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

<p>$$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$$</p> <p>$$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$$</p> <p>As given equation is identity</p> <p>Coefficient of ${x^2} =$ coefficient of ${y^2} =$ coefficient of $xy = 0$</p> <p>$$ \Rightarrow |\overrightarrow a {|^2} = 3|\overrightarrow b {|^2} \Rightarrow |\overrightarrow b | = 3\sqrt 3 $$</p> <p>and $\overrightarrow a .\overrightarrow b = 0$</p> <p>$$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a ||\overrightarrow b |\sin \theta $$</p> <p>$= 9.\,3\sqrt 3 .1 = 27\sqrt 3$</p>