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ત્રિપરિમાણીય ભૂમિતિ

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

(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.

આપેલી રેખાઓ x + 3 = 2y = - 12z એટલે કે, $fraction numerator straight x space minus space open parentheses negative 3 close parentheses over denominator 1 end fraction space equals space fraction numerator y space minus space 0 over denominator begin display style 1 half end style end fraction space equals space fraction numerator z space minus space 0 over denominator negative begin display style 1 over 12 end style end fraction$

તથા $fraction numerator straight x space minus space 0 over denominator 2 end fraction space equals space fraction numerator y space minus space open parentheses negative 6 close parentheses over denominator 2 end fraction space equals space fraction numerator z space minus space 3 over denominator begin display style 1 third end style end fraction$ માટે,

$straight a with bar on top space equals space open parentheses negative 3 comma space 0 comma space 0 close parentheses comma space straight b with bar on top space equals space open parentheses 0 comma space minus 6 comma space 3 close parentheses$

$straight l with bar on top space equals space open parentheses 1 comma space 1 half comma space minus 1 over 12 close parentheses space space bold અન ે bold space space straight m with bar on top space equals space open parentheses 2 comma space 2 comma space 1 half close parentheses$ મળે.

માગેલી રેખા એ આપેલી બંને રેખાઓને લંબ હોવાથી માગેલી રેખાની દિશા આપેલી બંને રેખાઓની દિશાને લંબ થાય.

જો માગેલી રેખાને દિશા $straight n with bar on top$ હોય, તો

$straight n with bar on top space equals space l with bar on top space cross times space m with bar on top space equals space open parentheses 1 comma space 1 half comma space minus 1 over 12 close parentheses space cross times space open parentheses 2 comma space 2. space 1 third close parentheses$

$equals space open parentheses open vertical bar table row cell 1 half end cell cell negative 1 over 12 end cell row 2 cell 1 third end cell end table close vertical bar comma space minus open vertical bar table row 1 cell negative 1 over 12 end cell row 2 cell 1 third end cell end table close vertical bar comma space open vertical bar table row 1 cell 1 half end cell row 2 2 end table close vertical bar close parentheses equals space open parentheses 1 third comma space minus 1 half comma space 1 close parentheses$

આમ, આપેલ બિંદુ $straight c with bar on top space equals space open parentheses 1 comma space minus 2 comma space 1 close parentheses$ માંથી પસાર થતી તથા $straight n with bar on top space equals space open parentheses 1 third comma space minus 1 half comma space 1 close parentheses$ દિશાવાળી રેખાન સદિશ સમીકરણ $straight r with bar on top space equals space straight c with bar on top space plus space straight k space straight n with bar on top space. space straight k space element of space straight R$ અનુસાર,

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

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

$fraction numerator straight x space minus space 1 over denominator begin display style 1 third end style end fraction space equals space fraction numerator y space minus space open parentheses negative 2 close parentheses over denominator negative begin display style 1 half end style end fraction space equals space fraction numerator z space minus space 1 over denominator 1 end fraction$

$bold therefore bold space fraction numerator bold 3 bold space open parentheses bold x bold space bold minus bold space bold 1 close parentheses over denominator bold 1 end fraction bold space bold equals bold space fraction numerator bold 2 bold space open parentheses bold y bold space bold plus bold space bold 2 close parentheses over denominator bold minus bold 1 end fraction bold space bold equals bold space fraction numerator bold z bold space bold minus bold space bold 1 over denominator bold 1 end fraction$ છે.

(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.

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