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NCERT Solutions for Class 11 Maths In Hindi Chapter 12 Introduction To Three Dimensional Geometry

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NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry in Hindi PDF Download

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Access NCERT Solutions for Class 12 Maths Chapter 12–त्रिविमीय ज्यामिति की परिभाषा

प्रश्नावली 12.1

1. एक बिन्दु \[x\,\,\, - \,\]अक्ष पर स्थित है। इस के \[y\,\,\, - \,\] निर्देशांक तथा \[z\,\,\, - \,\] निर्देशांक क्या है ?

उत्तर: \[x\,\,\, - \,\] अक्ष पर किसी बिन्दु के निर्देशांक $\left({\,\,x\,\,,\,\,0\,\,,\,\,0\,\,} \right)$ होते हैं जिसमें $y\,\, = \,\,0\,\,,\,\,z\,\, = \,\,0\,\,.$


2. एक बिन्दु $x$ तल में है। इसके $y\,\, - $ निर्देशांक के बारे में आप क्या कह सकते हैं?

उत्तर: $x\,\,z$ तल $y\,\, - $ निर्देशांक $0$ होता है। इस तल का बिन्दु $\left( {\,x\,\,,\,\,0\,\,,\,\,z\,} \right)$ के रूप में होता है।


3. अष्टाशों के नाम बताइए, जिनमें निम्नलिखित बिन्दु स्थित हैं:

$\mathbf{(\,\,1\,\,,\,\,2\,\,,\,\,3\,\,)\,\,,\,\,(\,\,4\,\,,\,\, - \,\,2\,\,,\,\,3\,\,)\,\,,\,\,(\,\,4\,\,,\,\, - \,\,2\,\,,\,\, - \,\,5\,\,)\,\,,\,\,(\,\,4\,\,,\,\,2\,\,,\,\, - \,\,5\,\,)\,\,,\,\,(\, - \,\,4\,\,,\,\,2\,\,,\,\, - \,\,5\,\,)\,\,,\,\,(\,\, - \,\,4\,\,,\,\,2\,\,,\,\,5\,\,)\,\,,\,\,}$

$\,athbf{(\,\, - \,\,3\,\,,\,\, - \,\,1\,\,,\,\,6\,\,)\,\,,\,\,(\,\,2\,\,,\,\, - \,\,4\,\,,\,\, - \,\,7\,\,)}$

उत्तर: दिए गए बिंदुओं के अंश है :

\[(i)\,\,\,\,(1,2,3)\,\, - \,\,XOYZ\,\, - \]  पहला 

\[(ii)\,\,\,\,(4, - 2,3)\,\, - \,\,{\text{XOYZ}}\,\, - \]  चौथा 

\[(iii)\,\,\,\,(4,2, - 5)\,\, - \,XOY'Z'\,\, - \]  आठवाँ 

\[(iv)\,\,\,\,(4,2, - 5)\,\, - \,\,XOYZ'\,\, - \]   पाचवाँ 

\[(v)\,\,\,\,\,( - 4,2, - 5)\,\, - \,\,XOYZ'\,\, - \]   छठा 

\[(vi)\,\,\,\,( - 4,2,5)\,\, - \,\,(XOYZ)\,\, - \]   दूसरा 

\[(vii)\,\,\,\,( - 3, - 1,6)\,\, - \,\,(XOY'Z)\,\, - \]  तीसरा 

\[(viii)\,\,\,(2, - 4. - 7)\,\, - \,\,(XOY'Z')\,\, - \]  आठवाँ 


4. रिक्त स्थानों की पूर्ति कीजिए:

(i)$x\,\, - $ अक्ष और $y\,\, - \,$ अक्ष दोनों एक साथ मिल कर एक तल बनाते हैं, उस तल को......कहते हैं।

उत्तर :  $X\,\,Y\,\, - $ तल  |

(ii). $X\,\,Y\,\, - $तल में एक बिन्दु के निर्देशांक........ रूप के होते हैं।

उत्तर :  $x\,\,,\,\,y\,\,,\,\,o$

(iii). निर्देशांक तल अंतरिक्ष को......... अष्टांश में विभाजित करते हैं।

उत्तर :  $8$


प्रश्नावली 12.2

1. निम्लिखित बिंदु -युग्मों के बीच की दूरी ज्ञात कीजिए:

(i). $\mathbf{\left( {\,\,2\,\,,\,\,3\,\,,\,\,5\,\,} \right)}$  और $\mathbf{\left( {\,4\,,\,\,3\,\,,\,\,1\,} \right)}$

उत्तर: $\left( {\,\,2\,\,,\,\,3\,\,,\,\,5\,\,} \right)$  और $\left( {\,4\,,\,\,3\,\,,\,\,1\,} \right)$

दो बिन्दु $\left( {\,{x_1},{y_1},{z_1}\,} \right)$ और  $\left( {\,{x_2},{y_2},{z_2}\,} \right)$ के बीच की दूरी 

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

बिन्दु $\left( {\,\,2\,\,,\,\,3\,\,,\,\,5\,\,} \right)$ और  $\left( {\,4\,,\,\,3\,\,,\,\,1\,} \right)$ के बीच की दूरी

$= \sqrt {{{(4 - 2)}^2} + {{(3 - 3)}^2} + {{(1 - 5)}^2}}$

$= \sqrt {{2^2} + {0^2} + {4^2}}$

$= \sqrt {4 + 16}$

$= \sqrt {20}$

$= 2\sqrt 5$ 

(ii). $\mathbf{\left( {\,\, - \,\,3\,\,,\,\,7\,\,,\,\,2\,\,} \right)}$ और $\mathbf{\left( {\,\,2\,\,,\,\,4\,\,,\,\, - \,\,1\,} \right)}$

उत्तर:   $\left( {\,\, - \,\,3\,\,,\,\,7\,\,,\,\,2\,\,} \right)$ और $\left( {\,\,2\,\,,\,\,4\,\,,\,\, - \,\,1\,} \right)$

दो बिन्दु $\left( {\,{x_1},{y_1},{z_1}\,} \right)$ और  $\left( {\,{x_2},{y_2},{z_2}\,} \right)$ के बीच की दूरी 

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

बिन्दु $\left( {\,\, - \,\,3\,\,,\,\,7\,\,,\,\,2\,\,} \right)$ और $\left( {\,\,2\,\,,\,\,4\,\,,\,\, - \,\,1\,} \right)$ के बीच की दूरी

$= \sqrt {{{(2 + 3)}^2} + {{(4 - 7)}^2} + {{(-1 - 2)}^2}}$

$= \sqrt {25 + 9 + 9}$

$= \sqrt {43}$ 

(iii). $\mathbf{\left( {\,\, - \,\,1\,\,,\,\,3\,\,,\,\, - \,4\,\,} \right)$ और $\left( {\,\,1\,\,,\,\, - \,3\,\,,\,\,4\,} \right)}$

उत्तर:  $\left( {\,\, - \,\,1\,\,,\,\,3\,\,,\,\, - \,4\,\,} \right)$ और  $\left( {\,\,1\,\,,\,\, - \,3\,\,,\,\,4\,} \right)$

