Given the vectorskjia 23  andkjib  52 . Find the unit vectorsperpendicular to the vectorsa andb .

The question seems to be a bit garbled, but I'll try to interpret it as best as I can. It seems like you're given two vectors, a and b, and you're asked to find the unit vectors that are perpendicular to both a and b.

The vectors a and b are given as:

a = -2i + 3j - k b = 5i + 2j - k

The cross product of two vectors is a vector that is perpendicular to both of the original vectors. So, we can find a vector that is perpendicular to both a and b by taking the cross product of a and b.

The cross product of a and b is given by:

a x b = (a2b3 - a3b2)i - (a1b3 - a3b1)j + (a1b2 - a2b1)k

Substituting the components of a and b into this equation gives:

a x b = (32 - (-1)(-2))i - ((-2)*2 - (-1)*5)j + ((-2)2 - 35)k = 4i - 9j - 11k

This is a vector that is perpendicular to both a and b.

To find the unit vector in the direction of a x b, we divide a x b by its magnitude. The magnitude of a vector is given by the square root of the sum of the squares of its components.

The magnitude of a x b is sqrt(4^2 + (-9)^2 + (-11)^2) = sqrt(16 + 81 + 121) = sqrt(218)

So, the unit vector in the direction of a x b is (4/sqrt(218))i - (9/sqrt(218))j - (11/sqrt(218))k.