{"id":32567,"date":"2021-08-28T15:49:18","date_gmt":"2021-08-28T10:19:18","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32567"},"modified":"2021-08-28T15:49:18","modified_gmt":"2021-08-28T10:19:18","slug":"ncert-solutions-for-class-12-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 11 Three Dimensional Geometry Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nShow that the three lines with direction cosines:
\n\\(\\frac { 12 }{ 13 } ,\\frac { -3 }{ 13 } ,\\frac { -4 }{ 13 } ,\\frac { 4 }{ 13 } ,\\frac { 12 }{ 13 } ,\\frac { 3 }{ 13 } ,\\frac { 3 }{ 13 } ,\\frac { -4 }{ 13 } ,\\frac { 12 }{ 13 } \\) are mutually perpendicular.
\nSolution:
\nConsider the lines with direction cosines
\n\\(\\frac{12}{13}, \\frac{-3}{13}, \\frac{-4}{13} \\text { and } \\frac{4}{13}, \\frac{12}{13}, \\frac{3}{13}\\)
\nThe lines with direction cosines l1<\/sub>, m1<\/sub>, n1<\/sub> and l2<\/sub>, m2<\/sub>, n2<\/sub> are perpendicular if
\nl1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub> + n1<\/sub>n2<\/sub> = 0
\nl1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub> + n1<\/sub>n2<\/sub>
\n= \\(\\left(\\frac{12}{13} \\times \\frac{4}{13}\\right)+\\left(\\frac{-3}{13} \\times \\frac{12}{13}\\right)+\\left(\\frac{-4}{13} \\times \\frac{3}{13}\\right)\\)
\n= \\(\\frac{48}{169}-\\frac{36}{169}-\\frac{12}{169}=0\\)
\n\u2234 The lines are perpendicular.
\nConsider the lines with direction cosines
\n\"NCERT
\nare mutually perpendicular.<\/p>\n

Question 2.
\nShow that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
\nSolution:
\nLet A (1, – 1, 2), B(3, 4, – 2), C (0, 3, 2) and D(3, 5, 6) be the points
\nDirection ratios of AB = 3 – 1, 4 – 1, – 2 – 2 = 2, 5, – 4
\nDirection ratios of CD = 3 – 0, 5 – 3, 6 – 2 = 3, 2, 4
\na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 2(3) + 5(2) + (- 4) (4) = 6 + 10 – 16 = 0
\nHence AB \u22a5 CD.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nShow that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1) and (1, 2, 5).
\nSolution:
\nLet A(4, 7, 8), B (2, 3, 4), C (- 1, – 2, 1) and D(1, 2, 5) be the points
\nDirection ratios of AB = 2 – 4, 3 – 7, 4 – 8 = – 2, – 4, – 4
\nDirection ratios of CD = 1 – 1, 2 – 2, 5 – 1 = 2, 4, 4
\n\"NCERT<\/p>\n

Question 4.
\nFind the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \\(3\\hat { i } +2\\hat { j } -2\\hat { k }\\)
\nSolution:
\nLet \\(\\vec{a}\\) be the position vector of the point (1, 2, 3) and \\(\\vec{b}\\) be the vector \\(3 \\hat{i}+2 \\hat{j}-2 \\hat{k}\\)
\n\u2234 \\(\\vec{a}=\\hat{i}+2 \\hat{j}+3 \\hat{k}\\)
\nThe vector equation of the line is \\(\\vec{r}\\) = \\(\\vec{a}\\) + \u03bb\\(\\vec{b}\\)
\n\\(\\vec{r}\\) = (\u2234 \\(\\vec{a}=\\hat{i}+2 \\hat{j}+3 \\hat{k}\\)) + \u03bb(\\(3\\hat { i } +2\\hat { j } -2\\hat { k }\\))<\/p>\n

Question 5.
\nFind the equation of the line in vector and in cartesian form that passes through the point with position vector \\(2\\hat { i } -\\hat { j } +4\\hat { k }\\) and is in the direction \\(\\hat { i } +2\\hat { j } -\\hat { k }\\).
\nSolution:
\nLet \\(\\vec{a}\\) be the position vector of the point \\(\\vec{a}\\) = \\(\\vec{a}=2 \\hat{i}-\\hat{j}+4 \\hat{k}\\)
\n\\(\\vec{b}=\\hat{i}+2 \\hat{j}-\\hat{k}\\)
\nThe vector equation of a line passing through the point with position vector \\(\\vec{a}\\) and parallel to \\(\\vec{b}\\) is \\(\\vec{r}\\) = \\(\\vec{a}+\\lambda \\vec{b}\\)
\nHence the equation of the line is
\n\\(\\vec{r}=(2 \\hat{i}-\\hat{j}+4 \\hat{k})+\\lambda(\\hat{i}+2 \\hat{j}-\\hat{k})\\)
\nThe cartesian equation is
\n\\(\\frac{x-2}{1}=\\frac{y+1}{2}=\\frac{z-4}{-1}\\)<\/p>\n