दो बिन्दु $\left( {\,{x_1},{y_1},{z_1}\,} \right)$ और  $\left( {\,{x_2},{y_2},{z_2}\,} \right)$ के बीच की दूरी 

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

बिन्दु $\left( {\,\, - \,\,1\,\,,\,\,3\,\,,\,\, - \,4\,\,} \right)$ और $\left( {\,\,1\,\,,\,\, - \,3\,\,,\,\,4\,} \right)$ के बीच की दूरी

$= \,\,\sqrt {\,\,{{(\,\,1\,\, + \,\,1\,\,\,)}^2}\,\, + \,\,\,{{(\,\, - \,\,3\,\, - \,\,3\,\,)}^2}\,\, + \,\,{{(\,4\, + \,4\,)}^2}}$

$= \,\,\sqrt {\,4\,\, + \,\,36\,\, + \,\,64}$

$= \,\,\sqrt {\,\,104}$

$= \,\,2\,\,\sqrt {\,\,26}$ 

(iv). $\mathbf{\left( {\,\,2\,\,,\,\, - \,\,1\,\,,\,\,3\,\,} \right)}$ और $\mathbf{\left( {\,\, - \,\,2\,\,,\,\,1\,\,,\,3\,\,} \right)}$

उत्तर: $\left( {\,\,2\,\,,\,\, - \,\,1\,\,,\,\,3\,\,} \right)$ और $\left( {\,\, - \,\,2\,\,,\,\,1\,\,,\,3\,\,} \right)$

दो बिन्दु $\left( {\,{x_1},{y_1},{z_1}\,} \right)$ और  $\left( {\,{x_2},{y_2},{z_2}\,} \right)$ के बीच की दूरी 

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

बिन्दु $\left( {\,\,2\,\,,\,\, - \,\,1\,\,,\,\,3\,\,} \right)$ और $\left( {\,\, - \,\,2\,\,,\,\,1\,\,,\,3\,\,} \right)$ के बीच की दूरी

$= \,\,\,\sqrt {\,{{(\, - 2 - 2\,)}^2}\,\, + \,\,{{(\,1 + 1\,)}^2}\,\, + \,\,{{(\,3 - 3\,)}^2}}$

$= \,\,\,\sqrt {\,16\,\, + \,\,4\,\, + \,\,0}$

$= \,\,\,\sqrt {\,20}$

$= \,\,\,2\,\sqrt {\,5}$ 


2. दर्शाइए कि \[\mathbf{{\text{(}}\,\,{\text{ - }}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{5}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{)}}}\] बिंदु और \[{\text{(}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{)}}\] संरेख हैं। 

उत्तर: मान लीजिए बिन्दु  \[{\text{A}}\,\,{\text{ = }}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{5}}\,\,{\text{)}}\,\] और  \[B\,\, = \,\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{)}}\] के बीच की दूरी 

$A\,\,B\,\, = \,\,\sqrt {\,\,{{(\,\,1\,\, + \,\,2\,\,)}^2}\,\, + \,\,{{(\,\,2\,\, - \,\,3\,\,)}^2}\,\, + \,\,{{(\,\,3\,\, - \,\,5\,\,)}^2}}$

$= \,\,\sqrt {\,\,9\,\, + \,\,1\,\, + \,\,4}$

$= \,\,\sqrt {\,\,14}$ 

बिन्दु   \[B\,\, = \,\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{)}}\] और  \[{\text{C}}\,\,{\text{ = }}\,\,{\text{(}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{)}}\]  के बीच की दूरी

$B\,\,C\,\, = \,\,\sqrt {\,\,{{(\,\,1\,\, + \,\,2\,\,)}^2}\,\, + \,\,{{(\,\,2\,\, - \,\,3\,\,)}^2}\,\, + \,\,{{(\,\,3\,\, - \,\,5\,\,)}^2}}$

$= \,\,\sqrt {\,36\,\, + \,\,4\,\, + \,\,16}$

$= \,\,\sqrt {\,56}$

$= \,\,2\,\,\sqrt {\,14}$ 

बिन्दु   \[{\text{A}}\,\,{\text{ = }}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{5}}\,\,{\text{)}}\,\] और  \[{\text{C}}\,\,{\text{ = }}\,\,{\text{(}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{)}}\]  के बीच की दूरी

$A\,\,C\,\, = \,\,\sqrt {\,\,{{(\,7 + 2\,)}^2}\,\, + \,\,{{(\,0 - 3\,)}^2}\,\, + \,\,{{(\, - 1 - 5\,)}^2}} $

$ = \,\,\sqrt {\,\,81\,\, + \,\,9\,\, + \,\,36} $

$ = \,\,\sqrt {\,\,126} $

$ = \,\,3\,\,\sqrt {\,\,14} $

अब $A\,B\,\, + \,\,B\,C$ को जोड़ने के बाद 

$A\,B\,\, + \,\,B\,C\,\, = \,\,\sqrt {\,\,14} \,\, + \,\,2\,\,\sqrt {\,\,14}$

 $= \,\,3\,\,\sqrt {\,\,14}$ 

जैसा हमें पता है कि ${\text{A}}\,\,{\text{C}}\,\,{\text{ = }}\,\,{\text{3}}\,\,\sqrt {\,\,14} $

यहाँ $A\,B\,\, + \,\,B\,C\,\, = \,\,A\,\,C$

अत: बिन्दु ${\text{A}}\,\,{\text{.}}\,\,{\text{B}}\,\,{\text{.}}\,\,{\text{C}}$संरेख हैं |


3. निम्लिखित को सत्यपति कीजिए:

(i). \[\mathbf{{\text{(}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{10}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{6}}\,\,{\text{)}}\,\]और \[\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{9}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{6}}\,\,{\text{)}}}\] एक समद्विबाहु त्रिभुज के शीर्ष हैं।

उत्तर: माना त्रिभुज $A\,\,B\,\,C$ के शीर्ष \[{\text{(}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{10}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{6}}\,\,{\text{)}}\,\]और\[\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{9}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{6}}\,\,{\text{)}}\] हैं |

अब , 

$A\,\,B\,\, = \,\,\sqrt {\,\,{{(\,1\,\, - \,\,0\,)}^{2\,}}\,\, + \,\,{{(\,6\,\, - \,\,7\,)}^2}\, + \,\,{{(\, - \,\,6\,\, + \,\,10\,)}^2}}$

$= \sqrt {\,\,1\,\, + \,\,1\,\, + \,\,16}$

$= \,\,\sqrt {\,\,18}$

$= \,\,3\,\,\sqrt {\,2} \,$ 

अब ,

$B\,\,C\,\, = \sqrt {\,{{(\,4\,\, - \,\,1\,)}^2}\,\, + \,\,{{(\,9\,\, - \,\,6\,)}^2}\,\, + \,\,{{(\, - \,\,6\,\, + \,\,6\,\,)}^2}}$

