For some non-zero vector \[\overrightarrow V \], if the sum of \[\overrightarrow V \] and the vector obtained from \[\overrightarrow V \] by rotating it by an angle \[2\alpha \] equals to the vector obtained from \[\overrightarrow V \] by rotating it by a then the value of \[\alpha \], is where \[n\] is a integer.
A.\[2n\pi = \dfrac{\pi }{3}\]
B.\[n\pi = \dfrac{\pi }{3}\]
C.\[2n\pi = \dfrac{{2\pi }}{3}\]
D.\[n\pi = \dfrac{{2\pi }}{3}\]
Answer
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Hint: In this problem, A nonzero vector is a vector with magnitude not equal to zero. The vector obtained by rotating the vector \[\overrightarrow V \] by an angle \[\alpha \] in the anticlockwise direction is defined by trigonometric formula. A vector is an object that has both a magnitude and a direction.
Complete step-by-step answer:
First, we have to sketch a diagram for some non-zero vector \[\overrightarrow V \], if the sum of \[\overrightarrow V \] and the vector obtained from \[\overrightarrow V \] by rotating it by an angle \[2\alpha \] equals to the vector obtained from \[\overrightarrow V \] by rotating it by a then the value of \[\alpha \], is where \[n\] is a integer.
According to the condition in this problem, the vector obtained by rotating the vector \[\overrightarrow V \] by an angle\[\alpha \] in the anticlockwise direction, then
\[vi - v\cos (\pi - 2\alpha )\hat i + v\sin (\pi - 2\alpha )\hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
We use the trigonometric identity $\cos (\pi - \theta ) = - \cos \theta $ and $\sin (\pi - \theta ) = \sin \theta $, we can get
Since, $\cos (\pi - 2\alpha ) = - \cos 2\alpha $ and $\sin (\pi - 2\alpha ) = \sin 2\alpha $
\[v\hat i - v( - cos2\alpha \hat i) + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\],
\[v\hat i + vcos2\alpha \hat i + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
Put common factor ‘v’ from bracket, we have
\[v(1 + cos2\alpha )\hat i + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
On separating i and j part on both sides of the equation, then
\[v(1 + \cos 2\alpha ) = v\cos \alpha \] and $v\sin 2\alpha = v\sin \alpha $
By dividing on both sides by ‘v’, we get
\[(1 + \cos 2\alpha ) = \cos \alpha \] and $\sin 2\alpha = \sin \alpha $
Here, by using this trigonometric identity , $\cos 2\alpha = 2{\cos ^2}\alpha - 1$ and $2\sin \alpha \cos \alpha = \sin \alpha $
\[1 + 2{\cos ^2}\alpha - 1 = \cos \alpha \] and $2\sin \alpha \cos \alpha = \sin \alpha $
On further simplification, we get
\[2{\cos ^2}\alpha = \cos \alpha \] and $2\cos \alpha = 1$
\[2{\cos ^2}\alpha - \cos \alpha = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Take out \[\cos \alpha \] commonly on the first function, then
\[\cos \alpha (2\cos \alpha - 1) = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Separate the factors to find the value of , then
\[\cos \alpha = 0\], \[(2\cos \alpha - 1) = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Expanding the equation from LHS to RHS, then
\[\cos \alpha = 0\], \[2\cos \alpha = 1 \Rightarrow \cos \alpha = \dfrac{1}{2}\] and \[\cos \alpha = \dfrac{1}{2}\]
\[\cos \alpha = 0\],\[\cos \alpha = \dfrac{1}{2}\] and $\cos \alpha = \dfrac{1}{2}$
We know that, the formula $\cos \dfrac{\pi }{2} = 0 \Rightarrow \alpha = 2n\pi \pm \dfrac{\pi }{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2} \Rightarrow \alpha = 2n\pi \pm \dfrac{\pi }{3}$,
$\alpha = 2n\pi \pm \dfrac{\pi }{2}$, $\alpha = 2n\pi \pm \dfrac{\pi }{3}$ and $\alpha = 2n\pi \pm \dfrac{\pi }{3}$
Therefore, $\alpha = 2n\pi \pm \dfrac{\pi }{3}$ satisfies both equations.
