Mathematics - Vector Question with Solution | TestHub

Given,

$\left|\overrightarrow{a}\right|=\sqrt{31}, 4\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=2$, $2\left(\overrightarrow{a}\times \overrightarrow{b}\right)=3\left(\overrightarrow{c}\times \overrightarrow{a}\right)$ and angle between $\overrightarrow{b} \& \overrightarrow{c}$ is given as $\frac{2\pi }{3}$

Now solving, $3\left(\overrightarrow{c}\times \overrightarrow{a}\right)+2\left(\overrightarrow{b}\times \overrightarrow{a}\right)=0$

$\Rightarrow \left(3\overrightarrow{c}\times 2\overrightarrow{b}\right)\times \overrightarrow{a}=0$

Means $\left(3\overrightarrow{c}\times 2\overrightarrow{b}\right) \& \overrightarrow{a}$ are parallel vector,

So, let $3\overrightarrow{c}\times 2\overrightarrow{b}=\lambda \overrightarrow{a}$

Now squaring both sides we get,
$9{\left|\overrightarrow{c}\right|}^2+4{\left|\overrightarrow{b}\right|}^2+12\left(\overrightarrow{b}·\overrightarrow{c}\right)={\lambda }^2{\left|\overrightarrow{a}\right|}^2$

$\Rightarrow 36+1+12\times \frac{1}{2}\times 2\left(\cos \left(2\frac{\pi }{3}\right)\right)={\lambda }^2\left(31\right)$

$\Rightarrow {\lambda }^2=1$

$\Rightarrow \lambda =\pm 1$

Now putting the value of $\lambda$ in $\left(3\overrightarrow{c}\times 2\overrightarrow{b}\right)=\lambda \overrightarrow{a}$ we get,

$3\overrightarrow{c}+2\overrightarrow{b}=\pm \overrightarrow{a} \dots \left(1\right)$

Now taking dot product with $\overrightarrow{b}$ in above equation we get,

$3\left(\overrightarrow{b}·\overrightarrow{c}\right)+2\left(\overrightarrow{b}·\overrightarrow{b}\right)=\pm \overrightarrow{a}·\overrightarrow{b}$

$\Rightarrow \overrightarrow{a}·\overrightarrow{b}=\pm \left(-\frac{3}{2}+\frac{1}{2}\right)=\pm \left(-1\right)$

$\Rightarrow {\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2=1$

Again taking $3\left(\overrightarrow{c}\times \overrightarrow{a}\right)=2\left(\overrightarrow{a}\times \overrightarrow{b}\right)$ and sqauring both side,

$\Rightarrow {\left(\overrightarrow{c}\times \overrightarrow{a}\right)}^2=\frac{4}{9}{\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2$

$\Rightarrow {\left(\overrightarrow{c}\times \overrightarrow{a}\right)}^2=\frac{4}{9}\left[{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2\right]$

$\Rightarrow {\left(\overrightarrow{c}\times \overrightarrow{a}\right)}^2=\frac{4}{9}\left[\frac{31}{4}-\left(1\right)\right]$

$\Rightarrow {\left(\overrightarrow{c}\times \overrightarrow{a}\right)}^2=\frac{4}{9}\times \frac{27}{4}=3$

Hence, the value of ${\left(\frac{\overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{a}·\overrightarrow{b}}\right)}^2=\frac{3}{1}=3$.