$= \,\,\sqrt {\,\,9\,\, + \,\,9}$

$= \,\,\sqrt {\,\,18}$

$= \,\,3\,\,\sqrt {\,2}$ 

हमें पता चला कि यहाँ ${\text{A}}\,\,{\text{B}}\,\,{\text{ = }}\,\,{\text{B}}\,\,{\text{C}}$ हैं |

अत: दिए गए शीर्ष समद्विबाहु त्रिभुज के हैं |

(ii). \[\mathbf{{\text{ (}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{10}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{)}}\,\]और \[\,{\text{(}}\,\,{\text{ - }}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{9}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{)}}}\] एक समकोण त्रिभुज के शीर्ष हैं।

उत्तर : माना त्रिभुज $P\,\,Q\,\,R$ के शीर्ष \[{\text{ (}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{10}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{)}}\,\]और\[\,{\text{(}}\,\,{\text{ - }}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{9}}\,\,{\text{,}}\,\,{\text{6}}\,\,{\text{)}}\] हैं|

अब , 

$P\,\,{Q^2}\,\, = \,\,{(\, - 1\, - 1\,)^2}\,\, + \,\,{(\,6\, - \,7\,)^2}\,\, + \,\,{(\,6 - 10\,)^2} $

$= \,\,1\,\, + \,\,1\,\, + \,\,16 $

$= 1\,\,8 $ 

अब ,

$Q\,\,{R^2}\,\, = \,\,{(\, - 4\,\, + \,\,1\,)^2}\,\, + \,\,{(\,9\,\, - \,\,6\,)^2}\,\, + \,\,{(\,6\,\, - \,\,6\,)^2}$

$= \,\,9\,\, + \,\,9\,\, + \,\,0\,\, = \,\,18$ 

यहाँ , 

$P\,\,{R^2}\,\, + \,\,Q\,\,{R^2}\,\, = \,\,18\,\, + \,\,18\,\,\, = \,\,36 $

$\therefore \,\,P\,\,R\,{\,^{2\,}}\,\, = \,\,36 $

 $P\,\,{Q^2}\,\, + \,\,Q\,\,{R^2}\,\, = \,\,P{R^2} $ 

अत: दिए गए शीर्ष समकोण त्रिभुज के हैं।

(iii).\[\mathbf{{\text{(}}\,\,{\text{-}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{1}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{5}}\,\,{\text{)}}\,\,{\text{,}}\,\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{8}}\,\,{\text{)}}\,}\]और \[\mathbf{\,{\text{(}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{4}}\,\,{\text{)}}}\] एक समांतर चतुर्भुज के शीर्ष हैं।

उत्तर :  माना त्रिभुज $A\,\,B\,\,C\,\,D$ के शीर्ष \[{\text{A}}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{1}}\,\,{\text{)}}\,\,{\text{,}}\,B\,\,{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{5}}\,\,{\text{)}}\,\,{\text{,}}\,C\,\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{7}}\,\,{\text{,}}\,\,{\text{8}}\,\,{\text{)}}\,\]और \[D\,\,\,{\text{(}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{4}}\,\,{\text{)}}\] हैं | अब ,

$A\,\,B{\,^2}\,\, = \,\,{(\,\,1\,\, + \,\,1\,\,)^2}\,\, + \,\,{(\,\, - \,\,2\,\, - \,\,2\,\,)^2}\,\, + \,\,{(\,\,5\,\, - \,\,1\,\,)^2} $

$= \,\,4\,\, + \,\,16\,\, + \,\,16\,\, $

$= \,\,36 $ 

अब ,

$B\,\,{C^2}\,\, = \,\,{(\,\,4\,\, - \,\,1\,\,)^2}\,\, + \,\,{(\,\, - \,\,2\,\, + \,\,2\,\,)^2}\,\, + \,\,{(\,\,8\,\, - \,\,5\,\,)^2} $

$= \,\,9\,\, + \,\,25\,\, + \,\,9 $

$= \,\,43 $ 

अब ,

$C\,\,{D^2}\,\, = \,\,{(\,\,2\,\, - \,\,4\,\,)^2}\,\, + \,\,{(\,\, - \,\,3\,\, + \,\,7\,\,)^2}\,\, + \,\,{(\,\,4\,\, - \,\,8\,\,)^2} $

$= \,\,4\,\, + \,\,16\,\, + \,\,16 $

$= \,\,36 $ 

अब ,

$A\,\,{D^2}\,\, = \,\,{(\,\,2\,\, + \,\,1\,\,)^2}\,\, + \,\,{(\,\, - \,\,3\,\, - \,\,2\,\,)^2}\,\, + \,\,{(\,\,4\,\, - \,\,1\,\,)^2} $

$= \,\,9\,\, + \,\,25\,\, + \,\,9 $

$= \,\,43 $ 

जैसा हमें पता चला कि 

$A\,\,{B^2} = \,\,C\,\,{D^2}$ और     $B\,\,{C^2}\,\, = \,\,A{D^2}$

$A\,\,B\,\, = \,\,C\,\,D$  और  $B\,\,C\,\, = \,\,AD$

अत: दिए गए बिन्दु एक समांतर चतुर्भुज के हैं |


4. ऐसे बिंदुओं के समुच्चय का समीकरण ज्ञात कीजिए जो \[\mathbf{{\text{(}}\,\,{\text{1}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{3}}\,\,{\text{)}}\,\,}\] और \[\mathbf{{\text{(}}\,\,{\text{3}}\,\,{\text{,}}\,\,{\text{2}}\,\,{\text{,}}\,\,{\text{ - }}\,\,{\text{1}}\,\,{\text{)}}}\] बिंदु से समदूरस्थ हैं।

उत्तर : माना बिन्दु बिन्दु और बिन्दु से समान दूरी पर है |

अत: \[P\,\,A\,\, = \,\,P\,\,B\]

अब , \[P\,\,{A^2}\,\, = \,\,P\,\,{B^2}\]

${(\,x - 1\,)^2} + {(\,y - 2\,)^2} + {(\,z - 3\,)^2}\,\, = \,\,{(\,x - 3\,)^2} + {(\,y - 2\,)^2} + {(\,z + 1\,)^2} $

$\Rightarrow \,\,\,\left( {\,{x^2} - 2x + 1\,} \right) + \left( {\,{z^2} - 6z + 9\,} \right)\,\, = \,\,\left( {\,{x^2} - 6x + 9\,} \right) + \left( {\,{z^2} + } \right.2z + 1\,) $

$\Rightarrow \,\,\, - 2x + 6x - 6z - 2z + 10 - 10\,\, = \,\,0 $

$\Rightarrow \,\,\,4x - 8z\,\, = \,\,0 $ 

अत: अभीष्ट समीकरण \[x\,\, - \,\,2\,\,z\,\, = \,\,0\]

5. बिंदुओं च् से बने समुच्चय का समीकरण ज्ञात कीजिए जिनकी बिंदुओं से \[\mathbf{{\text{A}}\,\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{)}}\] और  \[{\text{B}}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{)}}\]दूरियों का योगफल \[{\text{10}}}\] है।

