JEE Mains · Physics · STD 11 - Vectors
- JEE Mains 2024
- Hard
- Numerical
If \(\vec{a}\) and \(\vec{b}\) makes an angle \(\cos ^{-1}\left(\frac{5}{9}\right)\) with each other, then \(|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|\) for \(|\vec{a}|=n|\vec{b}|\) The integer value of \(n\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(\cos \theta=\frac{5}{9}\)
\(\frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9}\)
\(\mid \vec{a}+\cdots(1)\)
\(a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b}\)
\(6 \vec{a} \cdot \vec{b}=a^2+b^2\)
\(6 \times \frac{5}{9} a b=a^2+b^2\)
\(\frac{10}{3} a b=a^2+b^2 \quad \& \quad a=n b\)
\(\frac{10}{3} n b^2=n^2 b^2+b^2\)
\(3 n^2-10 n+3=0\)
\(n=\frac{1}{3} \text { and } n=3\)
\(\text { integer value } n=3\)
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