{"id":27937,"date":"2019-10-19T12:06:49","date_gmt":"2019-10-19T06:36:49","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27937"},"modified":"2019-10-25T12:12:17","modified_gmt":"2019-10-25T06:42:17","slug":"three-dimensional-geometry-class-12-maths-important-questions","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/","title":{"rendered":"Three Dimensional Geometry Class 12 Maths Important Questions"},"content":{"rendered":"

Three Dimensional Geometry Class 12 Maths Important Questions. <\/strong>myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide<\/a>\u00a0<\/strong>website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 11 Dimensional Geometry Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus<\/a><\/strong>\u00a0and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.<\/p>\n

Class 12 Chapter 11 Maths Extra Questions<\/strong><\/p>\n

Download as PDF<\/a><\/strong><\/p>\n

Class 12 Chapter 11 Maths Practice Questions<\/h2>\n

Chapter 11 Three Dimensional Geometry<\/strong><\/p>\n

\n
    \n
  1. Write the vector equation of a line that passes through the given point whose position vector is {tex}\\vec a{\/tex}<\/span> and parallel to a given vector {tex}\\vec b{\/tex}<\/span> .\n
      \n
    1. {tex}\\vec r = \\vec a – \\lambda \\vec b{\/tex}<\/span>,{tex} \\lambda \\in R{\/tex}<\/span><\/li>\n
    2. {tex}\\vec r = \\vec a + \\lambda \\vec b{\/tex}<\/span>, {tex} \\lambda \\in R{\/tex}<\/span><\/li>\n
    3. {tex}\\vec r = – \\vec a + \\lambda \\vec b{\/tex}<\/span>,{tex} \\lambda \\in R{\/tex}<\/span><\/li>\n
    4. {tex}\\vec r = – \\vec a – \\lambda \\vec b{\/tex}<\/span>, {tex} \\lambda \\in R{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
    5. If a line has the direction ratios \u2013 18, 12, \u2013 4, then what are its direction cosines ?\n
        \n
      1. {tex}\\frac{9}{{11}},\\;\\frac{6}{{11}},\\frac{{ – 2}}{{11}}{\/tex}<\/span><\/li>\n
      2. {tex}\\frac{{ – 9}}{{11}},\\;\\frac{6}{{11}},\\frac{{ – 2}}{{11}}{\/tex}<\/span><\/li>\n
      3. {tex}\\frac{{ – 9}}{{11}},\\;\\frac{6}{{11}},\\frac{2}{{11}}{\/tex}<\/span><\/li>\n
      4. {tex}\\frac{{ – 7}}{{11}},\\;\\frac{6}{{11}},\\frac{{ – 3}}{{11}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
      5. In the Cartesian form two lines {tex}\\frac{{x – {x_1}}}{{{a_1}}} = \\frac{{y – {y_1}}}{{{b_1}}} = \\frac{{z – {z_1}}}{{{c_1}}}{\/tex}<\/span>and {tex}\\frac{{x – {x_2}}}{{{a_2}}} = \\frac{{y – {y_2}}}{{{b_2}}} = \\frac{{z – {z_2}}}{{{c_2}}}{\/tex}<\/span>are coplanar if\n
          \n
        1. {tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ { – {a_2}}&{{b_2}}&{{c_2}} \\end{array}} \\right| = 0{\/tex}<\/span><\/li>\n
        2. {tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ {{a_2}}&{{b_2}}&{ – {c_2}} \\end{array}} \\right| = 0{\/tex}<\/span><\/li>\n
        3. {tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ {{a_2}}&{{b_2}}&{{c_2}} \\end{array}} \\right| = 0{\/tex}<\/span><\/li>\n
        4. {tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ {{a_2}}&{ – {b_2}}&{{c_2}} \\end{array}} \\right| = 0{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
        5. Express the Cartesian equation of a line that passes through two points{tex}\\left( {{x_1},{\\text{ }}{y_1},{\\text{ }}{z_1}} \\right){\/tex}<\/span> and {tex}\\left( {{x_2},{\\text{ }}{y_2},{\\text{ }}{z_2}} \\right){\/tex}<\/span> .\n
            \n
          1. {tex}\\frac{{x + {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} + {y_1}}} = \\frac{{z – {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span><\/li>\n
          2. {tex}\\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \\frac{{z + {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span><\/li>\n
          3. {tex}\\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} + {y_1}}} = \\frac{{z – {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span><\/li>\n
          4. {tex}\\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \\frac{{z – {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
          5. Two lines {tex}\\vec r = \\overrightarrow {{a_1}} + \\lambda \\overrightarrow {{b_1}} \\;and\\;\\vec r = \\overrightarrow {{a_2}} + \\mu \\overrightarrow {{b_2}} \\;{\/tex}<\/span> are coplanar if\n
              \n
            1. {tex}\\left( {\\overrightarrow {{a_2}} – \\overrightarrow {{a_1}} } \\right).\\left( { – \\overrightarrow {{b_1}} \\times \\overrightarrow { – {b_2}} } \\right) = 0{\/tex}<\/span><\/li>\n
            2. {tex}\\left( {\\overrightarrow {{a_2}} – \\overrightarrow {{a_1}} } \\right).\\left( {\\overrightarrow {{b_1}} \\times \\overrightarrow {{b_2}} } \\right) = 0{\/tex}<\/span><\/li>\n
            3. {tex}\\left( {\\overrightarrow {{a_2}} – \\overrightarrow {{a_1}} } \\right).\\left( { – \\overrightarrow {{b_1}} \\times \\overrightarrow {{b_2}} } \\right) = 0{\/tex}<\/span><\/li>\n
            4. {tex}\\left( {\\overrightarrow {{a_2}} – \\overrightarrow {{a_1}} } \\right).\\left( {\\overrightarrow {{b_1}} \\times – \\overrightarrow {{b_2}} } \\right) = 0{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