उत्तर: दिए गए बिन्दु \[{\text{A}}\,\,{\text{(}}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{)}}\] और \[{\text{B}}\,\,{\text{(}}\,\,{\text{ - }}\,\,{\text{4}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{,}}\,\,{\text{0}}\,\,{\text{)}}\] इस प्रकार है कि 

$P\,\,A\,\, + \,\,P\,\,B\,\, = \,\,10$

$\sqrt {{{(x - 4)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} \,\,\, + \,\,\sqrt {{{(x + 4)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} \,\,\,\, = \,\,\,\,\,10 $

 $\Rightarrow \,\,\sqrt {{x^2} + {y^2} + {y^2} - 8x + 16} \,\,\,\,\,\,\,\,\,\, = \,\,\,10 - \sqrt {{x^2} + {y^2} + {y^2} + 8x + 16}$ 

दोनों पक्षों का वर्ग करने पर ,

$\Rightarrow \,\,\,{x^2} + {y^2} + {y^2} - 8x + 16\,\,\, = \,\,\,100 + \left( {{x^2} + {y^2} + {y^2} - 8x + } \right.{\text{16) }} - 20\sqrt {{x^2} + {y^2} + {y^2} - 8x + 16}$

 $\Rightarrow \,\,\, - 16x - 100\,\,\, = \,\,\, - 20\sqrt {{x^2} + {y^2} + {y^2} + 8x + 16}$

 $\Rightarrow \,\,\,4x + 25\,\,\, = \,\,\,5\sqrt {{x^2} + {y^2} + {y^2} + 8x + 16}$ 

अत:  फिर से दोनों पक्षों का वर्ग करने पर ,

${(4x + 25)^2}\,\,\, = \,\,\,25\left( {{x^2} + {y^2} + {y^2} + 8x + 16} \right) $

 $\Rightarrow \,\,\,16{x^2} + 200x + 625\,\,\, = \,\,\,25{x^2} + 25{y^2} + 25{y^2} + 200x + 400 $

 $\Rightarrow \,\,\,9{x^2} + 25{y^2} + 25{z^2}\,\,\, = \,\,\,625 - 400 = 225 $ 

अत:  अभीष्ट समीकरण $9\,\,{x^2}\,\, + \,\,25\,\,{y^2}\,\, + \,\,25\,\,{z^2}\,\,\, = \,\,\,225$


प्रश्नावली 12.3

1. बिंदुओं $\mathbf{\left( {\,\, - \,\,2\,\,,\,\,3\,\,,\,\,5\,\,} \right)}$ और $\mathbf{\left( {\,\,1\,\,,\,\, - \,\,4\,\,,\,\,6\,\,} \right)}$ को मिलाने से बने रेखा खंड को अनुपात (i)  $\mathbf{2\,\,:\,\,3}$ में अंत: (ii) $\mathbf{2\,\,:\,\,3}$ में बाह्यतः विभाजित करने वाले बिंदु के निर्देशांक ज्ञात कीजिए।

उत्तर : (i) माना बिन्दुओं $A\,\,\left( {\,\, - \,\,2\,\,,\,\,3\,\,,\,\,5\,\,} \right)$ और $B\,\,\left( {\,\,1\,\,,\,\, - \,\,4\,\,,\,\,6\,\,} \right)$ को मिलाने से बने रेखाखण्ड $A\,\,B$ को $P\,\,\left( {\,\,X\,\,,\,\,Y\,\,,\,\,Z\,\,} \right)$ अनुपात $2\,\,:\,\,3$$$ में अंत विभाजित करता हो, तब बिन्दु  के निर्देशांक इस प्रकार-

$X = \,\,\,\dfrac{{(\,2 \times 1 + 3x\,\,(\, - 2\,)\,)}}{{2 + 3}}\,\,\,\,\, = \,\,\,\,\,\dfrac{{(\,2 - 6\,)}}{5}\,\,\,\,\, = \,\,\,\,\,\dfrac{{ - 4}}{5} $

$Y\,\,\, = \,\,\,\dfrac{{(\,2x\,\,(\, - 4\,\,) + 3 \times 3\,)}}{{2 + 3}}\,\,\,\,\, = \,\,\,\,\,\dfrac{{(\, - 8 + 9\,)}}{5}\,\,\,\,\,\, = \,\,\,\,\,\,\dfrac{1}{5} $

 $Z\,\,\, = \,\,\,\dfrac{{(\,2 \times 6 + 3 \times 5\,)}}{{2 + 3\,}}\,\,\,\,\, = \,\,\,\,\,\,\dfrac{{(\,12 + 15\,)}}{5}\,\,\,\,\, = \,\,\,\,\,\,\dfrac{{27}}{5} $ 

अत: बिन्दु ${\text{P}}$ के निर्देशांक $\left( {\,\,\dfrac{{ - 4}}{5}\,\,,\,\,\dfrac{1}{5}\,\,,\,\,\dfrac{{27}}{5}\,\,} \right)$ हैं |

(ii)  जब बिन्दु $P\,\,\left( {\,\,X\,\,,\,\,Y\,\,,\,\,Z\,\,} \right)$ रेखाखण्ड $A\,\,B$ के बाह्यतः ; विभाजित करता हो , तो निर्देशांक इस प्रकार होगे

$X\,\,\, = \,\,\,\dfrac{{(\,2 \times 1 - 3x\,(\, - 2\,)\,)}}{{2 - 3}}\,\,\,\,\, = \,\,\,\,\dfrac{{(\,2 + 6\,)}}{{ - 1}}\,\,\,\,\,\,\,\,\,\,\, =  - 8 $

  $Y\,\,\, = \,\,\,\dfrac{{(\,2x\,(\, - 4\,) - 3 \times 3\,)}}{{2 - 3}}\,\,\,\,\, = \,\,\,\,\dfrac{{(\, - 8 - 9\,)}}{{ - 1}}\,\,\,\,\,\,\,\, = 17 $

  $Z\,\,\, = \,\,\,\dfrac{{(\,2 \times 6 - 3 \times 5\,)}}{{2 - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\dfrac{{(\,12 - 15\,)}}{{ - 1}}\,\,\,\,\,\,\,\, = 3 $ 

अत: बिन्दु ${\text{P}}$ के निर्देशांक $\left( {\,\, - 8\,\,,\,\,17\,\,,\,\,3\,\,} \right)$ हैं |


2. दिया गया है कि बिंदु $\mathbf{P\,\,\left( {\,\,3\,\,,\,\,2\,\,,\,\, - \,\,4\,\,} \right)\,\,,\,\,Q\,\,\left( {\,\,5\,\,,\,\,4\,\,,\,\, - \,\,6\,\,} \right)}$और $\mathbf{R\,\,\left( {\,\,9\,\,,\,\,8\,\,,\,\, - \,\,10\,\,} \right)\,}$ संरेख है। वह अनुपात ज्ञात कीजिए जिसमें को विभाजित करता है।

उत्तर : $x$ निर्देशांक , 

${\text{5}}\,\,\,{\text{ = }}\,\,\,\dfrac{{\left( {\,\,k\,\,*\,\,9\,\, + \,\,1\,\,*\,\,3\,\,} \right)}}{{k\,\, + \,\,1}}$

