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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.

અહીં, $straight a with bar on top space equals space open parentheses x subscript 1 comma space y subscript 1 comma space z subscript 1 close parentheses space equals space open parentheses 2 comma space minus 1 comma space 3 close parentheses$

તથા $straight l with bar on top space equals space open parentheses l subscript 1 comma space l subscript 2 comma space l subscript 3 close parentheses space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top space equals space open parentheses 2 comma space minus 3 comma space 4 close parentheses$

સદિશ સ્વરૂપે રેખાના સમીકરણ $straight r with bar on top space equals space straight a with bar on top space plus space straight k space straight l with bar on top comma space straight k space element of space straight R$ અનુસાર,

$bold r with bold bar on top bold space bold equals bold space open parentheses bold 2 bold comma bold space bold minus bold 1 bold comma bold space bold 3 close parentheses bold space bold plus bold space bold k bold space open parentheses bold 2 bold comma bold space bold minus bold 3 bold comma bold space bold 4 close parentheses bold comma bold space bold k bold space bold element of bold space bold R$ થાય.

તથા કાર્તેઝીય સ્વરૂપે રેખાના સમીકરણ

$fraction numerator straight x space minus space straight x subscript 1 over denominator l subscript 1 end fraction space equals space fraction numerator y space minus space y subscript 1 over denominator l subscript 2 end fraction space equals space fraction numerator z space minus space z subscript 1 over denominator l subscript 3 end fraction$ અનુસાર

$fraction numerator straight x space minus space 2 over denominator 2 end fraction space equals space fraction numerator y space minus space open parentheses negative 1 close parentheses over denominator negative 3 end fraction space equals space fraction numerator z space minus space 3 space over denominator 4 end fraction$

$bold therefore bold space fraction numerator bold x bold space bold minus bold space bold 2 over denominator bold 2 end fraction bold space bold equals bold space fraction numerator bold y bold space bold plus bold space bold 1 over denominator bold minus bold 3 end fraction bold space bold equals bold space fraction numerator bold z bold space bold minus bold space bold 3 over denominator bold 4 end fraction$ થાય.

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

(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.

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

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

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ત્રિપરિમાણીય ભૂમિતિ

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