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JEE Mains · Physics · STD 11 - 3.1 vectors

यदि \(\vec{a}\) और \(\vec{b}\) के बीच का कोण \(\cos ^{-1}\left(\frac{5}{9}\right)\) है, तो \(|\vec{a}|=n|\vec{b}|\) के लिए \(|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|\) है। \(n\) का पूर्णांक मान _______ है।

  1. A \(3\)
  2. B \(5\)
  3. C \(4\)
  4. D \(6\)
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Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

\(\cos \theta=\frac{5}{9}\) \(\frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9}\) \(\mid \vec{a}+\cdots(1)\) \(a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b}\) \(6 \vec{a} \cdot \vec{b}=a^2+b^2\) \(6 \times \frac{5}{9} a b=a^2+b^2\)…
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