
Assertion:If $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$, then the angle between $\vec{A}$ and $\vec{B}$ is $90{}^\circ $.
Reason: $\vec{A}+\vec{B}=\vec{B}+\vec{A}$
A. Both assertion and reason are correct and reason is the correct explanation for assertion
B. Both assertion and reason are correct but reason is not the correct explanation for assertion.
C. Assertion is correct but reason is incorrect.
D. Both assertion and reason are incorrect.
Answer
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Hint: We are given an assertion and by following the assertion we are given the reason for that assertion. We are asked to find out whether the assertion and reason are correct and incorrect. Given $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. We first show that the angle between the two is $90{}^\circ $. Then we find out whether the reason is the correct explanation of assertion or not and choose the option which satisfies our answer.
Formula used:
$|\vec{A}+\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$
$|\vec{A}-\vec{B}|= \sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos (\pi -\theta )}$
Here, $A$ and $B$ are vectors and $\theta$ is the angle between them.
Complete step by step solution:
Given $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
We take square root on both sides, we get
$\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos (\pi -\theta )}$
As the square root cancel from both sides, we get
${{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}+2AB(-\cos \theta )$
We know $-(\cos \theta )=-\cos \theta $
${{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}-2AB\cos \theta $
By solving the above equation, we get
$4AB\cos \theta =0$
Hence, $\cos \theta =0$
We know $\cos 90{}^\circ =0$
Then $\cos \theta =\cos 90{}^\circ $
Which gives $\theta =90{}^\circ $
Also, we know in the properties of vector addition that vector addition is commutative.
Then $\vec{A}+\vec{B}=\vec{B}+\vec{A}$
Hence the assertion is correct but reason is not the correct explanation of the assertion.
Thus, option B is correct.
Note: In these types of questions, remember that while adding two vectors, do not only consider the magnitude of vectors but also consider the direction of the vectors. Students must know the basic trigonometric functions for finding out the correct option.
Formula used:
$|\vec{A}+\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$
$|\vec{A}-\vec{B}|= \sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos (\pi -\theta )}$
Here, $A$ and $B$ are vectors and $\theta$ is the angle between them.
Complete step by step solution:
Given $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
We take square root on both sides, we get
$\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos (\pi -\theta )}$
As the square root cancel from both sides, we get
${{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}+2AB(-\cos \theta )$
We know $-(\cos \theta )=-\cos \theta $
${{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}-2AB\cos \theta $
By solving the above equation, we get
$4AB\cos \theta =0$
Hence, $\cos \theta =0$
We know $\cos 90{}^\circ =0$
Then $\cos \theta =\cos 90{}^\circ $
Which gives $\theta =90{}^\circ $
Also, we know in the properties of vector addition that vector addition is commutative.
Then $\vec{A}+\vec{B}=\vec{B}+\vec{A}$
Hence the assertion is correct but reason is not the correct explanation of the assertion.
Thus, option B is correct.
Note: In these types of questions, remember that while adding two vectors, do not only consider the magnitude of vectors but also consider the direction of the vectors. Students must know the basic trigonometric functions for finding out the correct option.
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