf{p}&=3(-5 \utilde{i}+2\utilde{j}) \\\\ &=-15 \utilde{i}+6\utilde{j}. \end{aligned}\)
\(\begin{aligned} \textbf{r}&=2 \utilde{i}-7\utilde{j} \\\\ 2\textbf{r}&=2(2 \utilde{i}-7\utilde{j}) \\\\ &=4\utilde{i}-14\utilde{j}. \end{aligned}\)
\(\begin{aligned} &3\textbf{p}-\textbf{q}+2\textbf{r} \\\\ &=&-15 \utilde{i}+6\utilde{j} \\ &(-)&\quad \utilde{i}+3\utilde{j} \\ \hline &&-15\utilde{i}+3\utilde{j} \\ &(+) &4 \utilde{i}-14\utilde{j}\\ \hline &&-11\utilde{i}-11\utilde{j} \end{aligned}\)
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.
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,
\(\begin{aligned} \text{Time taken}&=\dfrac{0.2}{2.795} \\\\ &=0.07156 \,\text{ hour}. \end{aligned}\)
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