$\Rightarrow \,\,\,5\,\,(\,k + 1\,)\,\, = \,\,9\,k + 3 $

$\Rightarrow \,\,\,4\,k\,\, = \,\,5 - 3\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k\,\, = \,\,\dfrac{2}{4}\, $

 $\Rightarrow \,\,\,k\,\, = \,\,\dfrac{1}{2} $ 

इसी प्रकार $y\,\, - $ निर्देशांक 

\[4\,\,\, = \,\,\dfrac{{(\,8\,k\, + \,2\,)}}{{k + \,1}}\]

$\Rightarrow \,\,\,\,4\,\,k\,\, + \,\,4\,\,\, = \,\,\,8\,\,k\,\, + \,\,2 $

$\Rightarrow \,\,\,\,8\,\,k\,\, - \,\,4\,\,k\,\,\, = \,\,\,4\,\,\, - \,\,2 $

$\Rightarrow \,\,\,\,4\,\,{\text{k}}\,\, = \,\,2 $

$\Rightarrow \,\,\,\,{\text{k}}\,\, = \,\,\dfrac{1}{2} $ 

अब इसी प्रकार $z\,\, - $ निर्देशांक

$- \,6\,\,\, = \,\,\,\dfrac{{(\, - 10\,\,k - 4\,)}}{{k + \,\,1}} $

$\Rightarrow \,\,\,6\,\,k\,\, + \,\,6\,\,\, = \,\,10\,\,k\,\, + \,\,4 $

$\Rightarrow \,\,\,10\,\,k\,\, - \,\,6\,\,k\,\,  = \,\,6\,\,\,- \,\,4 $

$\Rightarrow \,\,\,4\,\,k\,\, = \,\, \,\,2\,\,\,\,\,\,\,\,\,$ 

$\Rightarrow \,\,\,k\,\, = \,\,\dfrac{1}{2} $ 

अत: बिन्दु $P\,\,,\,\,Q\,\,,\,\,R$ संरेख हैं और $Q\,\,,\,\,P\,\,,\,\,R$  को  $1\,\,:\,\,2$ के अनुपात में विभाजित करता है |


3. बिंदुओं $\mathbf{\left( {\,\, - \,\,2\,\,,\,\,4\,\,,\,\,7\,\,} \right)}$और $\mathbf{\left( {\,\,3\,\,,\,\, - \,\,5\,\,,\,\,8\,\,} \right)}$ को मिलाने वाली रेखा खंड, $\mathbf{Y\,\,2\,\, -} $ तल द्वारा किस अनुपात में विभक्त होता है, उसे ज्ञात कीजिए।

उत्तर: मान लीजिए बिन्दु $P$  पर तल $Y\,\,Z$ रेखाखण्ड $A\,\,B$ को $k\,\,:\,\,1$ के अनुपात में प्रतिच्छेद करता है, तब $Y\,\,Z\,\, - $ तल पर प्रत्येक बिन्दु $\left( {\,\,o\,,\,\,y\,,\,\,z\,\,} \right)$के रूप में होगा।

$A\,,\,B$ के निर्देशांक क्रमश:  $\left( {\,\, - \,\,2\,\,,\,\,4\,\,,\,\,7\,\,} \right)$ और  $\left( {\,\,3\,\,,\,\, - \,\,5\,\,,\,\,8\,\,} \right)$है |

$0\,\,\, = \,\,\,\dfrac{{\left[ {\,k\,{\,^ \star }\,\,3\,\, + \,\,1\,{\,^{*\,\,}}(\,\, - \,\,2\,\,)\,} \right]\,}}{{k + 1}}\,\,\, = \,\,\,\dfrac{{(\,\,3\,\,k\,\, - \,\,2\,\,)}}{{k + 1}} $

$\Rightarrow \,\,\,\,3\,\,k\,\, - \,\,2\,\, = \,\,0{\text{ }}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,k\,\, = \dfrac{2}{3} $ 

अत: $A\,\,B$ को $Y\,\,Z\,\, - $ तल $2\,\,:\,\,0$ के अनुपात में विभक्त करता है |

4. विभाजन सूत्र का प्रयोग करके दिखाइए कि बिंदु $\mathbf{{\text{A}}\,\,(\,\,2\,,\, - 3\,,\,\,4\,\,)\,\,,\,\,{\text{B}}\,\,(\, - 1\,,\,\,2\,\,,\,\,1\,\,)\,}$ तथा $\mathbf{C\,\,(\,\,\,0\,,\,\,1/3\,\,,\,\,2\,\,\,)}$ संरेख है।

उत्तर : माना  $A\,\,,\,\,B\,\,,\,\,C\,\,,$  सरेख हैं $B$, रेखाखण्ड $A\,\,C$को $k\,\,:\,\,1$ में विभाजित करता है। 

$- 1\,\, = \,\,\,\dfrac{{\left( {\,{k^ \star }0 + {1^*}2\,} \right)}}{{k + 1}} $

$\Rightarrow \,\,\,\, - k - 1\,\, = \,\,2 $

$ \Rightarrow \,\,\,\,k\,\, = \,\, - 3 $

$\bullet \,\,\,2 = \dfrac{{\left[ {{k^*}1/3 + {1^*}( - 3)} \right]}}{{k + 1}} $

$\Rightarrow \,\,\,\,2k + 2 = \dfrac{k}{{3 - 3}} $

$\Rightarrow \,\,\,\dfrac{{\,2k - k}}{3} =  - 3 - 2 $

$\Rightarrow \,\,\,\,\dfrac{5}{{3k}} =  - 5 $

$\Rightarrow \,\,\,\,k\,\, = \,\, - 3 $ 

और 

अत: $A\,\,,\,\,B\,\,,\,\,C\,\,,$ बिन्दु संरेख हैं |


5. \[\mathbf{{\text{P}}\,\,{\text{( 4 , 2 ,  -  6 )}}}\] और \[\mathbf{{\text{Q  (  10 ,  -  16 , 6 )}}}\]  के मिलाने वाली रेखा खंड \[\mathbf{{\text{P Q}}}\] को सम त्रि-माजित करने वाले बिंदुओं के निर्देशांक ज्ञात कीजिए |

उत्तर :


Line Segment PQ


माना बिन्दु \[A\,\,,\,\,B\] रेखाखण्ड $P\,\,Q$ को $3$ समान भागों में विभाजित करती है।

बिन्दु $A$ , रेखाखण्ड $P\,\,Q$ को $1\,\,:\,\,2$ के अनुपात में विभाजित करता है। 

${\text{A}}\,\, = \,\,\left[ {\,\dfrac{{(\,1 \times 10 + 2 \times 4\,)}}{{1 + 2}}\,\,,\,\,\dfrac{{(\,1 \times (\, - 16\,) + 2 \times 2\,)}}{{1 + 2}}\,\,,\,\,\dfrac{{(\,1 \times 6 + 2{\text{x}}(\, - 6\,)\,)}}{{1 + 2}}\,} \right]\,\,\,$

