Question

Given the vectors `u=<2,2>`, `v=<-3,4>`, and `w=<1,-2>`, how do you find `(u*2v)w`?

Answer 1

`(vecu*2vecv)vecw = <<4,-8>>`

Explanation:

Inner Product Definition
If `vecA = <<(a_1, a_2, a_3)>>`, and `vecB = <<(b_1, b_2, b_3)>>`, then the inner product (or dot product), a scaler quantity, is given by:
`vecA * vecB = a_1b_1 + a_2b_2 + a_3b_3`

Inner Product = 0 `hArr` vectors are perpendicular

So, `vecu=<<2,2>>`, and `vecv=<<-3,4>>`, and `vecw=<<1,-2>>`

Then;
`(vecu*2vecv)vecw = 2(vecu*vecv)vecw`
`:. (vecu*2vecv)vecw = 2{(2)(-3)+(2)(4)}vecw`
`:. (vecu*2vecv)vecw = 2(-6+8)vecw`
`:. (vecu*2vecv)vecw = 2(2)vecw`
`:. (vecu*2vecv)vecw = 4vecw`
`:. (vecu*2vecv)vecw = 4<<1,-2>>`
`:. (vecu*2vecv)vecw = <<4,-8>>`