For any vector $\vec r, (\vec r.\hat i)^2 + (\vec r.\hat j)^2 + (\vec r.\hat k)^2$ is equal to
Answer & explanation
Correct answer: option 4
Let $\vec r = x\hat i+y\hat j+z\hat k$. Then,
$\vec r.\hat i=x,\vec r.\hat j=y$ and $\vec r.\hat k=z$
$∴(\vec r.\hat i)^2 + (\vec r.\hat j)^2 + (\vec r.\hat k)^2=x^2+y^2+z^2=|\vec r|^2$