
If a be a constant vector and b be a vector of constant magnitude such that $∣\vec {b}∣$=$\dfrac{a}{2}$ and $∣\vec {a}∣ \neq 0$. then maximum value of angle between $\vec a$ and $\vec a + \vec b$ is:
A) $\dfrac{{5\pi }}{6}$
B) $\dfrac{{5\pi }}{3}$
C) $\dfrac{{2\pi }}{3}$
D) $\dfrac{\pi }{6}$
Answer
148.5k+ 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
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Uniform Acceleration

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Electrical Field of Charged Spherical Shell - JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
