Question

If $\left|\overrightarrow{a}\right|=5$, $|\overrightarrow{a}-\overrightarrow{b}|=8$ and $|\overrightarrow{a}+\overrightarrow{b}|=10$ then $\left|\overrightarrow{b}\right|$ is

• A. $1$
• B. $\sqrt{57}$
• C. $3$
• D. None of these

Answer 1

The correct option is B $\sqrt{57}$

$|\overrightarrow{a}|=5,|\overrightarrow{a}-\overrightarrow{b}|=8,|\overrightarrow{a}+\overrightarrow{b}|=10$

$\Rightarrow a^2+b^2-2ab\left(\cos \theta \right)=64$ & $a^2+b^2+2ab\left(\cos \theta \right)=100$

Adding the two, we get $a^2+b^2=\frac{100+64}{2}=82$
$\Rightarrow b^2=82-25=57$
$\Rightarrow |\overrightarrow{b}|=\sqrt{57}$