Let a vector vec{a} have a magnitude 9. Let a | vedclass

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(B) Given $|\vec{a}| = 9$. Since $(x \vec{a} + y \vec{b}) \perp (6y \vec{a} - 18x \vec{b})$,their dot product is zero:$(x \vec{a} + y \vec{b}) \cdot (

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$27 \sqrt{3}$

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(B) Given $|\vec{a}| = 9$. Since $(x \vec{a} + y \vec{b}) \perp (6y \vec{a} - 18x \vec{b})$,their dot product is zero:$(x \vec{a} + y \vec{b}) \cdot (6y \vec{a} - 18x \vec{b}) = 0$$6xy |\vec{a}|^2 - 18x^2 (\vec{a} \cdot \vec{b}) + 6y^2 (\vec{a} \cdot \vec{b}) - 18xy |\vec{b}|^2 = 0$$6xy (|\vec{a}|^2 - 3|\vec{b}|^2) + (\vec{a} \cdot \vec{b})(6y^2 - 18x^2) = 0$For this to hold for all $(x, y)$,the coefficients of $xy$,$x^2$,and $y^2$ must be zero:$|\vec{a}|^2 - 3|\vec{b}|^2 = 0 \implies |\vec{b}|^2 = \frac{|\vec{a}|^2}{3} = \frac{81}{3} = 27$$\vec{a} \cdot \vec{b} = 0$Now,$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 81 	imes 27 - 0 = 2187$$|\vec{a} 	imes \vec{b}| = \sqrt{2187} = \sqrt{81 	imes 27} = 9 	imes 3 \sqrt{3} = 27 \sqrt{3}$.

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