Question

Question #55131

Answer 1

See below.

Explanation:

Calling

`vec a = {a_1, a_2, a_3}`
`hat i = {1, 0, 0}`
`hat j = {0, 1, 0}`
`hat k = {0, 0, 1}`

we have

`vec a xx hat i = {0, a_3, -a_2}`
`vec a xx hat j ={-a_3, 0, a_1}`
`vec a xx hat k ={a_2, -a_1, 0}`

and finally

`hat i xx (vec a xx hat i )+hat j xx (vec a xx hat j)+hat k xx(vec a xx hat k ) = 2{a_1,a_2,a_3} = 2 vec a`