Answer
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Hint: Plane motion is a form of motion that occurs in a plane. Motion in two dimensions is also known as motion in a plane. Circular motion, projectile motion, and so on. The reference point for such motion analysis would be made up of an origin and the two coordinate axes X and Y.
Complete step-by-step solution:
A vector is a quantity that has both magnitude and direction in physics. It's normally represented by an arrow that points in the same direction as the quantity and has the same length as the quantity's magnitude. Even though it has magnitude and direction, a vector does not have a position. If a vector is displaced parallel to itself, it is unaffected as long as its length is not increased.
A quantity with magnitude and direction must also follow certain combination rules to qualify as a vector. Vector addition, written symbolically as\[A{\text{ }} + {\text{ }}B{\text{ }} = {\text{ }}C\], is one of them (vectors are conventionally written as boldface letters). The vector number can be visualised geometrically by putting the tail of vector B at the head of vector A and drawing vector C from the tail of A to the head of B, completing the triangle. If A, B, and C are vectors, the same operation must be possible in reverse order to obtain the same result \[\left( C \right),{\text{ }}B{\text{ }} + {\text{ }}A{\text{ }} = {\text{ }}C.\]
Now, coming to the question;
$\overrightarrow C = \overrightarrow A + \overrightarrow B $
$\overrightarrow D = \overrightarrow A - \overrightarrow B $
$\angle \overrightarrow C \overrightarrow D = \theta $
$\angle \overrightarrow A \overrightarrow B = \varphi $
\[R = \sqrt {{C^2} + {D^2} + 2CD\,\cos \theta \,} \]$R = \sqrt {\left( {{A^2} + {B^2} + 2AB\cos \varphi } \right) + \left( {{A^2} + {B^2} - 2AB\cos \varphi } \right) + 2\left( {{A^2} - {B^2}} \right)\cos \theta } $$R = \sqrt {2{A^2} + 2{B^2} + 2\left( {{A^2} - {B^2}} \right)\cos \theta } $
$R = \sqrt {3{A^2} + {B^2}} $(given)
$ \Rightarrow \cos \theta = \dfrac{1}{2}$
$ \Rightarrow \theta = \dfrac{\pi }{3}$
$\therefore $The angle between $\overrightarrow A + \overrightarrow B $ and $\overrightarrow A - \overrightarrow B $ is $\dfrac{\pi }{3}$
Hence the correct option is: (B) $\dfrac{\pi }{3}$.
Note:While vectors are mathematically simple and extremely useful in physics discussions, they were not established in their modern form until the late nineteenth century, when Josiah Willard Gibbs and Oliver Heaviside (of the United States and England, respectively) used vector analysis to help articulate James Clerk Maxwell's new electromagnetism laws.
Complete step-by-step solution:
A vector is a quantity that has both magnitude and direction in physics. It's normally represented by an arrow that points in the same direction as the quantity and has the same length as the quantity's magnitude. Even though it has magnitude and direction, a vector does not have a position. If a vector is displaced parallel to itself, it is unaffected as long as its length is not increased.
A quantity with magnitude and direction must also follow certain combination rules to qualify as a vector. Vector addition, written symbolically as\[A{\text{ }} + {\text{ }}B{\text{ }} = {\text{ }}C\], is one of them (vectors are conventionally written as boldface letters). The vector number can be visualised geometrically by putting the tail of vector B at the head of vector A and drawing vector C from the tail of A to the head of B, completing the triangle. If A, B, and C are vectors, the same operation must be possible in reverse order to obtain the same result \[\left( C \right),{\text{ }}B{\text{ }} + {\text{ }}A{\text{ }} = {\text{ }}C.\]
Now, coming to the question;
$\overrightarrow C = \overrightarrow A + \overrightarrow B $
$\overrightarrow D = \overrightarrow A - \overrightarrow B $
$\angle \overrightarrow C \overrightarrow D = \theta $
$\angle \overrightarrow A \overrightarrow B = \varphi $
\[R = \sqrt {{C^2} + {D^2} + 2CD\,\cos \theta \,} \]$R = \sqrt {\left( {{A^2} + {B^2} + 2AB\cos \varphi } \right) + \left( {{A^2} + {B^2} - 2AB\cos \varphi } \right) + 2\left( {{A^2} - {B^2}} \right)\cos \theta } $$R = \sqrt {2{A^2} + 2{B^2} + 2\left( {{A^2} - {B^2}} \right)\cos \theta } $
$R = \sqrt {3{A^2} + {B^2}} $(given)
$ \Rightarrow \cos \theta = \dfrac{1}{2}$
$ \Rightarrow \theta = \dfrac{\pi }{3}$
$\therefore $The angle between $\overrightarrow A + \overrightarrow B $ and $\overrightarrow A - \overrightarrow B $ is $\dfrac{\pi }{3}$
Hence the correct option is: (B) $\dfrac{\pi }{3}$.
Note:While vectors are mathematically simple and extremely useful in physics discussions, they were not established in their modern form until the late nineteenth century, when Josiah Willard Gibbs and Oliver Heaviside (of the United States and England, respectively) used vector analysis to help articulate James Clerk Maxwell's new electromagnetism laws.
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