            5. Direction ratios of two _________ lines are proportional.<\/li>\n
            6. If l, m, n are the direction cosines of a line, then l2\u00a0<\/sup>+ m2<\/sup>\u00a0+ n2<\/sup>\u00a0= ________.<\/li>\n
            7. The distance of a point P(a, b, c) from x-axis is ________.<\/li>\n
            8. Find the vector equation for the line passing through the points (-1,0,2) and (3,4,6).<\/li>\n
            9. Write the vector equation of the plane passing through the point (a, b, c) and parallel to the plane {tex}\\vec { r } \\cdot ( \\hat { i } + \\hat { j } + \\hat { k } ) = 2.{\/tex}<\/span><\/li>\n
            10. Write the equation of a plane which is at a distance of {tex}5 \\sqrt { 3 }{\/tex}<\/span> units from origin and the normal to which is equally inclined to coordinate axes.<\/li>\n
            11. Find angle between lines {tex}\\frac{x}{2} = \\frac{y}{2} = \\frac{z}{1},\\frac{{x – 5}}{4} = \\frac{{y – 2}}{1} = \\frac{{z – 3}}{8}{\/tex}<\/span>.<\/li>\n
            12. The x – coordinate of a point on the line joining the points Q(2, 2, 1) and R(5, 1, -2) is 4. Find its z – coordinate.<\/li>\n
            13. Find the vector and Cartesian equation of the line through the point (5, 2,-4) and which is parallel to the vector {tex}3\\hat i + 2\\hat j – 8\\hat k{\/tex}<\/span>.<\/li>\n
            14. Write the vector equations of following lines and hence find the distance between them.
              \n{tex} \\frac { x – 1 } { 2 } = \\frac { y – 2 } { 3 } = \\frac { z + 4 } { 6 },{\/tex}<\/span> {tex} \\frac { x – 3 } { 4 } = \\frac { y – 3 } { 6 } = \\frac { z + 5 } { 12 }{\/tex}<\/span><\/li>\n
            15. The points A(4, 5,10), B(2, 3,4) and C(1, 2, -1) are three vertices of parallelogram ABCD. Find the vector equations of sides A and BC and also find coordinates of point D.<\/li>\n
            16. Find the shortest distance between the lines whose vector equations are
              \n{tex}\\overrightarrow r = (1 – t)\\hat i + (t – 2)\\hat j + (3 – 2t)\\hat k{\/tex}<\/span>
              \n{tex}\\overrightarrow r = (s + 1)\\hat i + (2s – 1)\\hat j – (2s + 1)\\hat k{\/tex}<\/span><\/li>\n
            17. Find the distance of the point (-1, -5, -10) from the point of intersection of the line {tex}\\vec r = \\left( {2\\hat i – \\hat j + 2\\hat k} \\right) + \\lambda \\left( {3\\hat i + 4\\hat j + 2\\hat k} \\right){\/tex}<\/span>and the plane {tex}\\vec r.\\left( {\\hat i – \\hat j + \\hat k} \\right) = 5{\/tex}<\/span>.<\/li>\n<\/ol>\n

              Chapter 11 Three Dimensional Geometry<\/strong><\/p>\n

              \n

              Solution<\/strong><\/p>\n

                \n
              1. \n
                  \n
                1. {tex}\\vec r = \\vec a + \\lambda \\vec b{\/tex}<\/span>, {tex} \\lambda \\in R{\/tex}<\/span>
                  \nExplanation:<\/strong> The vector equation of a line that passes through the given point whose position vector is{tex}\\vec a{\/tex}<\/span> and parallel to a given vector {tex}\\vec b{\/tex}<\/span> is given by : {tex}\\vec r = \\vec a + \\lambda \\vec b{\/tex}<\/span>
                  \n{tex}\\lambda \\in R{\/tex}<\/span>
                  \nWhere, {tex}\\overrightarrow r = x\\hat i + y\\hat j + z\\hat k{\/tex}<\/span>
                  \n{tex}\\overrightarrow a = {a_1}\\hat i + {b_1}\\hat j + {c_1}\\hat k{\/tex}<\/span>
                  \n{tex}\\overrightarrow b = {a_1}\\hat i + {b_1}\\hat j + {c_1}\\hat k{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                2. \n
                    \n
                  1. {tex}\\frac{{ – 9}}{{11}},\\;\\frac{6}{{11}},\\frac{{ – 2}}{{11}}{\/tex}<\/span>
                    \nExplanation:<\/strong> If a line has the direction ratios -18, 12, -4, then its direction cosines are given by:
                    \n{tex}l = \\frac{{-18}}{{\\sqrt {{{( – 18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{\/tex}<\/span>
                    \n{tex}\\frac{{ – 18}}{{\\sqrt {324 + 144 + 16} }} = \\frac{{ – 18}}{{\\sqrt {484} }}{\/tex}<\/span>
                    \n{tex} = \\frac{{ – 18}}{{22}} = \\frac{{ – 9}}{{11}}{\/tex}<\/span>
                    \n{tex}m = \\frac{{12}}{{\\sqrt {{{( – 18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{\/tex}<\/span>
                    \n{tex}=\\frac{{ 12}}{{\\sqrt {324 + 144 + 16} }} = \\frac{{ 12}}{{\\sqrt {484} }}{\/tex}<\/span>
                    \n{tex} = \\frac{{ 12}}{{22}} = \\frac{{ 6}}{{11}}{\/tex}<\/span>
                    \n{tex}n = \\frac{{-4}}{{\\sqrt {{{( -18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{\/tex}<\/span>
                    \n{tex}=\\frac{{- 4}}{{\\sqrt {324 + 144 + 16} }} = \\frac{{ -4}}{{\\sqrt {484} }}{\/tex}<\/span>
                    \n{tex} = \\frac{{ -4}}{{22}} = \\frac{{-2}}{{11}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                  2. \n
                      \n
                    1. {tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ {{a_2}}&{{b_2}}&{{c_2}} \\end{array}} \\right| = 0{\/tex}<\/span>.