$ \Rightarrow {\text{A}}\,\,\, = \,\,\left( {\,\dfrac{{18}}{3}\,,\, - \dfrac{{12}}{3}\,,\, - \dfrac{6}{3}\,} \right)$        या        $ \Rightarrow {\text{A}}\,(6, - 4, - 2)$

बिन्दु $B$ , रेखाखण्ड $P\,\,Q$ को $2\,\,:\,\,1$ के अनुपात में विभाजित करता है। 

$B$ के निर्देशांक

$B\,\,\left[ {\,\dfrac{{(\,2 \times 10 + 1 \times 4\,)}}{{2 + 1}},\dfrac{{(\,2x\,(\, - 16\,) + 1 \times 2\,)}}{{2 + 1}},\dfrac{{(\,2 \times 6 + 1x\,(\, - 6\,)\,)}}{{2 + 1}}\,} \right] $

$ = B\,\,(\,\,8\,, - 10\,,2\,\,) $ 

अत: $4$ तथा $B$ के निर्देशांक क्रमश: $\left( {\,6\,\,,\,\, - 4\,\,,\,\, - 2\,} \right)$ और  $\left( {\,\,8\,\,,\,\, - \,\,10\,\,,\,\,2\,\,} \right)$ हैं |


प्रश्नावली A1.2

1. समांतर चतुर्भुज के तीन शीर्ष \[\mathbf{A\,\,\left( {\,\,3\,\,,\,\, - 1\,\,,\,\,2\,\,} \right)\,\,,{\text{ }}B\,\,\left( {\,\,1\,\,,\,\,2\,\,,\,\, - 4\,} \right)\] व  \[C{\text{ }}\left( {\, - \,1\,\,,\,\,1\,\,,\,\,2\,\,} \right)}\] है। चौथे शीर्ष के निर्देशांक ज्ञात कीजिए ।

उत्तर: शीर्ष ${\text{A}}$ और $C$ क्रमश: \[\left( {\,\,3\,\,,\,\, - 1\,\,,\,\,2\,\,} \right)\,\,,\] \[\left( {\, - \,1\,\,,\,\,1\,\,,\,\,2\,\,} \right)\]  है |

${\text{A}}$ और $C$ मध्य बिन्दु के निर्देशांक $ = \,\,\,\left( {\,\dfrac{{3 - 1}}{2}\,\,,\,\,\dfrac{{ - 1 + 1}}{2}\,\,,\,\,\dfrac{{2 + 2}}{2}\,\,} \right)$

$ = \,\,\left( {\,\,1\,\,,\,\,0\,\,,\,\,2\,\,} \right)$

माना बिन्दु $D$ के निर्देशांक $\left( {\,\,x\,\,,\,\,y\,\,,\,\,z\,\,} \right)$ और बिन्दु $B$ के निर्देशांक $\left( {\,\,1\,\,,\,\,2\,\,,\,\, - \,\,4\,\,} \right)$ है |

$\therefore \,\,D\,B$ का मध्य बिन्दु $\left( {\,\,\dfrac{{x + 1}}{2}\,\,,\,\,\dfrac{{y + 2}}{2}\,\,,\,\,\dfrac{{z - 4}}{2}\,\,} \right)$

समांतर चतुर्भुज के विकरण एक दूसरे के $2$ समान भागों में बांटते है 

इसलिए 

$\left( {\,\,\dfrac{{x + 1}}{2} = 1\,\,,\,\,\dfrac{{y + 2}}{2} = 0\,\,,\,\,\dfrac{{z - 4}}{2} = 2\,} \right) $

$x\,\, = \,\,1\,\,,\,\,y\,\, = \,\, - 2\,\,,\,\,z\,\, = \,\,8 $ 

अत: बिन्दु के निर्देशांक $\left( {\,\,1\,\,,\,\, - 2\,\,,\,\,8\,\,} \right)$ है |


2. एक त्रिभुज के शीर्षों के निर्देशां क्रमशः \[\mathbf{A\,\,\left( {\,\,0,0,6\,\,} \right),B\,\,\left( {\,\,0,4,0\,\,} \right)}\] तथा \[\mathbf{C\,\,\left( {\,\,6,0,0\,\,} \right)}\]हैं। त्रिभुज की माध्यिकाओं की लंबाई ज्ञात कीजिए ।

उत्तर :


Triangle ABC


बिन्दु $B\,\,\left( {\,\,0\,\,,\,\,4\,\,,\,\,0\,} \right)$ और $C\,\,\,\left( {\,\,6\,\,,\,\,0\,\,,\,\,0\,\,} \right)$ को मिलाने वाला रेखाखण्ड का मध्य बिन्दु $D\,\,\left( {\,\,\dfrac{{0 + 6}}{2}\,\,,\,\,\dfrac{{4 + 0}}{2}\,\,,\,\,\dfrac{{0 + 0}}{2}\,\,} \right)$ है 

बिन्दु के $A$ निर्देशांक $\left( {\,\,0\,\,,\,\,0\,\,,\,\,6\,\,} \right)$ है 

त्रिभुज $A\,\,B\,\,$ के माध्यिका $A\,\,D$ की लम्बाई 

$ = \,\,\sqrt {\,\,{{(\,\,3 - 0\,\,)}^2}\,\, + \,\,\,{{(\,\,2 - 0\,\,)}^2}\,\, + \,\,{{(\,\,0 - 6\,\,)}^2}} $

$= \,\,\sqrt {\,\,9\,\, + \,\,4\,\, + \,\,36}$

$= \,\,\,\sqrt {\,\,49}$

$= \,\,7 $ 

$C$  और  $A$ के निर्देशांक $\left( {\,\,6\,\,,\,\,0\,\,,\,\,0\,} \right)$ और  \[\left( {\,\,0,0,6\,\,} \right)\] हैं |

$A\,\,C$ का मध्य बिन्दु $E$ \[\,\left( {\,\,\dfrac{{0 + 6}}{2}\,\,,\,\,\dfrac{{0 + 0}}{2}\,\,,\,\,\dfrac{{0 + 6}}{2}\,\,} \right)\]

$E\,\,\,\, = \,\,\,\,(\,\,3\,\,,\,\,0\,\,,\,\,3\,\,) $

$B\,\,\,\, = \,\,\,\,(\,\,0\,\,,\,\,4\,\,,\,\,0\,\,) $ 

त्रिभुज $A\,\,B\,\,C$ के माध्यिका $B\,\,E$ की लंबाई 

$= \,\,\,\sqrt {\,\,{{(\,\,3\,\, - \,\,0\,\,)}^2}\,\, + \,\,{{(\,\,0\,\, - \,\,4\,\,)}^2}\,\, + \,\,{{(\,\,3\,\, - \,\,0\,\,)}^2}}$

$= \,\,\,\sqrt {\,\,9\,\, + \,\,16\,\, + \,\,9}$

$= \,\,\,\sqrt {\,\,34}$ 

$A$  और  $B$ के निर्देशांक $\left( {\,\,0\,\,,\,\,0\,\,,\,\,6\,} \right)$ और  \[\left( {\,\,0,4,0\,\,} \right)\] है |

