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 Let \mathrm{a} vector \mathrm{\overrightarrow{a}} has magnitude 9. Let a vector \mathrm{\overrightarrow{b}} be such that for every \mathrm{(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}}, the vector \mathrm{(x \overrightarrow{a}+y \overrightarrow{b})} is perpendicular to the vector \mathrm{(6 y \overrightarrow{a}-18 x \overrightarrow{b})}. Then the value of \mathrm{|\overrightarrow{a} \times \overrightarrow{b}|} is equal to:

Option: 1

9 \sqrt{3}


Option: 2

27 \sqrt{3}


Option: 3

9


Option: 4

81


Answers (1)

best_answer

\begin{aligned} & \mathrm{|\vec{a}|=9 \quad \&(x \vec{a}+y \vec{b}) \cdot(6 y \vec{a}-18 x \vec{b})=0} \\ & \mathrm{\Rightarrow 6 x y|\vec{a}|^{2}-18 x^{2}(\vec{a} \cdot \vec{b})+6 y^{2}(\vec{a} \cdot \vec{b})-\left.|8 x y| \vec{b}\right|^{2}=0 }\\ & \mathrm{\Rightarrow 6 xy \left(\left|\vec{a}\right|^{2}-3\left|\vec{b}\right|^{2}\right)+(\vec{a} \cdot \vec{b})\left(y^{2}-3 x^{2}\right)=0} \end{aligned}\text{This should hold}\; \forall \mathrm{x, y \in R X R}\\\begin{aligned} &\mathrm{\therefore|\vec{a}|^{2}=3|\vec{b}|^{2} \&(\vec{a} \cdot \vec{b})=0 }\\ &\text { Now }\mathrm{|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}} \\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mathrm{\quad=|\vec{a}|^{2} \cdot \frac{|\vec{a}|^{2}}{3} }\\ & \mathrm{\therefore|\vec{a} \times \vec{b}|=\frac{|\vec{a}|^{2}}{\sqrt{2}}=\frac{81}{\sqrt{3}}=27 \sqrt{3} }\end{aligned}

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himanshu.meshram

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