                      \nExplanation:<\/strong> In the Cartesian form two lines
                      \n{tex}\\frac{{x – {x_1}}}{{{a_1}}} = \\frac{{y – {y_1}}}{{{b_1}}} = \\frac{{z – {z_1}}}{{{c_1}}} {\/tex}<\/span>
                      \nand
                      \n{tex}\\frac{{x – {x_2}}}{{{a_2}}} = \\frac{{y – {y_2}}}{{{b_2}}} = \\frac{{z – {z_2}}}{{{c_2}}}{\/tex}<\/span>
                      \nare coplanar if
                      \n{tex}\\left| {\\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\\\ {{a_1}}&{{b_1}}&{{c_1}} \\\\ {{a_2}}&{{b_2}}&{{c_2}} \\end{array}} \\right| = 0{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                    2. \n
                        \n
                      1. {tex}\\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \\frac{{z – {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span>
                        \nExplanation:<\/strong> The Cartesian equation of a line that passes through two points {tex}\\left( {{x_1},{\\text{ }}{y_1},{\\text{ }}{z_1}} \\right){\/tex}<\/span> and {tex}\\left( {{x_2},{\\text{ }}{y_2},{\\text{ }}{z_2}} \\right){\/tex}<\/span> is given by : {tex}\\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \\frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \\frac{{z – {z_1}}}{{{z_2} – {z_1}}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                      2. \n
                          \n
                        1. {tex}\\left( {\\overrightarrow {{a_2}} – \\overrightarrow {{a_1}} } \\right).\\left( {\\overrightarrow {{b_1}} \\times \\overrightarrow {{b_2}} } \\right) = 0{\/tex}<\/span>
                          \nExplanation:<\/strong> In vector form: Two lines {tex}\\vec r = \\overrightarrow {{a_1}} + \\lambda \\overrightarrow {{b_1}} \\;and\\;\\vec r = \\overrightarrow {{a_2}} + \\mu \\overrightarrow {{b_2}} \\;{\/tex}<\/span>are coplanar if<\/li>\n<\/ol>\n<\/li>\n
                        2. Parallel<\/li>\n
                        3. 1<\/li>\n
                        4. {tex}\\sqrt{b^2 + c^2}{\/tex}<\/span><\/li>\n
                        5. Let {tex}\\overrightarrow a {\/tex}<\/span> and {tex}\\overrightarrow b {\/tex}<\/span> be the p.v of the points A (-1,0,2) and B (3, 4, 6)
                          \n{tex}\\vec r = \\vec a + \\lambda \\left( {\\vec b – \\vec a} \\right){\/tex}<\/span>
                          \n{tex} = \\left( { – \\hat i + 2\\hat k} \\right) + \\lambda \\left( {4\\hat i + 4\\hat j + 4\\hat k} \\right){\/tex}<\/span><\/li>\n
                        6. According to the question, The required plane is passing through the point {tex}(a, b, c){\/tex}<\/span> whose position vector is {tex}\\vec { p } = a \\hat { i } + b \\hat { j } + c \\hat { k }{\/tex}<\/span> and is parallel to the plane {tex}\\vec { r } \\cdot ( \\hat { i } + \\hat { j } + \\hat { k } ) = 2{\/tex}<\/span>
                          \n{tex}\\therefore{\/tex}<\/span> it is normal to the vector
                          \n{tex}\\vec { n } = \\hat { i } + \\hat { j } + \\hat { k }{\/tex}<\/span>
                          \nRequired equation of plane is
                          \n{tex}( \\vec { r } – \\vec { p } ). \\vec { n } = 0 \\Rightarrow \\vec { r } .\\vec { n } = \\vec { p } .\\vec { n }{\/tex}<\/span>
                          \n{tex}\\Rightarrow{\/tex}<\/span> {tex}\\vec { r } .( \\hat i + \\hat { j } + \\hat { k } ) = ( a \\hat { i } + \\hat { b } + c \\hat { k } ) \\cdot (\\hat { i } + \\hat { \\jmath } + \\hat { k } ){\/tex}<\/span>
                          \n{tex}\\therefore \\quad \\vec { r }. ( \\hat { i } + \\hat { j } + \\hat { k } ) = a + b + c{\/tex}<\/span><\/li>\n
                        7. According to the question, the normal to the plane is equally inclined with coordinates axes, and the distance of the plane from origin is {tex}5\\sqrt3{\/tex}<\/span> units
                          \n{tex}\\therefore{\/tex}<\/span> the direction cosines are {tex}\\frac { 1 } { \\sqrt { 3 } } , \\frac { 1 } { \\sqrt { 3 } } \\text { and } \\frac { 1 } { \\sqrt { 3 } }{\/tex}<\/span>
                          \nThe required equation of plane is
