Категория:Задачник Кузнецова Аналитическая геометрия Задача 13

В этой категории варианты Задачи 13, раздела "Аналитическая геометрия" в задачнике Кузнецова Л.А.

Варианты[править]

Найти точку пересечения прямой и плоскости.

Задача	Условие
13-1	${\displaystyle {\frac {x-2}{-1}}={\frac {y-3}{-1}}={\frac {z+1}{4}}}$	${\displaystyle x+2y+3z-14=0}$
13-2	${\displaystyle {\frac {x+1}{3}}={\frac {y-3}{-4}}={\frac {z+1}{5}}}$	${\displaystyle x+2y-5z+20=0}$
13-3	${\displaystyle {\frac {x-1}{-1}}={\frac {y+5}{4}}={\frac {z-1}{2}}}$	${\displaystyle x-3y+7z-24=0}$
13-4	${\displaystyle {\frac {x-1}{1}}={\frac {y}{0}}={\frac {z+3}{2}}}$	${\displaystyle 2x-y+4z=0}$
13-5	${\displaystyle {\frac {x-5}{1}}={\frac {y-3}{-1}}={\frac {z-2}{0}}}$	${\displaystyle 3x+y-5z-12=0}$
13-6	${\displaystyle {\frac {x+1}{-3}}={\frac {y+2}{2}}={\frac {z-3}{-2}}}$	${\displaystyle x+3y-5z+9=0}$
13-7	${\displaystyle {\frac {x-1}{-2}}={\frac {y-2}{1}}={\frac {z+1}{-1}}}$	${\displaystyle x-2y+5z+17=0}$
13-8	${\displaystyle {\frac {x-1}{2}}={\frac {y-2}{0}}={\frac {z-4}{1}}}$	${\displaystyle x-2y+4z-19=0}$
13-9	${\displaystyle {\frac {x+2}{-1}}={\frac {y-1}{1}}={\frac {z+4}{-1}}}$	${\displaystyle 2x-y+3z+23=0}$
13-10	${\displaystyle {\frac {x+2}{1}}={\frac {y-2}{0}}={\frac {z+3}{0}}}$	${\displaystyle 2x-3y-5z-7=0}$
13-11	${\displaystyle {\frac {x-1}{2}}={\frac {y-1}{-1}}={\frac {z+2}{3}}}$	${\displaystyle 4x+2y-z-11=0}$
13-12	${\displaystyle {\frac {x-1}{1}}={\frac {y+1}{0}}={\frac {z-1}{-1}}}$	${\displaystyle 3x-2y-4z-8=0}$
13-13	${\displaystyle {\frac {x+2}{-1}}={\frac {y-1}{1}}={\frac {z+3}{2}}}$	${\displaystyle x+2y-z-2=0}$
13-14	${\displaystyle {\frac {x+3}{1}}={\frac {y-2}{-5}}={\frac {z+2}{3}}}$	${\displaystyle 5x-y+4z+3=0}$
13-15	${\displaystyle {\frac {x-2}{2}}={\frac {y-2}{-1}}={\frac {z-4}{3}}}$	${\displaystyle x+3y+5z-42=0}$
13-16	${\displaystyle {\frac {x-3}{-1}}={\frac {y-4}{5}}={\frac {z-4}{2}}}$	${\displaystyle 7x+y+4z-47=0}$
13-17	${\displaystyle {\frac {x+3}{2}}={\frac {y-1}{3}}={\frac {z-1}{5}}}$	${\displaystyle 2x+3y+7z-52=0}$
13-18	${\displaystyle {\frac {x-3}{2}}={\frac {y+1}{3}}={\frac {z+3}{2}}}$	${\displaystyle 3x+4y+7z-16=0}$
13-19	${\displaystyle {\frac {x-5}{-2}}={\frac {y-2}{0}}={\frac {z+4}{-1}}}$	${\displaystyle 2x-5y+4z+24=0}$
13-20	${\displaystyle {\frac {x-1}{8}}={\frac {y-8}{-5}}={\frac {z+5}{12}}}$	${\displaystyle x-2y-3z+18=0}$
13-21	${\displaystyle {\frac {x-3}{1}}={\frac {y-1}{-1}}={\frac {z+5}{0}}}$	${\displaystyle x+7y+3z+11=0}$
13-22	${\displaystyle {\frac {x-5}{-1}}={\frac {y+3}{5}}={\frac {z-1}{2}}}$	${\displaystyle 3x+7y-5z-11=0}$
13-23	${\displaystyle {\frac {x-1}{7}}={\frac {y-2}{1}}={\frac {z-6}{-1}}}$	${\displaystyle 4x+y-6z-5=0}$
13-24	${\displaystyle {\frac {x-3}{1}}={\frac {y+2}{-1}}={\frac {z-8}{0}}}$	${\displaystyle 5x+9y+4z-25=0}$
13-25	${\displaystyle {\frac {x+1}{-2}}={\frac {y}{0}}={\frac {z+1}{3}}}$	${\displaystyle x+4y+13z-23=0}$
13-26	${\displaystyle {\frac {x-1}{6}}={\frac {y-3}{1}}={\frac {z+5}{3}}}$	${\displaystyle 3x-2y+5z-3=0}$
13-27	${\displaystyle {\frac {x-2}{4}}={\frac {y-1}{-3}}={\frac {z+3}{-2}}}$	${\displaystyle 3x-y+4z=0}$
13-28	${\displaystyle {\frac {x-1}{2}}={\frac {y+2}{-5}}={\frac {z-3}{-2}}}$	${\displaystyle x+2y-5z+16=0}$
13-29	${\displaystyle {\frac {x-1}{1}}={\frac {y-3}{0}}={\frac {z+2}{-2}}}$	${\displaystyle 3x-7y-2z+7=0}$
13-30	${\displaystyle {\frac {x+3}{0}}={\frac {y-2}{-3}}={\frac {z+5}{11}}}$	${\displaystyle 5x+7y+9z-32=0}$
13-31	${\displaystyle {\frac {x-7}{3}}={\frac {y-3}{1}}={\frac {z+1}{-2}}}$	${\displaystyle 2x+y+7z-3=0}$