Two vectors have magnitudes 2 m and 3m. The angle between them is `60^0`. Find a the scalar product of the two vectors b. the magnitude of their vector product.

Answer

`(a)3 m^2 (b)3sqrt(3) m^2`

Explanation

`|veca|=2 m, |vecb|=3 m
` Angle between them
` theta = 60^@`
(a)`veca=|veca|.|vecb| cos 60^@`
` =2xx3xx 1/2 = 3
m^2`
`(b) |vecaxxvecb|=|veca||vecb| sin 60^0`
`=2xx3xx sqrt(3)/2`
`=3sqrt(3)m^2`