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ત્રિપરિમાણીય ભૂમિતિ

(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.

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

અહીં, $straight a with bar on top space equals space open parentheses straight x subscript 1 comma space straight y subscript 1 comma space straight z subscript 1 close parentheses space equals space open parentheses 2 comma space 3 comma space minus 9 close parentheses space space bold તથ ા space space straight b with bar on top space equals space open parentheses straight x subscript 2 comma space straight y subscript 2 comma space straight z subscript 2 close parentheses space equals space open parentheses 4 comma space 3 comma space minus 5 close parentheses$ છે.

સંમિત સ્વરૂપે રેખાના સમીકરણ

$fraction numerator straight x space minus space straight x subscript 1 over denominator straight x subscript 2 space minus space straight x subscript 1 end fraction space equals space fraction numerator straight y space minus space straight y subscript 1 over denominator straight y subscript 2 space minus space straight y subscript 1 end fraction space equals space fraction numerator straight z space minus space straight z subscript 1 over denominator straight z subscript 2 space minus space straight z subscript 1 end fraction$ અનુસાર,

$fraction numerator straight x space minus space 2 over denominator 4 space minus space 2 end fraction space equals space fraction numerator y space minus space 3 over denominator 3 space minus space 3 end fraction space equals space fraction numerator z space minus space open parentheses negative 9 close parentheses over denominator negative 5 space minus space open parentheses negative 9 close parentheses end fraction therefore bold space fraction numerator bold x bold space bold minus bold space bold 2 over denominator bold 2 end fraction bold space bold equals bold space fraction numerator bold z bold space bold plus bold space bold 9 over denominator bold 4 end fraction bold comma bold space bold italic y bold space bold minus bold space bold 3 bold space bold equals bold space bold 0$

( y - 3 ના છેદમાં 3 - 3 = 0 )

તથા સદિશ સ્વરૂપે રેખાના સમીકરણ

$straight r with bar on top space equals space straight a with bar on top space plus space straight k space open parentheses straight b with bar on top space minus space straight a with bar on top close parentheses comma space straight k space element of space straight R$ અનુસાર

$straight r with bar on top space equals space open parentheses 2 comma space 3 comma space minus 9 close parentheses space plus space straight k space open parentheses 4 comma space 3 comma space minus 5 close parentheses space minus space open parentheses 2 comma space 3 comma space minus 9 close parentheses bold therefore bold r with bold bar on top bold space bold equals bold space open parentheses bold 2 bold comma bold space bold 3 bold comma bold space bold minus bold 9 close parentheses bold space bold plus bold space bold k bold space open parentheses bold 2 bold comma bold space bold 0 bold comma bold space bold 4 close parentheses bold comma bold space bold k bold space bold element of bold space bold R bold space bold થ ા ય bold.$

(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.

(0, 1, 1), (0, 4, 4) અને (2, 0, 1) સમરેખ છે ? શા માટે ?

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

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ત્રિપરિમાણીય ભૂમિતિ

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