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>\nNCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2<\/h2>\n <\/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 \nare mutually perpendicular.<\/p>\nClass 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 <\/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 <\/p>\nQuestion 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 <\/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>\nQuestion 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 <\/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
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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 <\/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 <\/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 <\/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 <\/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 <\/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 <\/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 <\/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 <\/p>\n
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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 <\/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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n