Diketahui tiga buah vektor:
A = i + 2j - k
B = 4i + 2j + 3k
C = 2j - 3k
Tentukan:
a. A ∙ ( B × C )
b. A ∙ ( B + C )
c. A × ( B + C )
Pembahasan:
a. B × C
= ( 4i + 2j + 3k ) × ( 2j -3k )
= ( 4i × 2j ) + (4i × (-3k)) +
= 8k + 12j - 6i - 6i
= -12i + 12j + 8k
A ∙ ( B × C )
= ( i + 2j - k ) ∙ ( -12i + 12j + 8k )
= ( 1 ∙ (-12)) + ( 2 ∙ 12 ) + ((-1) ∙ (8))
= -12 + 24 - 8
= 4
b. B + C
= ( 4i + 2j + 3k ) + ( 2j -3k )
= 4i + 4j
A ∙ ( B + C )
= ( i + 2j - k ) ∙ ( 4i + 4j )
= ( 1 ∙ 4 ) + ( 2 ∙ 4 ) + ((-1) ∙ 0 )
= 4 + 8
= 12
c. A × ( B + C )
= ( i + 2j - k ) × ( 4i + 4j )
=
= 4k - 8k - 4j + 4i
= 4i - 4j - 4k
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