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Question

Resultant of two vectors $$\vec{A}$$ and $$\vec{B}$$ is of magnitude P. If $$\vec{B}$$ is reversed, then resultant is of magnitude Q. What is the value of P$$^{2}$$ + Q$$^{2}$$ ?


A
2(A2 + B2)
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B
2(A2-B2)
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C
A2-B2
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D
A2 + B2
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Solution

The correct option is B 2(A$$^{2}$$ + B$$^{2}$$)
Let $$\theta$$ be angle between $$\vec{A}$$ and $$\vec{B}$$.$$\therefore$$ Resultant of $$\vec{A}$$ and $$\vec{B}$$ is 
$$P =\sqrt{A^{2}+B^{2}+2ABcos\theta}$$   ---(i)

When $$\vec{B}$$ is reversed, then the angle between $$\vec{A}$$ and - $$\vec{B}$$ is (180$$^{\circ}$$ - $$\theta$$).Resultant of $$\vec{A}$$ and - $$\vec{B}$$ is

Q = $$\sqrt{A^{2}+B^{2}+2ABcos(180^ {\circ} - \theta)}$$

Q = $$\sqrt{A^{2}+B^{2}+2ABcos\theta}$$      --(ii)

Squaring and adding (i) and (ii), we get$$P^{2} + Q^{2} = 2(A^{2} + B^{2})$$

Physics

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