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For any vector $\vec r, (\vec r.\hat i)^2 + (\vec r.\hat j)^2 + (\vec r.\hat k)^2$ is equal to |
1 $|\vec r|$ $\vec r$ $|\vec r|^2$ |
$|\vec r|^2$ |
Let $\vec r = x\hat i+y\hat j+z\hat k$. Then, $\vec r.\hat i=x,\vec r.\hat j=y$ and $\vec r.\hat k=z$ $∴(\vec r.\hat i)^2 + (\vec r.\hat j)^2 + (\vec r.\hat k)^2=x^2+y^2+z^2=|\vec r|^2$ |
