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please solve

Asked by swayamagarwal2114 | 21 Jul, 2022, 17:18: PM

Expert Answer

Let space straight l space be space the space line space of space intersection space of space the space planes. Let space straight P subscript 1 colon space straight x minus 2 straight y plus 3 straight z equals 0 space space and space space straight P subscript 2 colon 2 straight x plus 3 straight y minus 4 straight z equals 0 Considering space straight a comma straight b comma space straight c space as space the space direction space ratios space we space have straight a minus 2 straight b plus 3 straight z equals 0 space space and space space 2 straight a plus 3 straight b minus 4 straight c equals 0 rightwards double arrow fraction numerator straight a over denominator open vertical bar table row cell negative 2 end cell 3 row 3 cell negative 4 end cell end table close vertical bar end fraction equals fraction numerator negative straight b over denominator open vertical bar table row 1 3 row 2 cell negative 4 end cell end table close vertical bar end fraction equals fraction numerator straight c over denominator open vertical bar table row 1 cell negative 2 end cell row 2 3 end table close vertical bar end fraction rightwards double arrow fraction numerator straight a over denominator negative 1 end fraction equals straight b over 10 equals straight c over 7 Required space line space be space straight L colon fraction numerator straight x minus 7 over denominator straight l end fraction equals fraction numerator straight y minus 2 over denominator straight m end fraction equals fraction numerator straight z plus 1 over denominator straight n end fraction Here comma space straight l comma space straight m comma space straight n space are space the space direction space ratios space of space the space required space line rightwards double arrow straight l minus 2 straight m plus 3 straight n equals 0 space space and space space minus straight l plus 10 straight m plus 7 straight n equals 0 rightwards double arrow fraction numerator straight l over denominator open vertical bar table row cell negative 2 end cell 3 row 10 7 end table close vertical bar end fraction equals fraction numerator negative straight m over denominator open vertical bar table row 1 3 row cell negative 1 end cell 7 end table close vertical bar end fraction equals fraction numerator straight n over denominator open vertical bar table row 1 cell negative 2 end cell row cell negative 1 end cell 10 end table close vertical bar end fraction rightwards double arrow fraction numerator straight l over denominator negative 44 end fraction equals fraction numerator negative straight m over denominator 10 end fraction equals straight n over 8 rightwards double arrow straight l over 22 equals straight m over 5 equals fraction numerator straight n over denominator negative 4 end fraction So comma space the space required space line space is space fraction numerator straight x minus 7 over denominator 22 end fraction equals fraction numerator straight y minus 2 over denominator 5 end fraction equals fraction numerator straight z plus 1 over denominator negative 4 end fraction

Answered by Renu Varma | 21 Jul, 2022, 18:30: PM

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