${\text{A}}\,\,{\text{B}}$ का मध्य बिन्दु $\left( {\,\,\dfrac{{0 + 0}}{2}\,\,,\,\,\dfrac{{0 + 4}}{2}\,\,,\,\,\dfrac{{6 + 0}}{2}\,\,} \right)$ है 

त्रिभुज $A\,\,B\,\,C$ के माध्यिका $C\,\,F$ की लंबाई 

$= \,\,\,\sqrt {\,\,{{(\,\,6\,\, - \,\,0\,\,)}^2}\,\, + \,\,{{(\,\,0\,\, - \,\,2\,\,)}^2}\,\, + \,\,{{(\,\,0\,\, - \,3\,\,)}^2}}$

$= \,\,\,\sqrt {\,\,36\,\, + \,\,4\,\, + \,\,9}$

$= \,\,\,\sqrt {\,\,49}$

$= \,\,\,7 $ 


3. यदि त्रिभुज का केंद्रक मूल बिंदु है और शीर्ष \[\mathbf{P\,\,\left( {2{\text{ }}a,{\text{ }}2,6} \right)}\] \[\mathbf{,\,Q\left( { - 4,3{\text{ }}B - 10} \right)}\] और \[\mathbf{R\left( {8,14,2{\text{ }}c} \right)\]हैं \[a,{\text{ }}b}\] तो और $c$ का मान ज्ञात कीजिए।

उत्तर:  त्रिभुज $P\,Q\,R$ के शीर्षों \[P\,\,(\,\,2{\text{ }}a\,\,,{\text{ }}2\,\,,\,\,6\,\,)\,\,,{\text{ }}Q\,\,(\,\, - \,\,4\,\,,\,\,3{\text{ }}b,\,\, - 10\,\,)\,\,,{\text{ }}R\,\,(\,\,8\,\,,\,\,14\,\,,\,\,2{\text{ }}c\,\,)\] 

$\Delta \,\,P\,\,Q\,\,R$ का केन्द्रक  $\left( {\,\,\dfrac{{{x_1}\,\, + \,\,{x_2}\,\, + \,\,{x_3}}}{3}\,\,,\,\,\dfrac{{{y_1}\,\, + \,\,{y_2}\,\, + \,\,{y_3}}}{3}\,\,,\,\,\dfrac{{{z_1}\,\, + \,\,{z_2}\,\, + \,\,{z_3}}}{3}\,\,} \right)$


4. $y$ अक्ष पर उस बिंदु के निर्देशांक ज्ञात कीजिए जिसकी बिंदु \[\mathbf{P\,\,\left( {\,3\,, - 2\,,\,5\,} \right)}\]से दूर $\mathbf{5\,\sqrt {\,2} }$ है।

उत्तर:  $y\,\, - $अक्ष पर किसी बिंदु के निर्देशांक $A\,\,\left( {\,\,0\,\,,\,\,{y_1}\,\,,\,\,0\,\,} \right)$  है। $A$ से \[P\,\,\left( {\,\,3\,\,,\,\, - 2\,\,,\,\,5\,\,} \right)\]के बीच की दूरी $ = \,\,\,5\,\,\sqrt {\,\,2} $

$A{P^2}\,\, = \,\,{(\,\,3 - 0\,\,)^2}\,\, + \,\,{\left( {\,\, - 2 - {y_1}\,} \right)^2} + \,\,\,{(\,\,5 - 0\,\,)^2} $

$= \,\,9 + {\left( {\,\, - 2 - {y_1}\,\,} \right)^2} + \,\,25 $

$= \,\,{\left( {\,\,{y_1}\,\, + \,\,2\,\,} \right)^2}\,\, + \,\,34 $ 

$AP\,\, = \,\,\sqrt {\,{{\left( {\,\,{y_1} + 2\,\,} \right)}^2}\,\, + \,\,34}  = 5\,\,\sqrt {\,\,2}$

$\therefore \,\,\,\,{\left( {\,{y_1} + 2\,} \right)^2} + 34\,\, = \,\,50 $

${\left( {\,\,{y_1} + 2\,\,} \right)^2}\,\, = \,\,50 - 34\,\, = \,\,16 $

${y_1} + 2\,\,\, = \,\,\, \pm 4 $ 

$ + \,\,v\,\,e$  चिन्ह लेने पर  ${y_1}\,\,\, = \,\,4\,\, - \,\,2\,\,\,\,\,\, = \,\,2$

$ - \,\,v\,\,e$  चिन्ह लेने पर  ${y_1}\,\,\, = \,\, - \,\,4\,\, - \,\,2\,\,\,\,\,\, = \,\, - \,\,6$

$y$ अक्ष पर अभीष्ट बिन्दु $\left( {\,\,0\,\,,\,\,2\,\,,\,\,0\,\,} \right)$ और $\left( {\,\,0\,\,,\,\, - \,\,6\,\,,\,\,0\,\,} \right)$ है |


5. \[P\left( {2, - 3,4} \right)\] और \[Q\,\,\left( {\,8,0,10\,} \right)\]को मिलाने वाली रेखा खंड पर स्थित एक बिंदु $R$ को $x$ निर्देशांक $4$  है। बिंदु $R$ के निर्देशांक ज्ञात काजिए | (संकेत मान लीजिए को \[k:{\text{ }}1\] में विभाजित करता है। बिंदु $R$ के निर्देशांक $\left( {\,\dfrac{{8k + 1}}{{k + 1}}\,\,,\,\,\dfrac{{ - 3}}{{k + 1}}\,\,,\,\,\dfrac{{10k + 4}}{{k + 1}}\,} \right)$ है ।)

उत्तर: माना $R\,\,,\,\,P\,\,,\,\,Q$ बिन्दु को $k\,\,:\,\,1$ में विभाजित करता है जबकि $P$ और $Q$ के निर्देशांक $P\,\,(\,\,2\,\,, - 3\,\,,\,\,4\,\,)$ और $Q\,\,(\,\,8\,\,,\,\,0\,\,,\,\,10\,\,)$ है |

बिन्दु $R$ के बिन्दु के निर्देशांक $\left( {\,\,\dfrac{{8\,\,k + 2}}{{k + 1}}\,\,\,,\,\,\,\dfrac{{ - 3}}{{\,\,k + 1}}\,\,\,,\,\,\dfrac{{10\,\,k + 4}}{{k + 1}}\,\,} \right)$

परंतु $x$ बिन्दु के निर्देशांक $4$ के समान है |

$\dfrac{{8k + 2}}{{k + 1}}\,\,\,\, = \,\,4\,\,\,\,\,\,\, \Rightarrow \,\,\,8\,\,k\,\, + \,\,2\,\,\,\,\, = \,\,4\,\,k\,\, + \,\,4 $

$4\,\,k\,\, = \,\,2\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,k\,\, = \,\,\dfrac{2}{4}\,\,\,\,\,\, \Rightarrow \,\,\,\,k = \dfrac{1}{2}\,\,\,\,\,\,\, = \,\,1\,\,:\,\,2 $ 

