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**Hint:**In this question we are asked to find the component form of the vector which is given in form $ v = \left| v \right|\cos \theta i + \left| v \right|\sin \theta j $ , where $ \left| v \right| $ is the magnitude and $ \theta $ is the angle made with the $ x $ -axis, now substituting the values given in the component form, and using the values of trigonometric ratios of cos and sin we will get the required component form.

**Complete step by step answer:**

The component form of the vector $ v $ is given by $ v = \left| v \right|\cos \theta i + \left| v \right|\sin \theta j $ , where $ \left| v \right| $ is the where $ \left| v \right| $ is the magnitude and\[\theta \]is the angle made with the $ x $ -axis.

Now in the given question, the data given is, $ \theta = {45^o} $ , and the magnitude i..e, $ \left| v \right| $ is equal to 1.

Now substituting the given values in the component form we get,

$ v = \left| v \right|\cos \theta i + \left| v \right|\sin \theta j $ ,

By substituting the values we get,

$ \Rightarrow v = \left( 1 \right)\cos \left( {{{45}^0}} \right)i + \left( 1 \right)\sin \left( {{{45}^0}} \right)j $ ,

Now from the trigonometric ratios we know that, $ \cos {45^0} = \dfrac{1}{{\sqrt 2 }} $ and $ \sin {45^0} = \dfrac{1}{{\sqrt 2 }} $ ,

Now substituting the cos and sin values we get,

\[ \Rightarrow v = \left( 1 \right)\dfrac{1}{{\sqrt 2 }}i + \left( 1 \right)\dfrac{1}{{\sqrt 2 }}j\],

Now simplifying we get,

$ \Rightarrow v = \dfrac{1}{{\sqrt 2 }}i + \dfrac{1}{{\sqrt 2 }}j $ ,

So, the component form is equal to $ v = \dfrac{1}{{\sqrt 2 }}i + \dfrac{1}{{\sqrt 2 }}j $ .

This can be represented using the graph,

The component form can also be written in the form $ \left( {\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right) $ .

$ \therefore $ The component form of the vector magnitude 1 and the angle it makes with the positive $ x $ -axis is $ \theta = {45^o} $ is equal to $ v = \dfrac{1}{{\sqrt 2 }}i + \dfrac{1}{{\sqrt 2 }}j $ , the component form can also be written as,\[\left( {\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right)\].

**Note:**For resolving a vector into its components, we can use the following formulas:

Consider $ a $ to be the magnitude of the vector $ \overrightarrow a $ and $ \theta $ to be the angle that is formed by the vector along the $ x $ -axis or to be the direction of the given vector. Then we will get,

\[{a_x} = \left| a \right| \times \cos \theta \]and $ {a_y} = \left| a \right| \times \sin \theta $ ,

Where $ {a_x} $ is called the magnitude of the $ x $ -component of the given vector $ \overrightarrow a $ ,

Where $ {a_y} $ is called the magnitude of the $ y $ -component of the given vector $ \overrightarrow a $ .

And the vector form is written as, $ \overrightarrow a = \left( {{a_x},{a_y}} \right) = \left| a \right|\cos \theta i + \left| a \right|\sin \theta j $ .

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