{"id":32567,"date":"2022-03-30T10:00:18","date_gmt":"2022-03-30T04:30:18","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32567"},"modified":"2022-03-30T10:22:35","modified_gmt":"2022-03-30T04:52:35","slug":"ncert-solutions-for-class-12-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/<\/p>\n

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

\"NCERT<\/p>\n

Ex 11.2 Class 12 NCERT Solutions Question 1.<\/strong>
\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\"Ex
\nare mutually perpendicular.<\/p>\n

Class 12 Math Ex 11.2 NCERT Solutions Question 2.<\/strong>
\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

Ex 11.2 12 Class NCERT Solutions Question 3.<\/strong>
\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\"Class<\/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\"Ex<\/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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/ NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 Ex 11.2 Class 12 NCERT Solutions Question 1. Show that the three lines with direction cosines: …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2\/ NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 Ex 11.2 Class 12 NCERT Solutions Question 1. 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