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How do you find the component form of $ v $ given its magnitude 1 and the angle it makes with the positive $ x $ -axis is $ \theta = {45^o} $ ?

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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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 $ .