Another Method:
\nThe point is (2, – 1, 4)
\nThe direction ratios of the line are 1, 2, – 1
\nThe equation of the line passing through (x1<\/sub>, y1<\/sub>, z1<\/sub>) and having direction ratios a, b,c is
\n\\(\\frac{x-x_{1}}{a}=\\frac{y-y_{1}}{b}=\\frac{z-z_{1}}{c}\\)
\nHence cartesian equation is
\n\\(\\frac{x-x_{1}}{a}=\\frac{y-y_{1}}{b}=\\frac{z-z_{1}}{c}\\)
\nThe vector equation is
\n\\(\\vec{r}=(2 \\hat{i}-\\hat{j}+4 \\hat{k})+\\lambda(\\hat{i}+2 \\hat{j}-\\hat{k})\\)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the cartesian equation of the line which passes through the point (- 2, 4, -5) and parallel to the line given by \\(\\frac{x+3}{3}=\\frac{y-4}{5}=\\frac{z+8}{6}\\)
\nSolution:
\nThe cartesian equation of a line passing through the point (x1<\/sub>, y1<\/sub>, z1<\/sub>) and parallel to the line with direction ratios a, b, c is
\n\\(\\frac{x-x_{1}}{a}=\\frac{y-y_{1}}{b}=\\frac{z-z_{1}}{c}\\)
\nHence the cartesian equation of the required line is \\(\\frac{x+2}{3}=\\frac{y-4}{5}=\\frac{z+5}{6}\\)<\/p>\n

Question 7.
\nThe cartesian equation of a line is \\(\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}\\) write its vector form.
\nSolution:
\nThe equation of the line is \\(\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}\\)
\nThe required line passes through the point (5, – 4, 6) and is parallel to the vector \\(3 \\hat{i}+7 \\hat{j}+2 \\hat{k}\\). Let \\(\\vec{r}\\) be the position vector of any point on the line, then the vector equation of the line is
\n\\(\\vec{r}=(5 \\hat{i}-4 \\hat{j}+6 \\hat{k})+\\lambda(3 \\hat{i}+7 \\hat{j}+2 \\hat{k})\\)<\/p>\n

Question 8.
\nFind the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).
\nSolution:
\nLet \\(\\vec{a}\\) and \\(\\vec{b}\\) be the position vectors of the point A (0, 0,0) and B (5, – 2, 3)
\n\"NCERT<\/p>\n

Question 9.
\nFind the vector and cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).
\nSolution:
\nLet \\(\\vec{a}\\) and \\(\\vec{b}\\) be the position vector of the point A (3, -2, -5) and B (3, -2, 6)
\n\\(\\vec{a}=3 \\hat{i}-2 \\hat{j}-5 \\hat{k}\\) and \\(\\vec{b}=3 \\hat{i}-2 \\hat{j}+6 \\hat{k}\\)
\n\u2234 \\(\\vec{b}-\\vec{a}=0 \\hat{i}+0 \\hat{j}+11 \\hat{k}\\)
\nLet \\(\\vec{r}\\) be the position vector of any point on the line. Then the vector equation of the line is \\(\\vec{r}\\) = \\(\\vec{a}\\) + \u03bb(\\(\\vec{b}\\) – \\(\\vec{a}\\))
\n\\(\\vec{r}\\) = \\((3 \\hat{i}-2 \\hat{j}-5 \\hat{k})+\\lambda(11 \\hat{k})\\)
\nThe cartesian equation is
\n\\(\\frac{x-3}{0}=\\frac{y+2}{0}=\\frac{z+5}{11}\\)<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nFind the angle between the following pair of lines
\n(i) \\(\\overrightarrow { r } =2\\hat { i } -5\\hat { j } +\\hat { k } +\\lambda (3\\hat { i } +2\\hat { j } +6\\hat { k } )\\)
\n\\(and\\quad \\overrightarrow { r } =7\\hat { i } -6\\hat { j } +\\mu (\\hat { i } +2\\hat { j } +2\\hat { k } )\\)
\n(ii) \\(\\overrightarrow { r } =3\\hat { i } +\\hat { j } -2\\hat { k } +\\lambda (\\hat { i } -\\hat { j } -2\\hat { k } )\\)
\n\\(\\overrightarrow { r } =2\\hat { i } -\\hat { j } -56\\hat { k } +\\mu (3\\hat { i } -5\\hat { j } -4\\hat { k } )\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 11.
\nFind the angle between the following pair of lines
\n(i) \\(\\frac{x-2}{2}=\\frac{y-1}{5}=\\frac{z+3}{-3}\\) and \\(\\frac{x+2}{-1}=\\frac{y-4}{8}=\\frac{z-5}{4}\\)
\n(ii) \\(\\frac{x}{2}=\\frac{y}{2}=\\frac{z}{1} \\text { and } \\frac{x-5}{4}=\\frac{y-2}{1}=\\frac{z-3}{8}\\)
\nSolution:
\ni. The direction ratios of the first line are 2, 5, – 3 and the direction ratios of the second line are -1, 8,4
\nLet \u03b8 be the acute angle between the lines, then
\n\"NCERT<\/p>\n

