
Find a unit vector in the direction of the given vector of \[v = (5, - 12)\]?
Answer
548.4k+ views
Hint: In the given question, we are given coordinates of a point and we are asked to find a unit vector in the direction of the point. So, we first find the position vector of the coordinates given to us. Then, we find the unit vector in the direction of the given vector by dividing the vector by the magnitude of the same vector.
Complete step by step solution:
The position vector of the given coordinates v is $ \vec v = 5\hat i - 12\hat j $ . Hence, we now have to find the unit vector in the direction of the above vector.
So, A unit vector in the direction of a given vector, say $ \vec a $ can be found as $ \hat a = \dfrac{{\vec a}}{{\left| a \right|}} $ .
Now, Let $ \vec a = x\hat i + y\hat j + z\hat k $ .
Then, $ \left| a \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
So, we get the unit vector in the direction of the vector $ \vec a = x\hat i + y\hat j + z\hat k $ by dividing the vector by it’s magnitude as: $ \hat a = \left( {\dfrac{{x\hat i + y\hat j + z\hat k}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) $ .
For the given vector, $ \vec v = 5\hat i - 12\hat j $ , we get the magnitude as \[\left| {\vec v} \right| = \sqrt {{{(5)}^2} + {{( - 12)}^2}} \].
Hence, \[\left| {\vec v} \right| = \sqrt {25 + 144} \]
\[ \Rightarrow \left| {\vec v} \right| = \sqrt {169} \]
\[ \Rightarrow \left| {\vec v} \right| = 13\]
So, $ \hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {{{(5)}^2} + {{( - 12)}^2}} }} $
$ \Rightarrow \hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {169} }} $
$ \Rightarrow \hat v = \left( {\dfrac{{5\hat i - 12\hat j}}{{13}}} \right) $
$ \Rightarrow \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $
Hence, the unit vector in the direction of the given vector of \[v = (5, - 12)\] is $ \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $ .
So, the correct answer is “ $ \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $ ”.
Note: The position vector is the vector joining the origin to the point given to us. The position vector is obtained by taking into consideration the definition of the position vector and the coordinates of the point given to us. The magnitude of the vector represents the length of the vector and is the real number.
Complete step by step solution:
The position vector of the given coordinates v is $ \vec v = 5\hat i - 12\hat j $ . Hence, we now have to find the unit vector in the direction of the above vector.
So, A unit vector in the direction of a given vector, say $ \vec a $ can be found as $ \hat a = \dfrac{{\vec a}}{{\left| a \right|}} $ .
Now, Let $ \vec a = x\hat i + y\hat j + z\hat k $ .
Then, $ \left| a \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
So, we get the unit vector in the direction of the vector $ \vec a = x\hat i + y\hat j + z\hat k $ by dividing the vector by it’s magnitude as: $ \hat a = \left( {\dfrac{{x\hat i + y\hat j + z\hat k}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) $ .
For the given vector, $ \vec v = 5\hat i - 12\hat j $ , we get the magnitude as \[\left| {\vec v} \right| = \sqrt {{{(5)}^2} + {{( - 12)}^2}} \].
Hence, \[\left| {\vec v} \right| = \sqrt {25 + 144} \]
\[ \Rightarrow \left| {\vec v} \right| = \sqrt {169} \]
\[ \Rightarrow \left| {\vec v} \right| = 13\]
So, $ \hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {{{(5)}^2} + {{( - 12)}^2}} }} $
$ \Rightarrow \hat v = \dfrac{{5\hat i - 12\hat j}}{{\sqrt {169} }} $
$ \Rightarrow \hat v = \left( {\dfrac{{5\hat i - 12\hat j}}{{13}}} \right) $
$ \Rightarrow \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $
Hence, the unit vector in the direction of the given vector of \[v = (5, - 12)\] is $ \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $ .
So, the correct answer is “ $ \hat v = \left( {\dfrac{5}{{13}}\hat i - \dfrac{{12}}{{13}}\hat j} \right) $ ”.
Note: The position vector is the vector joining the origin to the point given to us. The position vector is obtained by taking into consideration the definition of the position vector and the coordinates of the point given to us. The magnitude of the vector represents the length of the vector and is the real number.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