                          \n{tex}\\frac { 1 } { \\sqrt { 3 } } \\cdot x + \\frac { 1 } { \\sqrt { 3 } } \\cdot y + \\frac { 1 } { \\sqrt { 3 } } \\cdot z = 5 \\sqrt { 3 }{\/tex}<\/span>
                          \n{tex}\\Rightarrow x+y+z=5\\times 3{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad x + y + z = 15{\/tex}<\/span>
                          \n[{tex}\\because{\/tex}<\/span> If l, m and n are direction cosines of normal to the plane and P is a distance of a plane from origin, then the equation of plane is given by {tex} lx + my+ nz = p{\/tex}<\/span>]<\/li>\n
                        8. {tex}\\frac{{x – 0}}{2} = \\frac{{y – 0}}{2} = \\frac{{z – 0}}{1}{\/tex}<\/span>
                          \n{tex}\\frac{{x – 5}}{4} = \\frac{{y – 2}}{1} = \\frac{{z – 3}}{8}{\/tex}<\/span>
                          \na1<\/sub> = 2, b1<\/sub> = 2, c1<\/sub> = 1
                          \na2<\/sub> = 4, b2<\/sub> = 1, c2<\/sub> = 8
                          \n{tex}\\cos \\theta = \\frac{{\\left| {{{\\vec b}_1}.{{\\vec b}_2}} \\right|}}{{\\left| {{{\\vec b}_1}} \\right|\\left| {{{\\vec b}_2}} \\right|}}{\/tex}<\/span>
                          \n{tex} = \\left| {\\frac{{2(4) + 2(1) + 1(8)}}{{\\sqrt {{2^2} + {2^2} + 1} \\sqrt {{4^2} + {1^2} + {8^2}} }}} \\right|{\/tex}<\/span>
                          \n{tex} = \\left| {\\frac{{8 + 2 + 8}}{{\\sqrt 9 \\sqrt {81} }}} \\right|{\/tex}<\/span>
                          \n{tex} = \\frac{{18}}{{27}}{\/tex}<\/span>
                          \n{tex} = \\frac{2}{3}{\/tex}<\/span>
                          \n{tex}\\theta = {\\cos ^{ – 1}}\\left( {\\frac{2}{3}} \\right){\/tex}<\/span><\/li>\n
                        9. Let the point P divide QR in the ratio {tex}\\lambda :1{\/tex}<\/span>, then the co-ordinate of P are
                          \n{tex}\\left( {\\frac{{5\\lambda + 2}}{{\\lambda + 1}},\\frac{{\\lambda + 2}}{{\\lambda + 1}},\\frac{{ – 2\\lambda + 1}}{{\\lambda + 1}}} \\right){\/tex}<\/span>
                          \nBut x – coordinate of P is 4. Therefore,
                          \n{tex}\\frac{{5\\lambda + 2}}{{\\lambda + 1}} = 4 \\Rightarrow \\lambda = 2{\/tex}<\/span>
                          \nHence, the z – coordinate of P is {tex}\\frac{{ – 2\\lambda + 1}}{{\\lambda + 1}} = – 1{\/tex}<\/span>.<\/li>\n
                        10. {tex}\\vec a = 5\\hat i + 2\\hat j – 4\\hat k,\\vec b = 3\\hat i + 2\\hat j – 8\\hat k{\/tex}<\/span>
                          \nVector equation of line is
                          \n{tex}\\vec r = \\vec a + \\lambda \\vec b{\/tex}<\/span>
                          \n{tex} = 5\\hat i + 2\\hat j – 4\\hat k + \\lambda (3\\hat i + 2\\hat j – 8\\hat k){\/tex}<\/span>
                          \nCartesian equation is
                          \n{tex}x\\hat i + y\\hat j + z\\hat k = 5\\hat i + 2\\hat j – 4\\hat k + \\lambda (3\\hat i + 2\\hat j – 8\\hat k){\/tex}<\/span>
                          \n{tex}\\Rightarrow x\\hat i + y\\hat j + z\\hat k = (5 + 3\\lambda )\\hat i + (2 +2\\lambda )\\hat j + ( – 4 – 8\\lambda )\\hat k{\/tex}<\/span>
                          \n{tex}\\Rightarrow x = 5 + 3\\lambda ,y = 2 + 2\\lambda ,z = – 4 – 8\\lambda {\/tex}<\/span>
                          \n{tex}\\Rightarrow\\frac{{x – 5}}{3} = \\frac{{y – 2}}{2} = \\frac{{z + 4}}{{ – 8}}{\/tex}<\/span>{tex}=\\lambda{\/tex}<\/span>
                          \nTherefore, required equation is,
                          \n{tex}\\frac{x-5}{3}=\\frac{y-2}{2}=\\frac{z+4}{-8}{\/tex}<\/span><\/li>\n
                        11. The given equations of lines are
                          \n{tex} \\frac { x – 1 } { 2 } = \\frac { y – 2 } { 3 } = \\frac { z + 4 } { 6 }{\/tex}<\/span>
                          \nand {tex} \\frac { x – 3 } { 4 } = \\frac { y – 3 } { 6 } = \\frac { z + 5 } { 12 }{\/tex}<\/span>
                          \nNow, the vector equation of given lines are
                          \n{tex} \\vec { r } = ( \\hat { i } + 2 \\hat { j } – 4 \\hat { k } ) + \\lambda ( 2\\hat { i } + 3 \\hat { j } + 6 \\hat { k } ){\/tex}<\/span>……(i)
