Let $\vec a,\vec b,\vec c$ be three vectors of equal magnitude such that the angle between each pair is $\frac{π}{3}$. If $|\vec a+\vec b+\vec c=\sqrt{6}$, then $|\vec a|=$
Answer & explanation
Correct answer: option 3
Let $|\vec a|=|\vec b|=|\vec c|=λ$. Then,
$\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=\frac{λ^2}{2}$
Now,
$|\vec a+\vec b+\vec c|=\sqrt{6}$
$⇒|\vec a+\vec b+\vec c|^2=6$
$⇒|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a.\vec b+\vec b.\vec a+\vec c.\vec a)=6$
$⇒6λ^2=1⇒λ=1⇒|\vec a|=1$