As u had asked perpendicular vector to A and B so first findA x B it will come - i-2j+2kAs it unit vector so Divide it by magnitude of lAxBl which is 3 So final answer is (-i-2j+2k)/3

Finding a unit vector perpendicular to the given vectors A=4i-j+3k and B=-2i+j-2k involves a few steps. Let's break down the process:

Step 1: Find the cross product of vectors A and B
To find a vector that is perpendicular to both A and B, we need to compute their cross product. The cross product of two vectors can be calculated using the following formula:

A x B = |i j k |
|4 -1 3 |
|-2 1 -2 |

Using the determinant method, we can calculate the cross product as follows:

= (4 * 1 - (-1) * 1) i - (4 * -2 - (-1) * -2) j + (4 * (-2) - 3 * -2) k
= 5i + 6j - 2k

Step 2: Normalize the cross product
To obtain a unit vector, we need to normalize the cross product vector. The normalization process involves dividing the vector by its magnitude. The magnitude of the cross product vector can be calculated using the formula:

|A x B| = sqrt((5)^2 + (6)^2 + (-2)^2)
= sqrt(25 + 36 + 4)
= sqrt(65)

Now, to normalize the cross product vector, we divide each component by the magnitude:

Unit vector = (5i + 6j - 2k) / sqrt(65)

Step 3: Simplify the unit vector
To simplify the expression, we divide each component by sqrt(65):

Unit vector = (5/sqrt(65))i + (6/sqrt(65))j + (-2/sqrt(65))k

Therefore, the unit vector perpendicular to vectors A=4i-j+3k and B=-2i+j-2k is:

Unit vector = (5/sqrt(65))i + (6/sqrt(65))j + (-2/sqrt(65))k

This unit vector satisfies the condition of being perpendicular to both A and B.