JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 5)
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 :
$$9 \sqrt{3}$$
$$27 \sqrt{3}$$
9
81
Explanation
$$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$$
$$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$$
As given equation is identity
Coefficient of $${x^2} = $$ coefficient of $${y^2} = $$ coefficient of $$xy = 0$$
$$ \Rightarrow |\overrightarrow a {|^2} = 3|\overrightarrow b {|^2} \Rightarrow |\overrightarrow b | = 3\sqrt 3 $$
and $$\overrightarrow a .\overrightarrow b = 0$$
$$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a ||\overrightarrow b |\sin \theta $$
$$ = 9.\,3\sqrt 3 .1 = 27\sqrt 3 $$
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