Find out the resultant of the following three displacement vectors : `vec(A) =10 m`, along north -west, `vec(B) =20 m, 30^@` north of east, `vecC=35m,` along south.

Explanation

Let us choose ,x -axis along east and y-axis along north . Taking into account the x-and y- components of the given vectors, they can be written as (in metre),
`vec(A)=hati(-10cos 45^@)+hatj 10 sin45^@=-5sqrt(2)hati+5sqrt(2)hatj`
`vec(B)=hati20 cos 30^@+hatj20 sin 30^@=10sqrt(3)hati+10hatj`
`vec(C)=-35hatj`
`therefore` Resultant , `vec(R)=vec(A)+vec(B)+vec(C)`
`=hati(-5sqrt(2)+10sqrt(3))+hatj(5sqrt(2)+10-35)`
`=hati5(2sqrt(3)-sqrt(2))-hatj5(5-sqrt(2))` [between east end south ]
`therefore R=|vec(R)|=sqrt{{5(2sqrt(3)-sqrt(2))}^2+ {-5(5-sqrt(2))}^2}`
`=5sqrt((2sqrt(3)-sqrt(2)^2)+(5-sqrt(2))^2)=20.65m`
If `vec(R)` makes an angle `theta` with the x-axis, then
`tan theta =(R_y)/(R_x)=(-5(5-sqrt(2)))/(5(2sqrt(3)-sqrt(2)))=-1.75 -tan(-60.2^@)`
so, `vec(R)` is inclined at `60.2^@` south of east.