If vec{a} and vec{b} make an angle cos^{-1}le | vedclass

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(A) Given that the angle $	heta$ between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}ight)$,so $\cos 	heta = \frac{5}{9}$.We are given t

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$3$

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(A) Given that the angle $	heta$ between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}ight)$,so $\cos 	heta = \frac{5}{9}$.We are given the condition $|\vec{a}+\vec{b}| = \sqrt{2}|\vec{a}-\vec{b}|$.Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = 2|\vec{a}-\vec{b}|^2$.Expanding the dot products: $a^2 + b^2 + 2\vec{a}\cdot\vec{b} = 2(a^2 + b^2 - 2\vec{a}\cdot\vec{b})$.$a^2 + b^2 + 2ab\cos	heta = 2a^2 + 2b^2 - 4ab\cos	heta$.Rearranging the terms: $6ab\cos	heta = a^2 + b^2$.Substituting $\cos	heta = \frac{5}{9}$: $6ab\left(\frac{5}{9}ight) = a^2 + b^2$.$\frac{10}{3}ab = a^2 + b^2$.Given $|\vec{a}| = n|\vec{b}|$,so $a = nb$. Substituting this into the equation:$\frac{10}{3}(nb)b = (nb)^2 + b^2$.$\frac{10}{3}nb^2 = n^2b^2 + b^2$.Dividing by $b^2$ (assuming $b 
eq 0$): $\frac{10}{3}n = n^2 + 1$.$3n^2 - 10n + 3 = 0$.Solving the quadratic equation: $(3n - 1)(n - 3) = 0$.Thus,$n = \frac{1}{3}$ or $n = 3$.The integer value of $n$ is $3$.

If $\vec{a}$ and $\vec{b}$ make 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 . . . . . . .

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