Answer:
Let the vectors be 
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have 
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
 c / | c |
Where | c |  = √ (7)² + (2)² + (-1)²  = 3√6
Therefore, the unit vector is 
![\frac{[7,2,-1]}{3\sqrt{6} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B7%2C2%2C-1%5D%7D%7B3%5Csqrt%7B6%7D%20%7D) 
 
or 
[  ,
 ,  ,
 ,  ]
 ]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[  ,
 ,  ,
 ,  ]
 ]
In conclusion, the two unit vectors are;
[  ,
 ,  ,
 ,  ]
 ] 
and 
[  ,
 ,  ,
 ,  ]
 ]
<em>Hope this helps!</em>