How would you find the unit vector having the same direction as $ v = - 5i + 12j? $
Answer
607.8k+ views
Hint: A vector can be expressed as the unit vector when it is divided by its vector magnitude. The unit vector is also known as the normalized vector. Here we will first find the magnitude of the given vector. Then find the values for the unit vector and then place the values in the standard formula for the resultant value.
Complete step by step solution:
Take the given expression:
$ v = - 5i + 12j $
Find the magnitude of the above vector.
$ \left| v \right| = \sqrt {{{( - 5)}^2} + {{(12)}^2}} $
Simplify the above equation.
$
\left| v \right| = \sqrt {25 + 144} \\
\left| v \right| = \sqrt {169} \\
$
Find the square root on the right hand side of the equation. When the same term is multiplied with itself then we get the square of that number. The above expression can be re-written as-
$ \left| v \right| = \sqrt {{{13}^2}} $
Square and square root cancel each other on the right hand side of the equation.
$ \left| v \right| = 13 $
Now, the unit vector can be expressed as:
$ \widehat v = \dfrac{{\overline v }}{{\left| {\overline v } \right|}} $
Place the values in the above equation:
$ \widehat v = \dfrac{{ - 5}}{{{\text{ 13}}}}\widehat i + \dfrac{{12}}{{13}}\widehat j $
This is the required solution.
Additional Information:
Be careful while simplifying between square and square-roots. Know the difference between the square and square-roots and apply accordingly. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $ 25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2} $ , generally it is denoted by n to the power two i.e. $ {n^2} $ . Whereas square-root is defined as $ \sqrt n $ , for example $ 25{\text{ = 5 }} \times {\text{ 5, }}\sqrt {{\text{25}}} {\text{ = }}\sqrt {{{\text{5}}^2}} = 5 $ F
Note :
Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative numbers or the positive numbers always give positive value but the magnitude of any vector is always positive.
Complete step by step solution:
Take the given expression:
$ v = - 5i + 12j $
Find the magnitude of the above vector.
$ \left| v \right| = \sqrt {{{( - 5)}^2} + {{(12)}^2}} $
Simplify the above equation.
$
\left| v \right| = \sqrt {25 + 144} \\
\left| v \right| = \sqrt {169} \\
$
Find the square root on the right hand side of the equation. When the same term is multiplied with itself then we get the square of that number. The above expression can be re-written as-
$ \left| v \right| = \sqrt {{{13}^2}} $
Square and square root cancel each other on the right hand side of the equation.
$ \left| v \right| = 13 $
Now, the unit vector can be expressed as:
$ \widehat v = \dfrac{{\overline v }}{{\left| {\overline v } \right|}} $
Place the values in the above equation:
$ \widehat v = \dfrac{{ - 5}}{{{\text{ 13}}}}\widehat i + \dfrac{{12}}{{13}}\widehat j $
This is the required solution.
Additional Information:
Be careful while simplifying between square and square-roots. Know the difference between the square and square-roots and apply accordingly. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $ 25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2} $ , generally it is denoted by n to the power two i.e. $ {n^2} $ . Whereas square-root is defined as $ \sqrt n $ , for example $ 25{\text{ = 5 }} \times {\text{ 5, }}\sqrt {{\text{25}}} {\text{ = }}\sqrt {{{\text{5}}^2}} = 5 $ F
Note :
Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative numbers or the positive numbers always give positive value but the magnitude of any vector is always positive.
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