Soal UN Matematika IPA 2015
16. Jika sudut antara vektor a = i + j - r . k , dan
vektor b = r . i - r . j - 2 . k yakni 60° maka nilai r yang memenuhi yakni ... (i, j, k yakni vektor)
A.√2
B. 1
C. 0
D. -1
E. -√2
Jawaban : A
a = ( 1, 1 , -r)
b = (r, -r , -2)
a . b = |a| x |b| x cos 60
(r - r + 2r) = |a| x |b| x 1/2
2r = √(2+r2) . √(4 + 2r2) . 1/2
4r = √(2+r2) . √2 √(2 + r2)
4r = (2+r2) .√2
4r = 2√2 + √2r2
0 = √2r2 - 4r + 2√2
0 = (√2r - 2) (r - √2)
√2r - 2 = 0
√2r = 2
r = 2/√2 = √2
r - √2 = 0
r = √2
16. Jika sudut antara vektor a = i + j - r . k , dan
vektor b = r . i - r . j - 2 . k yakni 60° maka nilai r yang memenuhi yakni ... (i, j, k yakni vektor)
A.√2
B. 1
C. 0
D. -1
E. -√2
Jawaban : A
a = ( 1, 1 , -r)
b = (r, -r , -2)
a . b = |a| x |b| x cos 60
(r - r + 2r) = |a| x |b| x 1/2
2r = √(2+r2) . √(4 + 2r2) . 1/2
4r = √(2+r2) . √2 √(2 + r2)
4r = (2+r2) .√2
4r = 2√2 + √2r2
0 = √2r2 - 4r + 2√2
0 = (√2r - 2) (r - √2)
√2r - 2 = 0
√2r = 2
r = 2/√2 = √2
r - √2 = 0
r = √2
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