Answer
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Hint: First of all, write the given quantities. For finding the magnitude of the change in the position vector, draw the relevant figure for sake of clear understanding and better understanding. Now, using this figure and given quantities, find out the change in magnitude in the position vector.
Complete step by step solution:
Given: ${{l}}$= length of the vector
And ${{\theta }}$ = angle of inclination
Two vectors represented in both magnitude and direction by two adjacent sides of a parallelogram drawn from a common point, then their resultant is completely represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.
Resultant vector is given by
$\Delta {{\vec r = r}}_2^2 - {{r}}_1^2$
According to the question, ${{{r}}_{{1}}}{{ = }}{{{r}}_{{2}}}{{ = l}}$
Now using parallelogram law of vector addition,
$\Delta {{r = }}\sqrt {{{{r}}_{{1}}}^{{2}}{{ + }}{{{r}}_{{2}}}^{{2}}{{ - 2}}{{{r}}_{{1}}}{{{r}}_{{2}}}{{cos\theta }}} $
Now, on substituting the value of${{{r}}_{{1}}}{{ and }}{{{r}}_{{2}}}$, we get
$\Delta {{r = }}\sqrt {{{{l}}^{{2}}}{{ + }}{{{l}}^{{2}}}{{ - 2}}{{{l}}^{{2}}}{{cos\theta }}} $
On further simplification, we get
$\Rightarrow \Delta {{r = }}\sqrt {{{2}}{{{l}}^{{2}}}{{(1 - cos\theta )}}} \\
\Rightarrow \Delta {{r = l}}\sqrt {{{2(2sin}}\dfrac{{{\theta }}}{{{2}}}{{)}}} \\
\therefore {{ \Delta r = 2lsin}}\left( {\dfrac{{{\theta }}}{{{2}}}} \right) $
Therefore, option (B) is the correct choice.
Note: A vector which gives the position of a point with reference to the origin of the coordinate system, is called position vector. Scalars can be added algebraically while the vectors don't because of the fact that vectors possess both the magnitude as well as direction. While adding two vectors with the help of parallelogram vector addition rule, we have to ensure that the two vectors act either towards a point or away from a point. However, the magnitude of resultant of two vectors is minimum when they act in opposite directions and the magnitude of the resultant of two vectors is maximum when they act in the same direction.
Complete step by step solution:
Given: ${{l}}$= length of the vector
And ${{\theta }}$ = angle of inclination
Two vectors represented in both magnitude and direction by two adjacent sides of a parallelogram drawn from a common point, then their resultant is completely represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.
Resultant vector is given by
$\Delta {{\vec r = r}}_2^2 - {{r}}_1^2$
According to the question, ${{{r}}_{{1}}}{{ = }}{{{r}}_{{2}}}{{ = l}}$
Now using parallelogram law of vector addition,
$\Delta {{r = }}\sqrt {{{{r}}_{{1}}}^{{2}}{{ + }}{{{r}}_{{2}}}^{{2}}{{ - 2}}{{{r}}_{{1}}}{{{r}}_{{2}}}{{cos\theta }}} $
Now, on substituting the value of${{{r}}_{{1}}}{{ and }}{{{r}}_{{2}}}$, we get
$\Delta {{r = }}\sqrt {{{{l}}^{{2}}}{{ + }}{{{l}}^{{2}}}{{ - 2}}{{{l}}^{{2}}}{{cos\theta }}} $
On further simplification, we get
$\Rightarrow \Delta {{r = }}\sqrt {{{2}}{{{l}}^{{2}}}{{(1 - cos\theta )}}} \\
\Rightarrow \Delta {{r = l}}\sqrt {{{2(2sin}}\dfrac{{{\theta }}}{{{2}}}{{)}}} \\
\therefore {{ \Delta r = 2lsin}}\left( {\dfrac{{{\theta }}}{{{2}}}} \right) $
Therefore, option (B) is the correct choice.
Note: A vector which gives the position of a point with reference to the origin of the coordinate system, is called position vector. Scalars can be added algebraically while the vectors don't because of the fact that vectors possess both the magnitude as well as direction. While adding two vectors with the help of parallelogram vector addition rule, we have to ensure that the two vectors act either towards a point or away from a point. However, the magnitude of resultant of two vectors is minimum when they act in opposite directions and the magnitude of the resultant of two vectors is maximum when they act in the same direction.
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