Question

Question #edb00

Answer 1

See proof below

Explanation:

`vecr=〈3,2,8〉`

`vecr.vecr=〈3,2,8〉.〈3,2,8〉=9+4+64=77`

`(∥r∥)^2=(∥〈3,2,8〉∥)^2=(sqrt(9+4+64))^2=9+4+64=77`

So, `vecr.vecr=(∥r∥)^2`