                          \n[{tex}\\because{\/tex}<\/span> vector form of equation of line is {tex} \\vec { r } = \\vec { a } + \\lambda \\vec { b } ]{\/tex}<\/span>
                          \nand {tex} \\vec { r } = ( 3 i + 3 \\hat { j} – 5 \\hat { k } ) + \\mu ( 4 \\hat { i } + 6 \\hat { j } + 12 \\hat { k } ){\/tex}<\/span>……………….(ii)
                          \nHere, {tex} \\vec { a _ { 1 } } = \\hat { i} + 2 \\hat { j } – 4 \\hat { k } , \\vec { b _ { 1 } } = 2 \\hat { i} + 3 \\hat { j } + 6 \\hat { k }{\/tex}<\/span>
                          \nand {tex} \\vec { a _ { 2 } } = 3 \\hat { i } + 3 \\hat { j } – 5 \\hat { k } , \\vec { b _ { 2 } } = 4 \\hat { i } + 6 \\hat { j } + 12 \\hat { k }{\/tex}<\/span>
                          \nNow, {tex} \\vec { a _ { 2 } } – \\vec { a _ { 1 } } = ( 3 \\hat { i } + 3 \\hat { j } – 5 \\hat { k } ) – ( \\hat { i} + 2 \\hat { j } – 4 \\hat { k } ){\/tex}<\/span>
                          \n{tex} = 2 \\hat { i } + \\hat { j } – \\hat { k }{\/tex}<\/span>……………..(iii)
                          \nand {tex} \\vec { b _ { 1 } } \\times \\vec { b _ { 2 } } = \\left| \\begin{array} { l l l } { \\hat { i } } & { \\hat { j } } & { \\hat { k } } \\\\ { 2 } & { 3 } & { 6 } \\\\ { 4 } & { 6 } & { 12 } \\end{array} \\right|{\/tex}<\/span>
                          \n{tex} = \\hat { i } ( 36 – 36 ) – \\hat { j } ( 24 – 24 ) + \\hat { k } ( 12 – 12 ){\/tex}<\/span>
                          \n{tex} = 0 \\hat { i} – \\hat { 0 } \\hat { j } + 0 \\hat { k } = \\vec { 0 }{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\vec { b } _ { 1 } \\times \\vec { b } _ { 2 } = \\vec { 0 },{\/tex}<\/span>
                          \ni.e. Vector b1 <\/sub>is parallel to {tex}\\vec { b } _ { 2 }{\/tex}<\/span>
                          \n{tex}[ \\because \\text { if } \\vec { a } \\times \\vec { b } = \\vec { 0 } , \\text { then } \\vec { a } \\| \\vec { b } ]{\/tex}<\/span>
                          \nThus, two lines are parallel.
                          \n{tex} \\therefore \\quad \\vec { b } = ( 2 \\hat { i } + 3 \\hat { j } + 6 \\hat { k } ){\/tex}<\/span>……………….(iv)
                          \n[since, DR’s of given lines are proportional]\nSince, the two lines are parallel, we use the formula for shortest distance between two parallel lines
                          \n{tex} d = \\left| \\frac { \\vec { b } \\times \\left( \\vec { a } _ { 2 } – \\vec { a _ { 1 } } \\right) } { | \\vec { b } | } \\right|{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad d = \\left| \\frac { ( 2 \\hat { i } + 3 \\hat { j} + 6 \\hat { k } ) \\times ( 2 \\hat { i } + \\hat { j } – \\hat { k } ) } { \\sqrt { ( 2 ) ^ { 2 } + ( 3 ) ^ { 2 } + ( 6 ) ^ { 2 } }} \\right|{\/tex}<\/span>…………..(v)
                          \n[from Eqs. (iii) and (iv) ]\nNow, {tex}( \\hat { 2 } i + 3 \\hat { j } + 6 \\hat { k } ) \\times ( 2 \\hat { i } + \\hat { j } – \\hat { k } ){\/tex}<\/span>
                          \n{tex}= \\left| \\begin{array} { c c c } { \\hat { i } } & { \\hat { j } } & { \\hat { k } } \\\\ { 2 } & { 3 } & { 6 } \\\\ { 2 } & { 1 } & { – 1 } \\end{array} \\right|{\/tex}<\/span>
                          \n{tex}= \\hat { i } ( – 3 – 6 ) – \\hat { j } ( – 2 – 12 ) + \\hat { k } ( 2 -6 ){\/tex}<\/span>
                          \n{tex}= -9 \\hat { i } + 14 \\hat { j } – 4 \\hat { k }{\/tex}<\/span>
                          \nFrom Eq, (v), we get
                          \n{tex} d = \\left| \\frac { – 9 \\hat { i } + 14 \\hat { j} – 4 \\hat { k } } { \\sqrt { 49 } } \\right| = \\frac { \\sqrt { ( – 9 ) ^ { 2 } + ( 14 ) ^ { 2 } + ( – 4 ) ^ { 2 } } } { 7 }{\/tex}<\/span>
                          \n{tex} \\therefore d = \\frac { \\sqrt { 81 + 196 + 16 } } { 7 } = \\frac { \\sqrt { 293 } } { 7 }{\/tex}<\/span>units<\/li>\n
                        12. The vector equation of a side of a parallelogram, when two points are given, is {tex}\\vec { r } = \\vec { a } + \\lambda ( \\vec { b } – \\vec { a } ).{\/tex}<\/span> Also, the diagonals of a parallelogram intersect each other at mid-point.