ii. The direction ratios of the first line are 2, 2, 1 and the direction ratios of the second line are 4, 1, 8 Let \u03b8 be the acute angle between the lines.
\n\"NCERT<\/p>\n

Question 12.
\nFind the values of p so that the lines
\n\\(\\frac{1-x}{3}=\\frac{7 y-14}{2 p}=\\frac{z-3}{2}\\) and \\(\\frac{7-7 x}{3 p}=\\frac{y-5}{1}=\\frac{6-z}{5}\\) are at right angles
\nSolution:
\nThe given lines are
\n\"NCERT<\/p>\n

Question 13.
\nShow that the lines \\(\\frac{x-5}{7}=\\frac{y+2}{-5}=\\frac{z}{1}\\) and \\(\\frac{x}{1}=\\frac{y}{2}=\\frac{z}{3}\\) are perpendicular to each other
\nSolution:
\nThe direction ratios of the first line are 7, – 5, 1 and the direction ratios of second line are 1, 2, 3
\na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = (7)(1) + (-5)(2) + (1)(3) = 7 – 10 + 3 = 0
\n\u2234 The lines are perpendicular to each other.<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nFind the shortest distance between the lines
\n\\(\\overrightarrow { r } =(\\hat { i } +2\\hat { j } +\\hat { k } )+\\lambda (\\hat { i } -\\hat { j } +\\hat { k } )\\) and \\(\\overrightarrow { r } =(2\\hat { i } -\\hat { j } -\\hat { k } )+\\mu (2\\hat { i } +\\hat { j } +2\\hat { k } )\\)
\nSolution:
\n\\(\\overrightarrow { r } =(\\hat { i } +2\\hat { j } +\\hat { k } )+\\lambda (\\hat { i } -\\hat { j } +\\hat { k } )\\) … (1)
\n\\(\\overrightarrow { r } =(2\\hat { i } -\\hat { j } -\\hat { k } )+\\mu (2\\hat { i } +\\hat { j } +2\\hat { k } )\\) … (2)
\n\"NCERT<\/p>\n

Question 15.
\nFind the shortest distance between the lines \\(\\frac{x+1}{7}=\\frac{y+1}{-6}=\\frac{z+1}{1}\\) and \\(\\frac{x-3}{1}=\\frac{y-5}{-2}=\\frac{z-7}{1}\\)
\nSolution:
\nShortest distance between the lines
\n\"NCERT<\/p>\n

Question 16.
\nFind the distance between die lines whose vector equations are:
\n\\(\\overrightarrow { r } =(\\hat { i } +2\\hat { j } +3\\hat { k) } +\\lambda (\\hat { i } -3\\hat { j } +2\\hat { k } )\\) and \\(\\overrightarrow { r } =(4\\hat { i } +5\\hat { j } +6\\hat { k) } +\\mu (2\\hat { i } +3\\hat { j } +\\hat { k } )\\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nFind the shortest distance between the lines whose vector equations are
\n\\(\\overrightarrow { r } =(1-t)\\hat { i } +(t-2)\\hat { j } +(3-2t)\\hat { k }\\) and \\(\\overrightarrow { r } =(s+1)\\hat { i } +(2s-1)\\hat { j } -(2s+1)\\hat { k }\\)
\nSolution:
\nThe given equation can be reduced as
\n\\(\\vec{r}=(\\hat{i}-2 \\hat{j}+3 \\hat{k})+t(-\\hat{i}+\\hat{j}-2 \\hat{k})\\)
\nand \\(\\vec{r}=(\\hat{i}-\\hat{j}-\\hat{k})+s(\\hat{i}+2 \\hat{j}-2 \\hat{k})\\)
\nComparing with the standard equation, we get
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 Question 1. Show that the three lines with direction cosines: are mutually perpendicular. Solution: Consider the lines …<\/p>\n

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NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 Question 1. Show that the three lines with direction cosines: are mutually perpendicular. Solution: Consider the lines … NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-08-28T10:19:18+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"11 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/","name":"NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 - 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