find the vector ->AB and its magnitude if it has initial point a (1,2,-1) nd final point B(3,2,2).

which of the followng forces has the maximum inclination with the X axis and what is its value

3i+4j ,    5i+3j      10j+12j

(a)

vector AB will be

AB = A - B = (i + 2j - k) - (3i + 2j + 2k)

or

AB = -2i +0k -3k

the magnitude will be

|AB| = (2² + 3²)^1/2

or

|AB| = 3.60

..

(b)

(i) the angle between x-axis and vector 3i + 4j will be

here

x = 1i + 0j + 0k

p = 3i + 4j

so,

cosθ₁ = x.p / |x| |p|

 = (1i + 0j + 0k).(3i + 4j) / [(1²)^1/2 x (3² + 4²)^1/2]

= 3 / 5 = 0.6

so,

θ₁ = 53.13 deg. 

similarly for other two cases, we get

θ₂ = 30.96 deg. 

θ₃ = 11.31 deg. 

so,

the force which has the maximum inclination would be p = 3i+4j, = 53.13 deg.