                          \nGiven points are A (4,5,10), B (2, 3,4) and C(1,2,-1).
                          \n\"Three
                          \nWe know that, two point vector form of line is
                          \ngiven by
                          \n{tex} \\vec { r } = \\vec { a } + \\lambda { ( \\vec b } – \\vec { a } ){\/tex}<\/span>…………………….. ……(i)
                          \nwhere, {tex} \\vec { a }{\/tex}<\/span> and {tex} \\vec { b }{\/tex}<\/span> are the position vector of points through which the line is passing through. Here, for line AB, position vectors are
                          \n{tex} \\vec { a } = \\vec { O A } = 4 \\hat { i } + 5 \\hat { j } + 10 \\hat { k }{\/tex}<\/span>
                          \nand {tex} \\vec { b } = \\vec { O B } = 2 \\hat { i } + 3 \\hat { j } + 4 \\hat { k }{\/tex}<\/span>
                          \nUsing Equation. (i), the required equation of line AB is
                          \n{tex}\\vec { r } = ( 4 \\hat { i } + 5 \\hat { j } + 10 \\hat { k } ) + \\lambda [ ( 2 \\hat { i} + 3 \\hat { j } + 4 \\hat { k } ){\/tex}<\/span>{tex}- ( 4\\hat{ i }+ 5 \\hat { j } + 10 \\hat { k } ) ]{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\vec { r } = ( 4 \\hat { i } + 5 \\hat { j } + 10 \\hat { k } ) + \\lambda ( – 2 \\hat { i } – 2 \\hat { j } – 6 \\hat { k } ){\/tex}<\/span>
                          \nSimilarly, vector equation of line BC, where B(2,3,4) and C (1, 2, -1) is
                          \n{tex}\\vec { r } = ( 2 \\hat { i } + 3 \\hat { j } + 4 \\hat { k } ) + \\mu (\\hat { i } + 2 \\hat { j } – \\hat { k } ){\/tex}<\/span>{tex}- ( 2 \\hat { i } + 3 \\hat { j } + 4 \\hat { k } ) ]{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\vec { r } = ( 2 \\hat { i } + 3 \\hat { j } + 4 \\hat { k } ) + \\mu ( – \\hat { i} – \\hat { j } – 5 \\hat { k } ){\/tex}<\/span>
                          \nWe know that, mid-point of diagonal BD
                          \n= Mid-point of diagonal AC
                          \n[{tex}\\therefore{\/tex}<\/span> diagonal of a parallelogram bisect each other]\n{tex} \\therefore \\left( \\frac { x + 2 } { 2 } , \\frac { y + 3 } { 2 } , \\frac { z + 4 } { 2 } \\right) = \\left( \\frac { 4 + 1 } { 2 } , \\frac { 5 + 2 } { 2 } , \\frac { 10 – 1 } { 2 } \\right){\/tex}<\/span>
                          \nTherefore, on comparing corresponding coordinates, we get
                          \n{tex} \\frac { x + 2 } { 2 } = \\frac { 5 } { 2 } , \\frac { y + 3 } { 2 } = \\frac { 7 } { 2 } \\text { and } \\frac { z + 4 } { 2 } = \\frac { 9 } { 2 }{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad x = 3 , y = 4 \\text { and } z = 5{\/tex}<\/span>
                          \nTherefore, coordinates of point D (x, y, z) is (3,4,5) and vector equations of sides AB and BC are
                          \n{tex}\\vec { r } = ( 4 \\hat { i } + 5 \\hat { j} + 10 \\hat { k } ) – \\lambda ( 2 \\hat { i } + 2 \\hat { j } + 6 \\hat { k } ){\/tex}<\/span> and
                          \n{tex}\\vec { r } = ({ 2 } \\hat {i} + 3 \\hat { j} + 4 \\hat { k } ) – \\mu \\hat { ( i } + \\hat { j } + \\ { 5 \\hat { k } } ),{\/tex}<\/span> respectively.
