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

1 Answer
Nov 1, 2016

Answer:

# (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>> #