If `vec(A)=vec(B)+vec(C )`, and the magnitudes of `vec(A)`,`vec(B)`,`vec(C )` are 5,4, and 3 units, then the angle between `vec(A)` and `vec(C )` is

Options

A. cos^(-1) ( 3/5)`
B. cos^(-1) (4/5)`
C. ` pi //2`
D. sin ^(-1) (3/5)`

Answer

A

Explanation

Herem ` | vec A| = 5 , | vec B|= 4, |vec C | = 3`
As ` vec A = vec B+ vec C. so vec B = vec A - vec C`
As ` vec B` is now resultant of ` vec A` and ` vec C`. So it is given
by ` `B^2 = A^2 + C^2 -2 AC cos theta`
` (4)^2= (5)^2 + (3)^2 - 2 xx 5 xx 3 cos theta`
` - 16 - 25 - 9 =- 30 cos theta`
` -18 =- 30 cos theta `
` cos theta = (-18)/(-30) = 3/2 theta = cos^(-1) ( 3/5)`.