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ત્રિપરિમાણીય ભૂમિતિ

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ગણિત ધોરણ 12 સિમેસ્ટર 4

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ગણિત

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ત્રિપરિમાણીય ભૂમિતિ

(1, -2, 1)માંથી પસાર થતી અને રેખાઓ straight x space plus space 3 space equals space 2 straight y space equals space minus 2 straight z space space bold તથ ા bold space space straight x over 2 space equals space fraction numerator straight y space plus space 6 over denominator 2 end fraction space equals space fraction numerator 3 straight z space minus space 9 over denominator 1 end fraction  બંનેને લંબરેખાનું સદિશ તેમજ કાર્તેઝીય સમીકરણ મેળવો.

Answer

    • (2, 3, -9) અને (4, 3, -5)માંથી પસાર થતી રેખાનું સંમિત સ્વરૂપે અને સદિશ સ્વરૂપે સમીકરણ મેળવો.

    Answer

    • (2, -1, 3) માંથી પસાર થતી અને 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top દિશાવાળી રેખાનું સદિશ તેમજ કાર્તેઝીય સમીકરણ મેળવો.

    Answer
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    રેખા x = 4z + 3, y = 2 - 3zની દિકકોસાઈન શોધો.

    straight x space equals space 4 straight z space plus space 3 space space space space space space space space space space space space space space space space space bold તથ ા bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space straight y space equals space 2 space minus space 3 straight z therefore space straight z space equals space fraction numerator straight x space minus space 3 over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space therefore space straight z space equals space fraction numerator 2 space minus space straight y over denominator 3 end fraction therefore space fraction numerator straight x space minus space 3 over denominator 4 end fraction space equals space fraction numerator 2 space minus space straight y over denominator 3 end fraction space equals space straight z therefore space fraction numerator straight x space minus space 3 over denominator 4 end fraction space equals space fraction numerator straight y space minus space 2 over denominator negative 3 end fraction space equals space fraction numerator straight z space minus space 0 over denominator 1 end fraction



    આ સમીકરણને રેખાના કાર્તેઝીય સમીકરણ,


    fraction numerator straight x space minus space straight x subscript 1 over denominator l subscript 1 end fraction space equals space fraction numerator y space minus space y subscript 1 over denominator l subscript 2 end fraction space equals space fraction numerator z space minus space z subscript 1 over denominator l subscript 3 end fraction સાથે સરખાવતાં,


    રેખાની દિશા l with italic bar on top space equals space open parentheses l subscript 1 comma space l subscript 2 comma space l subscript 3 close parentheses space equals space open parentheses 4 comma space minus 3 comma space 1 close parentheses થાય.


    therefore space open vertical bar space stack l italic space with bar on top close vertical bar space equals space square root of 4 squared space plus space open parentheses negative 3 close parentheses squared space plus space 1 squared end root space equals space square root of 26


    therefore space l with bar on top ની દિશામાં એકમ સદિશ equals space fraction numerator l with bar on top over denominator open vertical bar space l with bar on top space close vertical bar end fraction


                                       equals space fraction numerator open parentheses 4 comma space minus 3 comma space 1 close parentheses over denominator square root of 26 end fraction equals space open parentheses fraction numerator 4 over denominator square root of 26 end fraction comma space minus fraction numerator 3 over denominator square root of 26 end fraction comma space plus fraction numerator 1 over denominator square root of 26 end fraction close parentheses



    આમ, રેખાની દિકકોસાઈન એટલે કે l with italic bar on top ની ધન દિશાનો એકમ સદિશ fraction numerator bold 4 over denominator square root of bold 26 end fraction bold comma bold space fraction numerator bold minus bold 3 over denominator square root of bold 26 end fraction bold comma bold space fraction numerator bold 1 over denominator square root of bold 26 end fraction  થશે.
    Answer
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    (0, 1, 1), (0, 4, 4) અને (2, 0, 1) સમરેખ છે ? શા માટે ?

    Answer
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