JEE Main 2022 — Vector Algebra Question with Solution

Given,

${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2 \& \overrightarrow{a}·\overrightarrow{b}=3$

We know that  

${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}·\overrightarrow{b}$

Now using the above formula in ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2$ we get,

${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}·\overrightarrow{b}={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2$

Now on comparing both side we get, ${\left|\overrightarrow{b}\right|}^2=2\overrightarrow{a}·\overrightarrow{b}=6 .........\left(1\right)$

Also given ${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=75$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\sin }^2\theta =75$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2\left(1-{\cos }^2\theta \right)=75$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\cos }^2\theta =75$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2=75$

Now putting the value of ${\left|\overrightarrow{b}\right|}^2=2\overrightarrow{a}·\overrightarrow{b}=6$ we get, $6{\left|\overrightarrow{a}\right|}^2-9=75\Rightarrow {\left|\overrightarrow{a}\right|}^2=14$