To determine the unit vector that is perpendicular to both vectors A= 2i + j - k and B= i - j, we can utilize the cross product of the two vectors. The cross product of two vectors results in a vector that is perpendicular to both of the original vectors.

Cross Product:
The cross product of two vectors A= ai + bj + ck and B= di + ej + fk is given by the following equation:
A × B = (bf - ce)i - (af - cd)j + (ae - bd)k

In this case, vector A= 2i + j - k and B= i - j, so we can substitute the values into the cross product equation as follows:

(2 * -1 - 1 * 1)i - (2 * 1 - 1 * 1)j + (1 * 1 - 2 * -1)k
= (-2 - 1)i - (2 - 1)j + (1 + 2)k
= -3i - 1j + 3k

Unit Vector:
The resulting vector from the cross product, -3i - j + 3k, is the perpendicular vector to both A and B. However, we need to find the unit vector, which has a magnitude of 1. To do this, we divide the vector by its magnitude.

The magnitude of a vector V= xi + yj + zk is given by the formula:
|V| = √(x^2 + y^2 + z^2)

In this case, the magnitude of the vector -3i - j + 3k is:
|V| = √((-3)^2 + (-1)^2 + 3^2)
= √(9 + 1 + 9)
= √19

To obtain the unit vector, we divide each component of the vector -3i - j + 3k by the magnitude √19:

Unit Vector = (-3/√19)i - (1/√19)j + (3/√19)k

So, the unit vector that is perpendicular to both A= 2i + j - k and B= i - j is (-3/√19)i - (1/√19)j + (3/√19)k.

May be 3/root 19 (i+j-k)