Hence, The final answer is option(A) \[2n\pi = \dfrac{\pi }{3}\]
So, the correct answer is “Option A”.
Note: Geometrically, we can define a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.
Complete step-by-step answer:
First, we have to sketch a diagram for some non-zero vector \[\overrightarrow V \], if the sum of \[\overrightarrow V \] and the vector obtained from \[\overrightarrow V \] by rotating it by an angle \[2\alpha \] equals to the vector obtained from \[\overrightarrow V \] by rotating it by a then the value of \[\alpha \], is where \[n\] is a integer.
According to the condition in this problem, the vector obtained by rotating the vector \[\overrightarrow V \] by an angle\[\alpha \] in the anticlockwise direction, then
\[vi - v\cos (\pi - 2\alpha )\hat i + v\sin (\pi - 2\alpha )\hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
We use the trigonometric identity $\cos (\pi - \theta ) = - \cos \theta $ and $\sin (\pi - \theta ) = \sin \theta $, we can get
Since, $\cos (\pi - 2\alpha ) = - \cos 2\alpha $ and $\sin (\pi - 2\alpha ) = \sin 2\alpha $
\[v\hat i - v( - cos2\alpha \hat i) + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\],
\[v\hat i + vcos2\alpha \hat i + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
Put common factor ‘v’ from bracket, we have
\[v(1 + cos2\alpha )\hat i + v\sin 2\alpha \hat j = v\cos \alpha \hat i + v\sin \alpha \hat j\]
On separating i and j part on both sides of the equation, then
\[v(1 + \cos 2\alpha ) = v\cos \alpha \] and $v\sin 2\alpha = v\sin \alpha $
By dividing on both sides by ‘v’, we get
\[(1 + \cos 2\alpha ) = \cos \alpha \] and $\sin 2\alpha = \sin \alpha $
Here, by using this trigonometric identity , $\cos 2\alpha = 2{\cos ^2}\alpha - 1$ and $2\sin \alpha \cos \alpha = \sin \alpha $
\[1 + 2{\cos ^2}\alpha - 1 = \cos \alpha \] and $2\sin \alpha \cos \alpha = \sin \alpha $
On further simplification, we get
\[2{\cos ^2}\alpha = \cos \alpha \] and $2\cos \alpha = 1$
\[2{\cos ^2}\alpha - \cos \alpha = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Take out \[\cos \alpha \] commonly on the first function, then
\[\cos \alpha (2\cos \alpha - 1) = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Separate the factors to find the value of , then
\[\cos \alpha = 0\], \[(2\cos \alpha - 1) = 0\] and \[\cos \alpha = \dfrac{1}{2}\]
Expanding the equation from LHS to RHS, then
\[\cos \alpha = 0\], \[2\cos \alpha = 1 \Rightarrow \cos \alpha = \dfrac{1}{2}\] and \[\cos \alpha = \dfrac{1}{2}\]
\[\cos \alpha = 0\],\[\cos \alpha = \dfrac{1}{2}\] and $\cos \alpha = \dfrac{1}{2}$
We know that, the formula $\cos \dfrac{\pi }{2} = 0 \Rightarrow \alpha = 2n\pi \pm \dfrac{\pi }{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2} \Rightarrow \alpha = 2n\pi \pm \dfrac{\pi }{3}$,
$\alpha = 2n\pi \pm \dfrac{\pi }{2}$, $\alpha = 2n\pi \pm \dfrac{\pi }{3}$ and $\alpha = 2n\pi \pm \dfrac{\pi }{3}$
Therefore, $\alpha = 2n\pi \pm \dfrac{\pi }{3}$ satisfies both equations.
Hence, The final answer is option(A) \[2n\pi = \dfrac{\pi }{3}\]
So, the correct answer is “Option A”.
Note: Geometrically, we can define a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.
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