Ikki chiziq orasidagi burchakning formulasi. Tekislikdagi chiziqlar orasidagi burchak

1-topshiriq Chiziqlar orasidagi burchakni hisoblang
$$ \\ frac (x + 3) (- \\ sqrt2) \u003d \\ frac (y) (\\ sqrt2) \u003d \\ frac (z-7) (- 2) \\; \\; va \\; \\; \\ frac (x) (\\ sqrt3) \u003d \\ frac (y + 1) (\\ sqrt3) \u003d \\ frac (z-1) (\\ sqrt6) $$
Chiziqlarning yo'naltiruvchi vektorlari quyidagi koordinatlarga ega:
a \u003d (-√2; √2; -2), b = (√3 ; √3 ; √6 ).
(1) formula bo'yicha topamiz
$$ cos \\ phi \u003d \\ frac (| - \\ sqrt6 + \\ sqrt6-2 \\ sqrt6 |) (\\ sqrt (2 + 2 + 4) \\ sqrt (3 + 3 + 6)) \u003d \\ frac (2 \\ sqrt6) ( 2 \\ sqrt2 \\ cdot 2 \\ sqrt3) \u003d \\ frac (1) (2) $$
Shuning uchun bu chiziqlar orasidagi burchak 60 ° dir.
2-topshiriq Chiziqlar orasidagi burchakni hisoblang
$$ \\ boshlash (holatlar) 3x-12z + 7 \u003d 0 \\\\ x + y-3z-1 \u003d 0 \\ end (case) va \\ begin (case) 4x-y + z \u003d 0 \\\\ y + z + 1 \u003d 0 \\ oxiri (holatlar) $$
Qo'llanma vektori orqasida lekin  birinchi satrda normal vektorlarning vektor mahsulotini olamiz n  1 \u003d (3; 0; -12) va n  2 \u003d (1; 1; -3) bu chiziqni belgilaydigan tekisliklar. \\ (\u003d \\ Start (vmatrix) formulasi bo'yicha i & j & k \\\\ x_1 & y_1 & z_1 \\\\ x_2 & y_2 & z_2 \\ end (vmatrix) \\) formulasi bo'yicha
$$ a \u003d\u003d \\ start (vmatrix) i & j & k \\\\ 3 & 0 & -12 \\\\ 1 & 1 & -3 \\ end (vmatrix) \u003d 12i-3i + 3k $$
Xuddi shunday, biz ikkinchi qatorning yo'nalish vektorini topamiz:
$$ b \u003d \\ start (vmatrix) i & j & k \\\\ 4 & -1 & 1 \\\\ 0 & 1 & 1 \\ end (vmatrix) \u003d - 2i-4i + 4k $$
Ammo (1) formulaga biz kerakli burchakning kosinusini hisoblaymiz:
$$ cos \\ phi \u003d \\ frac (| 12 \\ cdot (-2) -3 (-4) +3 \\ cdot 4 |) (\\ sqrt (12 ^ 2 + 3 ^ 2 + 3 ^ 2) \\ sqrt (2 ^ 2 + 4 ^ 2 + 4 ^ 2)) \u003d 0 $$

Shuning uchun bu chiziqlar orasidagi burchak 90 ° dir.
3-topshiriq. MABC uchburchagi piramidasida MA, MB va MS qovurg'alari o'zaro perpendikulyar (207-rasm);

ularning uzunligi mos ravishda 4, 3, 6 bo'ladi. D nuqta - bu [MA] ning o'rtasi. CA va DB chiziqlari orasidagi φ burchakni toping.
CA va DB chiziqlari CA va DB yo'nalishlarining vektorlari bo'lsin.
M nuqtasini kelib chiqishi sifatida oling. Vazifaning shartiga ko'ra bizda A (4; 0; 0), B (0; 0; 3), C (0; 6; 0), D (2; 0; 0). Shuning uchun, \\ (\\ haddan tashqari yuk (CA) \\) \u003d (4; - 6; 0), \\ (\\ haddan tashqari yuk (DB) \\) \u003d (-2; 0; 3). Biz (1) formuladan foydalanamiz:
$$ cos \\ phi \u003d \\ frac (| 4 \\ cdot (-2) + (- 6) \\ cdot 0 + 0 \\ cdot 3 |) (\\ sqrt (16 + 36 + 0) \\ sqrt (4 + 0 + 9) )) $$
Kozinalar jadvaliga ko'ra, biz CA va DB chiziqlari orasidagi burchak taxminan 72 ° ekanligini aniqlaymiz.
Ohhhhhh ... yaxshi, qalay, go'yo hukmni o'zingiz o'qigandek \u003d) Ammo, keyin dam olish yordam beradi, ayniqsa bugungi kunga kelib men mos aksessuarlarni sotib oldim. Shuning uchun biz birinchi qismga o'tamiz, umid qilamanki, maqolaning oxiriga qadar men quvnoq kayfiyatni saqlab qolaman.