If a and b makes an angle ^ -1 ((5)/(9)) with each other, then | a + b |= 2 | a - b | for | a |=n| b | The integer value of n is ____

Correct answer: 0 aligned & θ=(5)/(9) & a · b a b =(5)/(9) & a^2+b^2+2 a · b =2 a^2+2 b^2-4 a · b & 6 a · b =a^2+b^2 aligned aligned & 6 × (5)/(9) a b=a^2+b^2 & (10)/(3) a b=a^2+b^2 & a=n b & (10)/(3) n b^2=n^2 b^2+b^2 & 3 n^2-10 n+3=0 & n=(1)/(3) and n=3 aligned integer value n=3

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2024 · 9 April · Morning Shift · Physics · Mathematics in Physics

If a and b makes an angle ^ -1 ((5)/(9)) with each other, then | a...

\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

\mathrm{n}

is ____

Correct answer

Step-by-step solution

\begin{aligned} & \cos \theta=\frac{5}{9} \\ & \frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9} \\ & 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\end{aligned}

\begin{aligned} & 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\end{aligned}

integer value

n=3$

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