Jawaban 1731130017 Dhetax Erbian Pujangga

Jawaban soal no 1

Vektor perpindahan (2,2) ke (6,5)

    R             = ( X_ujung – X_pangkal)i + ( Y_ujung – Y_pangkal)j

                  = ( 6 – 2) i + ( 5 – 2 ) j

                  = 4i +3j

  Jarak Perpindahan

  R             =  √(4² + 3²)

                  = √(9 +16 )

                  = √25

                  = 5

Jawaban soal no 2

A = 4i + 3j + 2k

B = -4i + 2j

A x B        = (4i + 3j + 2k) x ( -4i + 2j)

                  = (4i x (-4i)) + (4i x 2j) + (3j x (-4i)) + (3j x 2j) + (2k x (-4i)) + (2k x 2j)

                  =  0 + 8k + 12k + 0 + (-8j) + (-4i)

                  = -4i -8j + 20k

A . B         = (4i + 3j + 2k) x ( -4i + 2j)

                  = (4i x (-4i)) + (4i x 2j) + (3j x (-4i)) + (3j x 2j) + (2k x (-4i)) + (2k x 2j)

                  =  -16 + 0 + 0 + 6 + 0 + 0

                  = -10

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