Sir physics or english ki to tantion khatam ho gyi 😊😊 lekin math ka kha se kre guru ji. Math ki bhi classes lga do 😢 PLEASE ❤ Kapil sir ki bhi class lagva do🥹
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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Yes sister
Sir aap bahut achhe physics padha ta hai ❤❤❤❤
Sir please math ki practice bhi yt par suru karva do please sir 🙏❤️💕❤❤❤
Thank you cardiologist ❤
Sir ek type ke kam se kam 2-3 saval laya karo sir please samjhne me time lag jata hai.🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Love you sir ❤❤❤❤
Majja aa gaya Guru gi 🎉🎉🎉🎉
Questions gjb h sir ese hi daily lao❤❤
Thanks sir ji
Nice session Sir ji ❤
video ko like bhi kia karo sab sir ko full support karain
Sir Y group k liye classes kb hogi
Homework ka answer aaaaa
😊😊😊😊😊😊😊😊
Y group mei kaisa question aate hai
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Sir physics or english ki to tantion khatam ho gyi 😊😊
lekin math ka kha se kre guru ji.
Math ki bhi classes lga do 😢
PLEASE ❤
Kapil sir ki bhi class lagva do🥹
Bhai physics ke concept kaha se kar rahe ho
@@syamdevstudio2235 RS sir se kar lo concept ❤❤
Bhai RB sir se pad lo ❤❤
@@Manjeet_defence_lover nhi bhai unke concept samajh nhi aaye hai vo content update nhi karte hai padha hu me unse batch bhi liya tha
@@Manjeet_defence_lover cadets defence me Kapil sir sabse achha padhate h
Sir pdf upload nhi hui
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}|.
Sir Agni 2.0 present here ,physics paar thi aapk ashirvaad se bt maths me better scoring nhi ho payi thi,due to lack of practice
Nice class sir mne phle physics me 14 no 😢but final ni hua is br hoga i promise myself😊
Bhai koi paid batch bi le rahe ho kya
Koi batayega vardi batch mai physics kitna ho gya hai 😢😢
Mai bhi purchase kar lu kya
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3 chapter complete approx 11th ka
@MdImtiyaz42 kaisa content hai please reply
@@syamdevstudio2235 bhai mujhe to top level ka lag raha hai ab tumhe kaisa lagega mujhe nahi pata
@@MdImtiyaz42 bhai syllabus kab tak karayege and practice ke liye questions dete hai kya
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Nice session sir
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Very very nice session sir❤❤❤❤❤❤❤❤❤❤❤❤
Nice session sir ❤❤
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Nice session sir ji ❤❤❤❤❤❤❤❤❤❤
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