Hobbs O'Quinn
05/31/2023 · Junior High School

5. Quyidagi to'g'ri chiziqlarning o'zaro vaziyatini aniqlang: \( \left\{\begin{array}{c}x-2 y+3 z+4=0 \\ x+2 y-5 z+1=0\end{array}\right. \) va \( \left\{\begin{array}{c}x+y+z+1=0 \\ y-3 z=0\end{array}\right. \)

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Berilgan to'g'ri chiziqlarni aniqlash uchun, har bir to'g'ri chiziqning parametrik tenglamalarini yozamiz va ularning o'zaro vaziyatini tahlil qilamiz. 1. **Birinchi to'g'ri chiziq:** \[ \begin{cases} x - 2y + 3z + 4 = 0 \\ x + 2y - 5z + 1 = 0 \end{cases} \] Bu tenglamalarni yechamiz. Birinchi tenglamadan \( x \) ni ifodalash: \[ x = 2y - 3z - 4 \] Ikkinchi tenglamaga qo'yamiz: \[ (2y - 3z - 4) + 2y - 5z + 1 = 0 \] \[ 4y - 8z - 3 = 0 \] \[ 4y = 8z + 3 \quad \Rightarrow \quad y = 2z + \frac{3}{4} \] Endi \( y \) ni \( z \) ga bog'lab, \( x \) ni ifodalashimiz mumkin: \[ x = 2(2z + \frac{3}{4}) - 3z - 4 \] \[ x = 4z + \frac{3}{2} - 3z - 4 = z - \frac{5}{2} \] Natijada, birinchi to'g'ri chiziqning parametrik tenglamalari: \[ \begin{cases} x = z - \frac{5}{2} \\ y = 2z + \frac{3}{4} \\ z = z \end{cases} \] Bu yerda \( z \) parametr sifatida olinadi. 2. **Ikkinchi to'g'ri chiziq:** \[ \begin{cases} x + y + z + 1 = 0 \\ y - 3z = 0 \end{cases} \] Ikkinchi tenglamadan \( y \) ni ifodalash: \[ y = 3z \] Birinchi tenglamaga qo'yamiz: \[ x + 3z + z + 1 = 0 \] \[ x + 4z + 1 = 0 \quad \Rightarrow \quad x = -4z - 1 \] Natijada, ikkinchi to'g'ri chiziqning parametrik tenglamalari: \[ \begin{cases} x = -4z - 1 \\ y = 3z \\ z = z \end{cases} \] 3. **O'zaro vaziyatni aniqlash:** Har ikkala to'g'ri chiziqning parametrik tenglamalarini ko'rib chiqamiz: - Birinchi to'g'ri chiziq: \[ \begin{cases} x = z - \frac{5}{2} \\ y = 2z + \frac{3}{4} \\ z = z \end{cases} \] - Ikkinchi to'g'ri chiziq: \[ \begin{cases} x = -4z - 1 \\ y = 3z \\ z = z \end{cases} \] Har ikkala to'g'ri chiziqning tenglamalarini tenglashtiramiz: \[ z - \frac{5}{2} = -4z - 1 \quad \Rightarrow \quad 5z = \frac{3}{2} \quad \Rightarrow \quad z = \frac{3}{10} \] \( z \) ni topgach, \( y \) va \( x \) ni hisoblaymiz: - Birinchi to'g'ri chiziqdan: \[ y = 2 \cdot \frac{3}{10} + \frac{3}{4} = \frac{6}{10} + \frac{30}{40} = \frac{6}{10} + \frac{15}{20} = \frac{6 + 15}{

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Birinchi to'g'ri chiziq va ikkinchi to'g'ri chiziq o'zaro kesishadi.
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