Answer
77.7k+ views
Hint: A vector is represented by a line with an arrowhead. The point where the arrow starts is called the origin of the vector and the point where the arrow ends is called the tip. A vector remains unchanged when it is displaced parallel to itself.
Complete step by step solution:
Consider the vector b makes an angle $\theta $ with the negative side of vector a. Then the resultant component in the x-direction is given by
$(a + b)_x$ $ = a - b\cos \theta $
$ \Rightarrow $ $(a + b)_x$ $ = a - \dfrac{a}{2}\cos \theta $ $\left( {\because b = \dfrac{a}{2}} \right)$
$ \Rightarrow $ $(a + b)_x $ = $\dfrac{a}{2}(2 - \cos \theta )$
Similarly, the resultant vector of the y component is given by
$(a + b)_y $ = $b\sin \theta $
$ \Rightarrow $ $(a + b)_y $ = $\dfrac{a}{2}\sin \theta $\[\]$\left( {\because b = \dfrac{a}{2}} \right)$
We know that according to the parallelogram law of vector addition, the angle between the resultant with the first vector is given by,
$\tan \alpha = \left( {\dfrac{{\dfrac{a}{2}\sin \theta }}{{\dfrac{a}{2}\left( {2 + \cos \theta } \right)}}} \right) = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
If α has to be maximum, then $\dfrac{{d\alpha }}{{d\theta }}$ must be equal to zero, thus we have $\dfrac{d}{{d\theta }}\left( {\tan \alpha } \right) = \dfrac{d}{{d\theta }}\left( {\dfrac{{\sin \theta }}{{2 + \cos \theta }}} \right)$
\[ \Rightarrow se{c^2}\alpha \left( {\dfrac{{d\alpha }}{{d\theta }}} \right) = \dfrac{{\sin \theta \left( { - \sin \theta } \right) - \left( {2 + \cos \theta } \right)\left( {\cos \theta } \right)}}{{{{(2 + cos\theta )}^2}}}\]
$
\Rightarrow - si{n^2}\theta - 2\cos \theta - co{s^2}\theta = 0 \\
\Rightarrow 2\cos \theta = - 1 \\
\Rightarrow \theta = \dfrac{{2\pi }}{3} \\
$
Thus for the above value $\theta $, the angle between the resultant and vector a is maximum.
$\therefore $The maximum value of the angle \[\alpha \]is,
$\tan \alpha = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
Substituting the value of $\theta $, we get
\[ \Rightarrow \tan \alpha = \dfrac{{\sin \left( {\dfrac{{2\pi }}{3}} \right)}}{{(2 + \cos \left( {\dfrac{{2\pi }}{3}} \right))}}\] \[(\because \theta = \dfrac{{2\pi }}{3})\]
$ \Rightarrow \tan \alpha = \dfrac{{\;\sqrt {\dfrac{3}{2}} \;}}{{(2 - \dfrac{1}{2})}}$
$ \Rightarrow \tan \alpha = \dfrac{1}{{\sqrt 3 }}$
\[ \Rightarrow \alpha = ta{n^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
\[ \Rightarrow \alpha = \dfrac{\pi }{6}\]
$\therefore $ The maximum value of the angle, \[\alpha = \dfrac{\pi }{6}\]\[\;\;\]
Hence the correct option is D.
Note: 1) When a vector is multiplied by a scalar quantity then we get a new vector in the same direction but having a new magnitude. The vector product is distributive and the addition is commutative. The vector has a definite direction. Vectors obey the law of parallelograms of addition.
2) Zero vector is the one that has zero magnitudes and also no specific direction.
3) If two vectors have equal magnitude and are in the same direction, then they are called equal vectors.
Complete step by step solution:
Consider the vector b makes an angle $\theta $ with the negative side of vector a. Then the resultant component in the x-direction is given by
$(a + b)_x$ $ = a - b\cos \theta $
$ \Rightarrow $ $(a + b)_x$ $ = a - \dfrac{a}{2}\cos \theta $ $\left( {\because b = \dfrac{a}{2}} \right)$
$ \Rightarrow $ $(a + b)_x $ = $\dfrac{a}{2}(2 - \cos \theta )$
Similarly, the resultant vector of the y component is given by
$(a + b)_y $ = $b\sin \theta $
$ \Rightarrow $ $(a + b)_y $ = $\dfrac{a}{2}\sin \theta $\[\]$\left( {\because b = \dfrac{a}{2}} \right)$
We know that according to the parallelogram law of vector addition, the angle between the resultant with the first vector is given by,
$\tan \alpha = \left( {\dfrac{{\dfrac{a}{2}\sin \theta }}{{\dfrac{a}{2}\left( {2 + \cos \theta } \right)}}} \right) = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
If α has to be maximum, then $\dfrac{{d\alpha }}{{d\theta }}$ must be equal to zero, thus we have $\dfrac{d}{{d\theta }}\left( {\tan \alpha } \right) = \dfrac{d}{{d\theta }}\left( {\dfrac{{\sin \theta }}{{2 + \cos \theta }}} \right)$
\[ \Rightarrow se{c^2}\alpha \left( {\dfrac{{d\alpha }}{{d\theta }}} \right) = \dfrac{{\sin \theta \left( { - \sin \theta } \right) - \left( {2 + \cos \theta } \right)\left( {\cos \theta } \right)}}{{{{(2 + cos\theta )}^2}}}\]
$
\Rightarrow - si{n^2}\theta - 2\cos \theta - co{s^2}\theta = 0 \\
\Rightarrow 2\cos \theta = - 1 \\
\Rightarrow \theta = \dfrac{{2\pi }}{3} \\
$
Thus for the above value $\theta $, the angle between the resultant and vector a is maximum.
$\therefore $The maximum value of the angle \[\alpha \]is,
$\tan \alpha = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
Substituting the value of $\theta $, we get
\[ \Rightarrow \tan \alpha = \dfrac{{\sin \left( {\dfrac{{2\pi }}{3}} \right)}}{{(2 + \cos \left( {\dfrac{{2\pi }}{3}} \right))}}\] \[(\because \theta = \dfrac{{2\pi }}{3})\]
$ \Rightarrow \tan \alpha = \dfrac{{\;\sqrt {\dfrac{3}{2}} \;}}{{(2 - \dfrac{1}{2})}}$
$ \Rightarrow \tan \alpha = \dfrac{1}{{\sqrt 3 }}$
\[ \Rightarrow \alpha = ta{n^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
\[ \Rightarrow \alpha = \dfrac{\pi }{6}\]
$\therefore $ The maximum value of the angle, \[\alpha = \dfrac{\pi }{6}\]\[\;\;\]
Hence the correct option is D.
Note: 1) When a vector is multiplied by a scalar quantity then we get a new vector in the same direction but having a new magnitude. The vector product is distributive and the addition is commutative. The vector has a definite direction. Vectors obey the law of parallelograms of addition.
2) Zero vector is the one that has zero magnitudes and also no specific direction.
3) If two vectors have equal magnitude and are in the same direction, then they are called equal vectors.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)