The answer is (c) |\vec{C}|>|\vec{A}-\vec{B}|. The magnitude of the resultant vector \vec{C} is greater than the magnitude of the difference vector \vec{A}-\vec{B}. This can be seen by using the law of cosines to calculate the magnitude of \vec{C}: |\vec{C}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}|\cos\theta} where \theta is the angle between \vec{A} and \vec{B}. In this case, \theta=120^\circ, so \cos\theta=-\frac{1}{2}. Substituting this into the equation above gives: |\vec{C}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2-|\vec{A}||\vec{B}|} The magnitude of \vec{A}-\vec{B} can be calculated similarly: |\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}||\vec{B}|\cos\theta} Substituting \theta=120^\circ into this equation gives: |\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+|\vec{A}||\vec{B}|} Comparing the two magnitudes, we see that |\vec{C}|>|\vec{A}-\vec{B}|.
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The answer is (c) |\vec{C}|>|\vec{A}-\vec{B}|.
The magnitude of the resultant vector \vec{C} is greater than the magnitude of the difference vector \vec{A}-\vec{B}.
This can be seen by using the law of cosines to calculate the magnitude of \vec{C}:
|\vec{C}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}|\cos\theta}
where \theta is the angle between \vec{A} and \vec{B}.
In this case, \theta=120^\circ, so \cos\theta=-\frac{1}{2}. Substituting this into the equation above gives:
|\vec{C}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2-|\vec{A}||\vec{B}|}
The magnitude of \vec{A}-\vec{B} can be calculated similarly:
|\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}||\vec{B}|\cos\theta}
Substituting \theta=120^\circ into this equation gives:
|\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+|\vec{A}||\vec{B}|}
Comparing the two magnitudes, we see that |\vec{C}|>|\vec{A}-\vec{B}|.
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Sir physics or english ki to tantion khatam ho gyi 😊😊
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