Question

Let a vector $\vec{a}$ has a magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in R \times R-\{(0,0)\}$, 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:

• A. $9 \sqrt{3}$
• B. $27 \sqrt{3}$
• C. 9
• D. 81

Answer 1

Answer: (B)

$|\vec{a}|=9 \&(x \vec{a}+y \vec{b}) \cdot(6 y \vec{a}-18 x \vec{b})=0$
$\Rightarrow 6 x y|\bar{a}|^2-18 x^2(\overline{ a } \cdot \overline{ b })+6 y ^2(\overline{ a } \cdot \overline{ b })-18 xy |\overline{ b }|^2=0$
$\Rightarrow 6 x y\left(|\overrightarrow{ a }|^2-3|\overline{ b }|^2\right)+(\overline{ a } \cdot \overline{ b })\left( y ^2-3 x ^2\right)=0 $
$ \text { This should hold } \forall x , y \in R \times R$
$\therefore|\overline{ a }|^2=3|\overline{ b }|^2 \&(\overline{ a } \cdot \overline{ b })=0 $
$ \text { Now }|\overline{ a } \times \overline{ b }|^2=|\overline{ a }|^2|\overline{ b }|^2-(\overline{ a } \cdot \overline{ b })^2 $
$=|\overline{ a }|^2 \cdot \frac{|\overline{ a }|^2}{3} $
$ \therefore |\overline{ a } \times \overline{ b }|=\frac{|\overline{ a }|^2}{\sqrt{3}}=\frac{81}{\sqrt{3}}=27 \sqrt{3}$