For any vector $\vec r$, $(\vec r.\hat i)\hat i+(\vec r.\hat j)\hat j+(\vec r.\hat k)\hat k=$

$\vec r$

Let $\vec r = x\hat i+y\hat j+z\hat k$ be an arbitrary vector. Then,

$\vec r.\hat i = (x\hat i+ y\hat j +z\hat k).\hat i=x(\hat i·\hat i)+y(\hat j·\hat i)+z(\hat k.\hat i)=x$

$\vec r.\hat j=(x\hat i+ y\hat j +z\hat k).\hat k=x(\hat i·\hat j)+y(\hat j.\hat j)+z(\hat k.\hat j)$

and, $\vec r.\hat k=(x\hat i+ y\hat j +z\hat k).\hat k=x(\hat i·\hat k)+y(\hat j.\hat k)+z(\hat k.\hat k)$

Putting the values of x, y, z in $\vec r = x\hat i+y\hat j+z\hat k$, we obtain

$\vec r =(\vec r.\hat i)\hat i+(\vec r.\hat j)\hat j+(\vec r.\hat k)\hat k$