Mathematics - Vector Question with Solution | TestHub

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

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

$\Rightarrow 6xy{\left|\overrightarrow{a}\right|}^2-18x^2\left(\overrightarrow{a}·\overrightarrow{b}\right)+6y^2\left(\overrightarrow{a}·\overrightarrow{b}\right)-18xy{\left|\overrightarrow{b}\right|}^2=0$

$\Rightarrow 6xy\left({\left|\overrightarrow{a}\right|}^2-3{\left|\overrightarrow{b}\right|}^2\right)+\left(\overrightarrow{a}·\overrightarrow{b}\right)\left(y^2-3x^2\right)=0$

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

So, ${\left|\overrightarrow{a}\right|}^2-3{\left|\overrightarrow{b}\right|}^2=0 \& \left(\overrightarrow{a}·\overrightarrow{b}\right)=0$

$\therefore {\left|\overrightarrow{a}\right|}^2=3{\left|\overrightarrow{b}\right|}^2 \& \left(\overrightarrow{a}·\overrightarrow{b}\right)=0\Rightarrow \cos \theta =0\Rightarrow \theta =90°$

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

$={\left|\overrightarrow{a}\right|}^2·\frac{{\left|\overrightarrow{a}\right|}^2}{3}$ $\left\{\text{as }{\left|\overrightarrow{b}\right|}^2=\frac{{\left|\overrightarrow{a}\right|}^2}{3}\right\}$

So, $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\frac{{\left|\overrightarrow{a}\right|}^2}{\sqrt{3}}=\frac{81}{\sqrt{3}}=27\sqrt{3}$