Let a vector → a has a magnitude 9 . Let a vector → b be such that...

Given, $\left|\overset{\to}{a}\right| = 9$

And \left(x\overset{\to}{a} + y\overset{\to}{b}\right) \cdot \left(6y\overset{\to}{a} - 18x\overset{\to}{b}\right) = 0$$\left\{\text{as they are perpendicular}\right\}$

$⇒ 6xy{\left|\overset{\to}{a}\right|}^{2} - 18{x}^{2}\left(\overset{\to}{a} \cdot \overset{\to}{b}\right) + 6{y}^{2}\left(\overset{\to}{a} \cdot \overset{\to}{b}\right) - 18xy{\left|\overset{\to}{b}\right|}^{2} = 0$

$⇒ 6xy\left({\left|\overset{\to}{a}\right|}^{2} - 3{\left|\overset{\to}{b}\right|}^{2}\right) + \left(\overset{\to}{a} \cdot \overset{\to}{b}\right)\left({y}^{2} - 3{x}^{2}\right) = 0$

This should hold $\forall x, y \in R \times R - \left\{\left(0, 0\right)\right\}$

So, ${\left|\overset{\to}{a}\right|}^{2} - 3{\left|\overset{\to}{b}\right|}^{2} = 0 \text{ and } \left(\overset{\to}{a} \cdot \overset{\to}{b}\right) = 0$

$∴ {\left|\overset{\to}{a}\right|}^{2} = 3{\left|\overset{\to}{b}\right|}^{2} \text{ and } \left(\overset{\to}{a} \cdot \overset{\to}{b}\right) = 0 ⇒ cos\theta = 0 ⇒ \theta = 90 °$

Now ${\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\sin^{2}\theta = {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\left\{\text{as}\theta = {90}^{°}\right\}$

$= {\left|\overset{\to}{a}\right|}^{2} \cdot \dfrac{{\left|\overset{\to}{a}\right|}^{2}}{3}$ \left\{\text{as}{\left|\overset{\to}{b}\right|}^{2} = \dfrac{{\left|\overset{\to}{a}\right|}^{2}}{3}\right\}$

So, \left|\overset{\to}{a} \times \overset{\to}{b}\right| = \dfrac{{\left|\overset{\to}{a}\right|}^{2}}{\sqrt{3}} = \dfrac{81}{\sqrt{3}} = 27\sqrt{3}