For the vector `veca = (3.0 m) hati + (4.0m) hatj` and `vecb = (5.0 m) hati + (-2.0 m) hatj,` give `veca + vecb` in (a) unite-vector notation, and as (b) a magnitude and (c ) an angle (relation to `hati`). Now give `vecb -veca` in (d) unit vector notation, as (e) a magnitude and (f) an angle.

Explanation

(a) `veca + vecb = (3.0 hati + 4.0 hatj) m + ( 5.0 hati - 2.0 hatj) m`
`= (8.0 m) hati + (2.0m) hatj.`
(b) The magnitude of `veca + vecb ` is
`|veca+ vecb| = sqrt ((8.0 m) ^(2) + (2.0m)^(2))=8.2m.`
(c ) The angle between this vector and the + x axis is `tan ^(-1) [(2.0m)//(8.0m)]=14 ^(@).`
(d) `vecb - veca= (5.0 hati-2.0 hatj) m - (3.0 hati + 4.0 hatj)m`
` = (2.0 m )hati - (6.0 m) hatj.`
(e) The magnitude of the difference vector `vecb -veca` is
`|vecb -veca| = sqrt ((2.0 m) ^(2) + ( -6.0 m) ^(2)) = 6.3m.`
(f) The angle between this vector and the +x axis is `tan ^(-1) (-6.0m) //(2.0m) ]=-72^(@).` The vector is `72^(@)` clockwise from teh axis defined by `hati.`