Answer
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Hint: In this particular problem, we need to find the unit vector for the given vector. Vector is that quantity which has both magnitude and direction. Unit vector is that vector whose unit is equal to 1. You should know that unit Vector is represented by the symbol ‘^’, which is called a cap or hat, such as: $\hat{a}$. It is given by:
$\Rightarrow \hat{a}=\dfrac{a}{|a|}$
Complete step by step answer:
Now, let’s solve the question.
Let us discuss vector algebra. Vector is nothing but a quantity which has both magnitude and direction. Vector having a magnitude of 1 is known as a unit vector. You should also know that unit Vector is represented by the symbol ‘^’, which is called a cap or hat, such as: $\hat{a}$. It is given by:
$\Rightarrow \hat{a}=\dfrac{\overrightarrow{a}}{|a|}$
If we wish to change any vector in the unit vector, divide it by the vector’s magnitude. Magnitude of a vector is represented by $\overrightarrow{|a|}$.
From this figure,
Formula to find vector’s magnitude is:
$\Rightarrow \overrightarrow{|a|}=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
Where x, y and z are the magnitude along the x, y and z axis.
Unit vectors can be given by $\dfrac{vector}{magnitude}$.
Formula is given by:
$\Rightarrow \hat{a}=\dfrac{\overrightarrow{a}}{|a|}$
$\Rightarrow \hat{a}=\dfrac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Along the x, y and z axis and i, j and k are the directions of the vector.
So for \[\overrightarrow{a}\text{ }=\text{ }\overrightarrow{2i}\text{ }+\text{ }\overrightarrow{3j}\]
$\Rightarrow \hat{a}=\dfrac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Here x = 2, y = 3, z = 0. Place all the values in the formula, we will get:
$\Rightarrow \hat{a}=\dfrac{2\hat{i}+3\hat{j}}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 0 \right)}^{2}}}}$
Solve the under root, we get:
$\begin{align}
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{4+9}}\left( 2\hat{i}+3\hat{j} \right) \\
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{13}}\left( 2\hat{i}+3\hat{j} \right) \\
\end{align}$
We cannot leave root value in denominator, so we will rationalise the denominator:
$\begin{align}
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{13}}\times \dfrac{\sqrt{13}}{\sqrt{13}}\left( 2\hat{i}+3\hat{j} \right) \\
& \therefore \hat{a}=\dfrac{\sqrt{13}}{13}\left( 2\hat{i}+3\hat{j} \right) \\
\end{align}$
So this is the unit vector for \[\overrightarrow{a}\text{ }=\text{ }\overrightarrow{2i}\text{ }+\text{ }\overrightarrow{3j}\].
Note:
To rationalise the denominator, we will multiply the numerator and denominator with the given root value of the denominator. Secondly you should know that the vector gives the direction which is represented by the arrow $\overrightarrow{{}}$ symbol. You should also know the values of the square roots of different numbers.
$\Rightarrow \hat{a}=\dfrac{a}{|a|}$
Complete step by step answer:
Now, let’s solve the question.
Let us discuss vector algebra. Vector is nothing but a quantity which has both magnitude and direction. Vector having a magnitude of 1 is known as a unit vector. You should also know that unit Vector is represented by the symbol ‘^’, which is called a cap or hat, such as: $\hat{a}$. It is given by:
$\Rightarrow \hat{a}=\dfrac{\overrightarrow{a}}{|a|}$
If we wish to change any vector in the unit vector, divide it by the vector’s magnitude. Magnitude of a vector is represented by $\overrightarrow{|a|}$.
From this figure,
Formula to find vector’s magnitude is:
$\Rightarrow \overrightarrow{|a|}=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
Where x, y and z are the magnitude along the x, y and z axis.
Unit vectors can be given by $\dfrac{vector}{magnitude}$.
Formula is given by:
$\Rightarrow \hat{a}=\dfrac{\overrightarrow{a}}{|a|}$
$\Rightarrow \hat{a}=\dfrac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Along the x, y and z axis and i, j and k are the directions of the vector.
So for \[\overrightarrow{a}\text{ }=\text{ }\overrightarrow{2i}\text{ }+\text{ }\overrightarrow{3j}\]
$\Rightarrow \hat{a}=\dfrac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Here x = 2, y = 3, z = 0. Place all the values in the formula, we will get:
$\Rightarrow \hat{a}=\dfrac{2\hat{i}+3\hat{j}}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 0 \right)}^{2}}}}$
Solve the under root, we get:
$\begin{align}
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{4+9}}\left( 2\hat{i}+3\hat{j} \right) \\
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{13}}\left( 2\hat{i}+3\hat{j} \right) \\
\end{align}$
We cannot leave root value in denominator, so we will rationalise the denominator:
$\begin{align}
& \Rightarrow \hat{a}=\dfrac{1}{\sqrt{13}}\times \dfrac{\sqrt{13}}{\sqrt{13}}\left( 2\hat{i}+3\hat{j} \right) \\
& \therefore \hat{a}=\dfrac{\sqrt{13}}{13}\left( 2\hat{i}+3\hat{j} \right) \\
\end{align}$
So this is the unit vector for \[\overrightarrow{a}\text{ }=\text{ }\overrightarrow{2i}\text{ }+\text{ }\overrightarrow{3j}\].
Note:
To rationalise the denominator, we will multiply the numerator and denominator with the given root value of the denominator. Secondly you should know that the vector gives the direction which is represented by the arrow $\overrightarrow{{}}$ symbol. You should also know the values of the square roots of different numbers.
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