Hint: The dot product is also called scalar product because it gives a scalar quantity as a result of the dot product. It is also the product of the magnitude of the two given vectors and the cosine of the angle between them. It is the projection of one vector onto another vector, for parallel, the value of the dot product is one and for perpendicular, it is zero.Formula used: Suppose \\[\\mathop A\\limits^ \\to \\] and \\[\\mathop B\\limits^ \\to \\] are two vectors with components \\[\\mathop A\\limits^ \\to = {A_x} + {A_y} + {A_z}\\] and \\[\\mathop B\\limits^ \\to = {B_x} + {B_y} + {B_z}\\]in space, then \\[\\mathop A\\limits^ \\to .\\mathop B\\limits^ \\to = \\left| A \\right|\\left| B \\right|\\cos \\theta \\]Where \\[\\theta \\] is the angle between the two vectorsComplete step-by-step solution:The dot product of two vectors \\[\\mathop A\\limits^ \\to \\] and \\[\\mathop B\\limits^ \\to \\] with an angle \\[\\theta \\] is called the scalar product and it is equal to the product of the magnitude of the two vectors and cosine of the angle between them or it is the product of the magnitude of one of the vector and the component of the second vector.So, \\[\\mathop A\\limits^ \\to .\\mathop B\\limits^ \\to = \\left| A \\right|\\left| B \\right|\\cos \\theta \\]Or \\[\\mathop A\\limits^ \\to .\\mathop B\\limits^ \\to = \\left| A \\right|(\\left| B \\right|\\cos \\theta )\\]The quantity \\[A(Bcos\\theta )\\] is scalar.Now in the given question, we are asked to find the dot product of \\[\\widehat {i.}\\] and \\[\\widehat j\\]Here \\[\\widehat {i.}\\] and \\[\\widehat j\\] are the unit vectors in the direction along the x-axis and y-axis respectively.So, they are mutually perpendicular to each other, that's why \\[\\theta = 90^\\circ \\].Therefore,\\[\\widehat {i.}\\widehat j = ij\\cos {90^ \\circ }\\]\\[\\widehat {i.}\\widehat j = 0\\]. Hence, the dot product of the two unit vectors \\[\\widehat {i.}\\] and \\[\\widehat j\\] is zero. Note: The dot product is the product of the magnitude of the two given vectors and the cosine of the angle between themsimilarly, \\[\\widehat j.\\widehat k = \\widehat k.\\widehat i = 0\\] as they are mutually perpendicular to each other.And \\[\\widehat i.\\widehat i = \\widehat j.\\widehat j = \\widehat k.\\widehat k = 1\\].

With a bit of reshaping, we can use matmul. The idea is to treat the first 2 dimensions as the 'batch' dimensions, and to the dot on the last:

In [278]: E = A[...,None,:]@B[...,:,None]                                       
In [279]: E.shape                                                               
Out[279]: (100, 100, 1, 1)
In [280]: E = np.squeeze(A[...,None,:]@B[...,:,None])                           
In [281]: np.allclose(C,E)                                                      
Out[281]: True
In [282]: timeit E = np.squeeze(A[...,None,:]@B[...,:,None])                    
130 µs ± 2.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [283]: timeit C = np.einsum("ijk,ijk->ij", A, B)                             
90.2 µs ± 1.53 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Comparative timings can be a bit tricky. In the current versions, einsum can take different routes depending on the dimensions. In some cases it appears to delegate the task to matmul (or at least the same underlying BLAS-like code). While it's nice that einsum is faster in this test, I wouldn't generalize that.

tensordot just reshapes (and if needed transposes) the arrays so it can apply the ordinary 2d np.dot. Actually it doesn't work here because you are treating the first 2 axes as a 'batch', where as it does an outer product on them.