Let a vector a has a magnitude 9. Let a vector b be such that for every (x,y)R× R-(0,0), the vector (x a+y b) is perpendicular to the vector (6y…

Given, $|\vec{a}|=9$∣a∣=9

And $(x\vec{a}+y\vec{b})\cdot (6y\vec{a}-18x\vec{b})=0$(xa+yb)⋅(6ya−18xb)=0${\text{as they are perpendicular}}$as they are perpendicular

$$\Rightarrow 6xy{|\vec{a}|}^{2}-18{x}^{2}(\vec{a}\cdot \vec{b})+6{y}^{2}(\vec{a}\cdot \vec{b})-18xy{|\vec{b}|}^{2}=0$$⇒6xy∣a∣2−18x2(a⋅b)+6y2(a⋅b)−18xy∣b∣2=0

$$\Rightarrow 6xy({|\vec{a}|}^{2}-3{|\vec{b}|}^{2})+(\vec{a}\cdot \vec{b})({y}^{2}-3{x}^{2})=0$$⇒6xy(∣a∣2−3∣b∣2)+(a⋅b)(y2−3x2)=0

This should hold $\forall x,y\in R\times R-{(0,0)}$∀x,y∈R×R−(0,0)

So, {|\vec{a}|}^{2}-3{|\vec{b}|}^{2}=0&(\vec{a}\cdot \vec{b})=0

\therefore {|\vec{a}|}^{2}=3{|\vec{b}|}^{2}&(\vec{a}\cdot \vec{b})=0\Rightarrow \mathrm{cos}\theta =0\Rightarrow \theta =90^{\circ}

Now ${|\vec{a}\times \vec{b}|}^{2}={|\vec{a}|}^{2}{|\vec{b}|}^{2}{\mathrm{sin}}^{2}\theta ={|\vec{a}|}^{2}{|\vec{b}|}^{2}{\text{as }\theta ={90}^{^{\circ}}}$∣a×b∣2=∣a∣2∣b∣2sin2θ=∣a∣2∣b∣2as θ=90∘

$={|\vec{a}|}^{2}\cdot \frac{{|\vec{a}|}^{2}}{3}$=∣a∣2⋅3∣a∣2​ ${\text{as }{|\vec{b}|}^{2}=\frac{{|\vec{a}|}^{2}}{3}}$as ∣b∣2=3∣a∣2​

So, $|\vec{a}\times \vec{b}|=\frac{{|\vec{a}|}^{2}}{\sqrt{3}}=\frac{81}{\sqrt{3}}=27\sqrt{3}$∣a×b∣=3​∣a∣2​=3​81​=273​