Fazoda vektorlar
Javoblar ota-onalar va o‘qituvchilar uchun.
1
A(2; −3; 1), B(−1; 1; 13) bo‘lsa, AB vektor koordinatalarini va uzunligini toping.
AB = (−3; 4; 12), |AB| = √(9 + 16 + 144) = 13.
2
a(1; 1; 0) va b(0; 1; 1) vektorlar orasidagi burchak kosinusini va burchakni toping.
a · b = 1, |a| = |b| = √2, cos φ = 1/2, φ = 60°.
3
a(m; 2; −3) va b(4; m; 2) vektorlar perpendikular bo‘ladigan m ni toping.
4m + 2m − 6 = 0 ⇒ m = 1.
4
a(3; −6; 9) va b(−1; 2; −3) kollinearmi? Agar ha bo‘lsa, a = λb dagi λ ni toping.
Ha: koordinatalar nisbati 3/(−1) = −6/2 = 9/(−3) = −3, ya’ni a = −3b.
5
|a| = 3, |b| = 4 va ular orasidagi burchak 60° bo‘lsa, a · b va |a + b|² ni toping.
a · b = 12 · 1/2 = 6. |a + b|² = 9 + 2 · 6 + 16 = 37.