If A vector = i^-j^+k^ then unit vector in the direction of A vector a)√3,j^
b)(i^-j^+k^)
c)(i^-j^-k^)/√2
d)(i^-j^+k^)/√3
plzz answer my question with explanation​

If A = i - j + k, then the magnitude of A is

||A|| = √(1² + (-1)² + 1²) = √3

Then the unit vector in the direction of A is 1/||A|| multiplied by A :

u = A/||A|| = (i - j + k)/√3

(choice D)

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards