# numpy.vdot() in Python

Prerequisite – numpy.dot() in Python
`numpy.vdot(vector_a, vector_b) ` returns the dot product of vectors a and b. If first argument is complex the complex conjugate of the first argument(this is where `vdot()` differs working of `dot()` method) is used for the calculation of the dot product. It can handle multi-dimensional arrays but working on it as a flattened array.

Parameters –

1. vector_a : [array_like] if a is complex its complex conjugate is used for the calculation of the dot product.
2. vector_b : [array_like] if b is complex its complex conjugate is used for the calculation of the dot product.

Return – dot Product of vectors a and b.

Code 1 :

 `# Python Program illustrating ` `# numpy.vdot() method ` ` `  `import` `numpy as geek ` ` `  `# 1D array ` `vector_a ``=` `2` `+` `3j` `vector_b ``=` `4` `+` `5j` ` `  `product ``=` `geek.vdot(vector_a, vector_b) ` `print``(``"Dot Product  : "``, product) `

Output :

```Dot Product  :  (23-2j)
```

How Code1 works ?
vector_a = 2 + 3j
vector_b = 4 + 5j

As per method, take conjugate of vector_a i.e. 2 – 3j

now dot product = 2(4 – 5j) + 3j(4 – 5j)
= 8 – 10j + 12j + 15
= 23 – 2j

Code 2 :

 `# Python Program illustrating ` `# numpy.vdot() method ` ` `  `import` `numpy as geek ` ` `  `# 1D array ` `vector_a ``=` `geek.array([[``1``, ``4``], [``5``, ``6``]]) ` `vector_b ``=` `geek.array([[``2``, ``4``], [``5``, ``2``]]) ` ` `  `product ``=` `geek.vdot(vector_a, vector_b) ` `print``(``"Dot Product  : "``, product) ` ` `  `product ``=` `geek.vdot(vector_b, vector_a) ` `print``(``"\nDot Product  : "``, product) ` ` `  `"""  ` `How Code 2 works :  ` `array is being flattened ` ` `  `1 * 2 + 4 * 4 + 5 * 5 + 6 * 2 = 55 ` `"""`

Output :

```Dot Product  :  55

Dot Product  :  55
```

.
This article is contributed by Mohit Gupta_OMG 😀. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.