Jawaban 1731130094 Muhammad Fahreza Nurhidayat
Soal 1 soal 1
Jawaban :
AB = harga absolut vektor AB =(3^2+4^2+5^2)^1/2
= 7,05 meter
AC = harga absolut vektor AC = (2^2+3^2+3^2)^1/2
= 4,69 meter
cos(teta) = (3)(2)+(4)(3)+(5)(3)/(7,05)(4,69)
cos(teta) = 33/33,06 = 0,998
cos^2(teta)=0,996
Dengan demikian kita peroleh
luas segitiga ABC = 1/2AB.AC.sin(teta)=1/2AB.AC.(1-cos^2(teta))^1/2
= (2,05)(4,69)(0,0634)/2 = 1,0455m^2
Soal 2 soal 2
Jawaban :
a) AxB = (4i+6j)x(5i-3j+k)
= (0-12k-j)x(-30k-0+6i)
= 6i-j+42k
b) AxB = -BxA
(4i+6j x 5i-3j+k) = -(5i-3j+k x 4i+6j)
(0-12k-j-30k-0+6i) = -(0+30k+12k-0+4j-6i)
6i+j+42k = 6i+4j+42k
Jawaban :
AB = harga absolut vektor AB =(3^2+4^2+5^2)^1/2
= 7,05 meter
AC = harga absolut vektor AC = (2^2+3^2+3^2)^1/2
= 4,69 meter
cos(teta) = (3)(2)+(4)(3)+(5)(3)/(7,05)(4,69)
cos(teta) = 33/33,06 = 0,998
cos^2(teta)=0,996
Dengan demikian kita peroleh
luas segitiga ABC = 1/2AB.AC.sin(teta)=1/2AB.AC.(1-cos^2(teta))^1/2
= (2,05)(4,69)(0,0634)/2 = 1,0455m^2
Soal 2 soal 2
Jawaban :
a) AxB = (4i+6j)x(5i-3j+k)
= (0-12k-j)x(-30k-0+6i)
= 6i-j+42k
b) AxB = -BxA
(4i+6j x 5i-3j+k) = -(5i-3j+k x 4i+6j)
(0-12k-j-30k-0+6i) = -(0+30k+12k-0+4j-6i)
6i+j+42k = 6i+4j+42k