If 3vec{a} - 5vec{b} and 2vec{a} + vec{b} are | vedclass

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(B) Given that $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + \vec{b}) = 0$. $6|\vec{a}|^2 + 3\vec{a} \cdot \vec{b} - 10\vec{a} \cdot \vec{b} - 5|\vec{b}|^2

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$\frac{19}{5\sqrt{43}}$

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(B) Given that $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + \vec{b}) = 0$. $6|\vec{a}|^2 + 3\vec{a} \cdot \vec{b} - 10\vec{a} \cdot \vec{b} - 5|\vec{b}|^2 = 0$ $6|\vec{a}|^2 - 5|\vec{b}|^2 - 7\vec{a} \cdot \vec{b} = 0 \quad \dots(i)$ Also,$(\vec{a} + 4\vec{b}) \cdot (-\vec{a} + \vec{b}) = 0$. $-|\vec{a}|^2 + \vec{a} \cdot \vec{b} - 4\vec{a} \cdot \vec{b} + 4|\vec{b}|^2 = 0$ $|\vec{a}|^2 - 4|\vec{b}|^2 + 3\vec{a} \cdot \vec{b} = 0 \quad \dots(ii)$ From $(i)$ and $(ii)$,we eliminate $\vec{a} \cdot \vec{b}$. Multiply $(ii)$ by $7$ and $(i)$ by $3$: $18|\vec{a}|^2 - 15|\vec{b}|^2 - 21\vec{a} \cdot \vec{b} = 0$ $7|\vec{a}|^2 - 28|\vec{b}|^2 + 21\vec{a} \cdot \vec{b} = 0$ Adding these: $25|\vec{a}|^2 - 43|\vec{b}|^2 = 0 \implies |\vec{a}|^2 = \frac{43}{25}|\vec{b}|^2 \implies |\vec{a}| = \frac{\sqrt{43}}{5}|\vec{b}|$. From $(ii)$,$\vec{a} \cdot \vec{b} = \frac{4|\vec{b}|^2 - |\vec{a}|^2}{3} = \frac{4|\vec{b}|^2 - \frac{43}{25}|\vec{b}|^2}{3} = \frac{100-43}{75}|\vec{b}|^2 = \frac{57}{75}|\vec{b}|^2 = \frac{19}{25}|\vec{b}|^2$. Now,$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\frac{19}{25}|\vec{b}|^2}{(\frac{\sqrt{43}}{5}|\vec{b}|)|\vec{b}|} = \frac{19}{25} \cdot \frac{5}{\sqrt{43}} = \frac{19}{5\sqrt{43}}$.

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

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