$y\,\, - $ निर्देशांक $\dfrac{{ - 3}}{{k + 1}}\,\,\,\,\, = \,\,\dfrac{{ - 3}}{{\dfrac{1}{2} + 1}}\,\,\,\,\, = \,\,\,\dfrac{{ - 3 \times 2}}{3}\,\,\,\,\,\, = \,\,\dfrac{{ - 6}}{3}\,\,\,\,\,\,\, = \,\, - 2$

$z\,\, - $ निर्देशांक  $ = \,\,\dfrac{{10k + 4}}{{k + 1}}\,\,\,\,\,\,\, = \,\,\,\dfrac{{10 \times \dfrac{1}{2} + 4}}{{\dfrac{1}{2} + 1}}\,\,\,\,\,\,\,\, = \,\,\,\dfrac{{5 + 4}}{{\dfrac{3}{3}}}\,\,\,\,\,\,\, = \,\,\dfrac{{9 \times 2}}{3}\,\,\,\,\,\, = \,\,\,\dfrac{{18}}{3}\,\,\,\,\,\,\,\,\, = \,\,\,6$

अत:  $R$ के निर्देशांक  $\left( {\,\,4\,\,,\,\, - 2\,\,,\,\,6\,\,} \right)$ है |


6. यदि बिंदु $A$और $B$ क्रमश: \[\mathbf{\left( {3,4,5} \right)}\] तथा \[\mathbf{\left( { - 1,3, - 7} \right)}\]  हैं। चर बिंदु $P$  द्वारा निर्मित समुच्चय से संबिधत समीकरण ज्ञात कीजिए, जहाँ $\mathbf{P{A^2}\,\, + \,\,P{B^2}\,\, = \,\,{k^2}}$ अचर है।

उत्तर:  माना बिन्दु ${\text{P}}$ के निर्देशांक $\left( {\,\,x\,\,,\,\,y\,\,,\,\,z\,\,} \right)$

बिन्दु $A\,\,\left( {\,\,3\,\,,\,\,4\,\,,\,\,5\,\,} \right)$ 

$P{A^2}\,\, = \,\,{(x - 3)^2} + {(y - 4)^2} + {(z - 5)^2}$

बिन्दु $B\,\,\left( {\,\, - \,\,1\,\,,\,\,3\,\,,\,\,7\,\,} \right)$

$P{B^2}\,\, = \,\,{(x + 1)^2} + {(y - 3)^2} + {(z + 7)^2}$

दिया है 

$P{A^2} + P{B^2} = {k^2} $

$\left[ {\,\,{{(x - 3)}^2} + {{(y - 4)}^2} + {{(z - 5)}^2}\,\,} \right]\,\,\, + \,\,\,\left[ {\,\,{{(x + 1)}^2} + {{(y - 3)}^2} + } \right.\left. {{{(z + 7)}^2}\,\,} \right]\,\, = \,\,{k^2} $

$\Rightarrow \,\,\,\,\left( {\,\,{x^2} + {y^2} + {z^2} - 6x - \,8y - 10z + 9 + 16 + 25\,\,} \right)\,\, + \,\,\left( {\,\,{x^2} + {y^2} + {z^2} + 2x - 6y + 14z + 1 + 9 + 49\,\,} \right)\,\, = \,\,{k^2} $

$\Rightarrow \,\,\,\,\,2\,\,\left( {\,\,{x^2} + {y^2} + {z^2}\,\,} \right)\,\, - \,\,4x - 14y + 4z + 50 + 59 - {k^2}\,\, = \,\,0 $

$\Rightarrow \,\,\,\,2\,\,\left( {\,\,{x^2} + {y^2} + {z^2}\,\,} \right)\,\, - \,\,4x - 14y + 4z + 109 - {k^2}\,\, = \,\,0 $

$\Rightarrow \,\,\,\,\,\,2\,\,\left( {\,\,{x^2} + {y^2} + {z^2}\,\,} \right)\,\, - \,\,4x - 14y + 4z = {k^2} - 109 $

$\Rightarrow \,\,\,\,{x^2} + {y^2} + {z^2} - 2x - 7y + 2z\,\,\,\, = \,\,\,\,\dfrac{{{k^2} - 109}}{2} $ 


NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 11 Maths chapter 12 solution Hindi mediums are created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 11 Maths chapter 12 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.


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NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 11 Maths Introduction to Three Dimensional Geometry in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 11 can download these solutions at any time as per their convenience for self-study purpose.


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FAQs on NCERT Solutions for Class 11 Maths In Hindi Chapter 12 Introduction To Three Dimensional Geometry

1. Is Class 11 Maths Chapter 12 important from an examination point of view?

Class 11 is a base class for Class 12 which makes it necessary for the students to score good grades in Class 11. Chapters included in Class 12 are an extension of chapters that are studied in Class 11. Therefore students need to study every chapter with equal focus and determination. Questions are often asked from Chapter 12 in exams. This chapter is also important for future careers in engineering, architecture etc. as this chapter talks about the coordinates in space, coordinates in between to points etc.

2.  Where can I get the NCERT Solutions for Class 11 Maths Chapter 12?

NCERT Solutions are a great source to acquire necessary study material for the preparation of exams. You can find NCERT Solutions for Chapter 12 on Vedantu’s official site or Vedantu’s learning app. You can find study material such as revision notes, important questions, extra questions to practice, etc., in a PDF format on Vedantu’s website free of cost. Vedantu is a teacher-recommended site to study and prepare exams from and has proven to improve students’ grades. The solutions for Chapter 12 are structured while adhering to CBSE guidelines.

3. How are NCERT solutions beneficial for Class 11 Maths Chapter 12?

The NCERT solutions provided by Vedantu are curated specially for the understanding of students. The concepts of coordinates can be difficult to understand therefore, NCERT solutions provide you with a graphical representation of each question as well for better understanding. These solutions are curated by the subject experts at Vedantu who provide solutions and explanations alongside each question. These solutions boost students' confidence to perform well in mathematics exams. The solutions follow a step by step format i.e. each step is visible to students until the derivation of the result.

4. What do you mean by Three-dimensional Geometry in Chapter 11 of Class 12 Maths?

Three-dimensional Geometry is that branch of Mathematics that talks about three-dimensional shapes and objects in three-dimensional space and consists of three coordinate points. These three coordinates are x- coordinate, y-coordinate, and a new coordinate z coordinate which is introduced in this chapter. For further explanation of three-dimensional geometry, you can refer to NCERT Solutions for Class 11 Maths Chapter 12. NCERT solutions are available at Vedantu’s Official site for students’ access.

5. What all concepts are covered in Class 11 Maths Chapter 12?

Class 11 Mathematics Chapter 12 introduces us to the new concept of Three-dimensional Geometry. Three-dimensional Geometry deals with a section of Mathematics that studies the points, solid shapes, and lines in the three-dimensional coordinate system. Students study the basics of coordinate planes. Students learn about the new Z coordinate along with the existing X and Y to determine the location of a point in the three-dimensional coordinate plane. This is a fundamental theory that is applied in higher mathematics and other science fields.