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 $(6y\vec{a}-18x\vec{b})$. 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)

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