                          \n{tex}\\vec{r}=\\hat{i}-2 \\hat{j}+3 \\hat{k}+t(-\\hat{i}+\\hat{j}-2 \\hat{k}){\/tex}<\/span>
                          \n{tex}\\vec{r}=\\hat{i}-\\hat{j}-\\hat{k}+s(\\hat{i}+2 \\hat{j}-2 \\hat{k}){\/tex}<\/span>
                          \n{tex}\\overrightarrow{a_{1}}=\\hat{i}-2 \\hat{j}+3 \\hat{k}{\/tex}<\/span>
                          \n{tex}\\overrightarrow{b_{1}}=-\\hat{i}+\\hat{j}-2 \\hat{k}{\/tex}<\/span>
                          \n{tex}\\vec{a}_{2}=\\hat{i}-\\hat{j}-\\hat{k}{\/tex}<\/span>
                          \n{tex}\\vec{b}_{2}=\\hat{i}+2 \\hat{j}-2 \\hat{k}{\/tex}<\/span>
                          \n{tex}\\vec{a}_{2}-\\overrightarrow{a_{1}}=\\hat{j}-4 \\hat{k}{\/tex}<\/span>
                          \n{tex}\\overrightarrow{b_{1}} \\times \\hat{b}_{2}=\\left|\\begin{array}{ccc}{\\hat{i}} & {\\hat{j}} & {\\hat{k}} \\\\ {-1} & {1} & {-2} \\\\ {1} & {2} & {-2}\\end{array}\\right|{\/tex}<\/span><\/li>\n
                        13. {tex}2 \\hat{i}-4 \\hat{j}-3 \\hat{k}{\/tex}<\/span>
                          \n{tex}\\left(\\vec{a}_{2}-\\vec{a}_{1}\\right) \\cdot\\left(\\vec{b}_{1} \\times \\vec{b}_{2}\\right){\/tex}<\/span>{tex}=(0 \\vec{i}+\\vec{j}-4 \\vec{k}) \\cdot(2 \\vec{i}-4 \\vec{j}-3 \\vec{k}){\/tex}<\/span>{tex}=0-4+12=8{\/tex}<\/span>
                          \n{tex}\\left|\\vec{b}_{1} \\times \\vec{b}_{2}\\right|=\\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}}{\/tex}<\/span>
                          \n{tex}=\\sqrt{29}{\/tex}<\/span>
                          \n{tex}d=\\left|\\frac{\\left(\\vec{a}_{2}-\\vec{a}_{1}\\right)\\left(\\vec{b}_{1} \\times \\vec{b}_{2}\\right)}{\\left|\\vec{b}_{1} \\times \\vec{b}_{2}\\right|}\\right|=\\frac{8}{\\sqrt{29}}{\/tex}<\/span><\/li>\n
                        14. {tex}\\vec r = \\left( {2\\hat i – \\hat j + 3\\hat k} \\right) + \\lambda \\left( {3\\hat i + 4\\hat j + 2\\hat k} \\right){\/tex}<\/span>
                          \n{tex}\\Rightarrow\\frac{{x – 2}}{3} = \\frac{{y + 1}}{4} = \\frac{{z – 2}}{2} = \\lambda {\/tex}<\/span> …(1)
                          \nAny point on line (1) is,
                          \n{tex}P(3\\lambda + 2,4\\lambda – 1,2\\lambda + 2){\/tex}<\/span>
                          \nNow, {tex}\\vec r.\\left( {\\hat i – \\hat j + \\hat k} \\right) = 5{\/tex}<\/span>
                          \n{tex}(x\\hat i + y\\hat j + z\\hat k).(\\hat i – \\hat j + \\hat k) = 5{\/tex}<\/span>
                          \n{tex}x – y + z = 5{\/tex}<\/span> …(2)
                          \nSince point P lies on (2), therefore, from (2), we have,
                          \n{tex}(3\\lambda+2)-(4\\lambda-1)+(2\\lambda+2)=5{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\lambda+5=5{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\lambda = 0{\/tex}<\/span>
                          \nWe get (2, -1, 2)
                          \nas the coordinate of the point of intersection of the given line and the plane
                          \nNow distance between the points (-1, -5, -10) and (2, -1, 2)
                          \nreq. distance {tex} = \\sqrt {{{\\left( {2 + 1} \\right)}^2} + {{\\left( { – 1 + 5} \\right)}^2} + {{\\left( {2 + 10} \\right)}^2}} {\/tex}<\/span>
                          \n= {tex}\\sqrt{9+16+144}{\/tex}<\/span>=13<\/li>\n<\/ol>\n

                          Chapter Wise Important Questions Class 12 Maths Part I and Part II<\/h2>\n
                            \n
                          1. Relations and Functions<\/a><\/li>\n
                          2. Inverse Trigonometric Functions<\/a><\/li>\n
                          3. Matrices<\/a><\/li>\n
                          4. Determinants<\/a><\/li>\n
                          5. Continuity and Differentiability<\/a><\/li>\n
                          6. Application of Derivatives<\/a><\/li>\n
                          7. Integrals<\/a><\/li>\n
                          8. Application of Integrals<\/a><\/li>\n
                          9. Differential Equations<\/a><\/li>\n
                          10. Vector Algebra<\/a><\/li>\n
                          11. Three Dimensional Geometry<\/a><\/li>\n
                          12. Linear Programming<\/a><\/li>\n
                          13. Probability<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                            Three Dimensional Geometry Class 12 Maths Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide\u00a0website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1432],"tags":[1867,1839,1838,1833,1832,1854],"class_list":["post-27937","post","type-post","status-publish","format-standard","hentry","category-cbse","category-mathematics-cbse-class-12","tag-cbse-class-12-mathematics","tag-extra-questions","tag-important-questions","tag-latest-exam-questions","tag-practice-questions","tag-practice-test"],"yoast_head":"\nThree Dimensional Geometry Class 12 Maths Important Questions<\/title>\n<meta name=\"description\" content=\"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Three Dimensional Geometry Class 12 Maths Important Questions\" \/>\n<meta property=\"og:description\" content=\"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2019-10-19T06:36:49+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2019-10-25T06:42:17+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"16 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519\"},\"headline\":\"Three