The vectors 3 vec{a}-5 vec{b} and 2 vec{a}+ve | vedclass

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(A) Given that $(3 \vec{a}-5 \vec{b}) \cdot (2 \vec{a}+\vec{b}) = 0$. Expanding this,we get $6|\vec{a}|^2 + 3\vec{a} \cdot \vec{b} - 10\vec{a} \cdot \

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$\cos ^{-1}\left(\frac{19}{5 \sqrt{43}}\right)$

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(A) Given that $(3 \vec{a}-5 \vec{b}) \cdot (2 \vec{a}+\vec{b}) = 0$. Expanding this,we get $6|\vec{a}|^2 + 3\vec{a} \cdot \vec{b} - 10\vec{a} \cdot \vec{b} - 5|\vec{b}|^2 = 0$,which simplifies to $6|\vec{a}|^2 - 7\vec{a} \cdot \vec{b} - 5|\vec{b}|^2 = 0$. Thus,$\vec{a} \cdot \vec{b} = \frac{6|\vec{a}|^2 - 5|\vec{b}|^2}{7} \dots (i)$. Also,$(\vec{a}+4 \vec{b}) \cdot (\vec{b}-\vec{a}) = 0$. Expanding this,we get $\vec{a} \cdot \vec{b} - |\vec{a}|^2 + 4|\vec{b}|^2 - 4\vec{a} \cdot \vec{b} = 0$,which simplifies to $-|\vec{a}|^2 - 3\vec{a} \cdot \vec{b} + 4|\vec{b}|^2 = 0$. Thus,$\vec{a} \cdot \vec{b} = \frac{4|\vec{b}|^2 - |\vec{a}|^2}{3} \dots (ii)$. Equating $(i)$ and $(ii)$,$\frac{6|\vec{a}|^2 - 5|\vec{b}|^2}{7} = \frac{4|\vec{b}|^2 - |\vec{a}|^2}{3}$. $18|\vec{a}|^2 - 15|\vec{b}|^2 = 28|\vec{b}|^2 - 7|\vec{a}|^2$,so $25|\vec{a}|^2 = 43|\vec{b}|^2$,which means $|\vec{a}| = \sqrt{\frac{43}{25}} |\vec{b}| = \frac{\sqrt{43}}{5} |\vec{b}|$. Substituting $|\vec{a}|^2 = \frac{43}{25} |\vec{b}|^2$ into $(ii)$,$\vec{a} \cdot \vec{b} = \frac{4|\vec{b}|^2 - \frac{43}{25}|\vec{b}|^2}{3} = \frac{100-43}{75} |\vec{b}|^2 = \frac{57}{75} |\vec{b}|^2 = \frac{19}{25} |\vec{b}|^2$. Since $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$,we have $\cos 	heta = \frac{\frac{19}{25} |\vec{b}|^2}{(\frac{\sqrt{43}}{5} |\vec{b}|) |\vec{b}|} = \frac{19}{25} \cdot \frac{5}{\sqrt{43}} = \frac{19}{5\sqrt{43}}$. Therefore,$	heta = \cos^{-1}\left(\frac{19}{5\sqrt{43}}ight)$.

The vectors $3 \vec{a}-5 \vec{b}$ and $2 \vec{a}+\vec{b}$ are mutually perpendicular and the vectors $\vec{a}+4 \vec{b}$ and $-\vec{a}+\vec{b}$ are also mutually perpendicular. Then the angle between the vectors $\vec{a}$ and $\vec{b}$ is:

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