Dimensional Geometry Class 12 Maths Important Questions\",\"datePublished\":\"2019-10-19T06:36:49+00:00\",\"dateModified\":\"2019-10-25T06:42:17+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\"},\"wordCount\":3230,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png\",\"keywords\":[\"CBSE Class 12 Mathematics\",\"Extra Questions\",\"important questions\",\"latest exam questions\",\"practice questions\",\"practice Test\"],\"articleSection\":[\"CBSE\",\"Mathematics\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\",\"name\":\"Three Dimensional Geometry Class 12 Maths Important Questions\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png\",\"datePublished\":\"2019-10-19T06:36:49+00:00\",\"dateModified\":\"2019-10-25T06:42:17+00:00\",\"description\":\"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter\",\"breadcrumb\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage\",\"url\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png\",\"contentUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mycbseguide.com\/blog\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"CBSE\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/cbse\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Three Dimensional Geometry Class 12 Maths Important Questions\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"name\":\"myCBSEguide\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\",\"name\":\"myCBSEguide\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"width\":180,\"height\":180,\"caption\":\"myCBSEguide\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/mycbseguide\/\",\"https:\/\/x.com\/mycbseguide\",\"https:\/\/www.linkedin.com\/company\/mycbseguide\/\",\"http:\/\/in.pinterest.com\/mycbseguide\/\",\"https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519\",\"name\":\"myCBSEguide\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Three Dimensional Geometry Class 12 Maths Important Questions","description":"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/","og_locale":"en_US","og_type":"article","og_title":"Three Dimensional Geometry Class 12 Maths Important Questions","og_description":"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter","og_url":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/","og_site_name":"myCBSEguide","article_publisher":"https:\/\/www.facebook.com\/mycbseguide\/","article_published_time":"2019-10-19T06:36:49+00:00","article_modified_time":"2019-10-25T06:42:17+00:00","og_image":[{"url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png","type":"","width":"","height":""}],"author":"myCBSEguide","twitter_card":"summary_large_image","twitter_creator":"@mycbseguide","twitter_site":"@mycbseguide","twitter_misc":{"Written by":"myCBSEguide","Est. reading time":"16 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#article","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/"},"author":{"name":"myCBSEguide","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519"},"headline":"Three Dimensional Geometry Class 12 Maths Important Questions","datePublished":"2019-10-19T06:36:49+00:00","dateModified":"2019-10-25T06:42:17+00:00","mainEntityOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/"},"wordCount":3230,"commentCount":0,"publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png","keywords":["CBSE Class 12 Mathematics","Extra Questions","important questions","latest exam questions","practice questions","practice Test"],"articleSection":["CBSE","Mathematics"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/","url":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/","name":"Three Dimensional Geometry Class 12 Maths Important Questions","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png","datePublished":"2019-10-19T06:36:49+00:00","dateModified":"2019-10-25T06:42:17+00:00","description":"Three Dimensional Geometry Class 12 Maths Important Questions There are around 4-5 set of solved Extra Questions from each and every chapter","breadcrumb":{"@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#primaryimage","url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png","contentUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AST7zu3.png"},{"@type":"BreadcrumbList","@id":"https:\/\/mycbseguide.com\/blog\/three-dimensional-geometry-class-12-maths-important-questions\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mycbseguide.com\/blog\/"},{"@type":"ListItem","position":2,"name":"CBSE","item":"https:\/\/mycbseguide.com\/blog\/category\/cbse\/"},{"@type":"ListItem","position":3,"name":"Three Dimensional Geometry Class 12 Maths Important Questions"}]},{"@type":"WebSite","@id":"https:\/\/mycbseguide.com\/blog\/#website","url":"https:\/\/mycbseguide.com\/blog\/","name":"myCBSEguide","description":"","publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/mycbseguide.com\/blog\/#organization","name":"myCBSEguide","url":"https:\/\/mycbseguide.com\/blog\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","width":180,"height":180,"caption":"myCBSEguide"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/mycbseguide\/","https:\/\/x.com\/mycbseguide","https:\/\/www.linkedin.com\/company\/mycbseguide\/","http:\/\/in.pinterest.com\/mycbseguide\/","https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ"]},{"@type":"Person","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/f67796d5f5c5a468e8c680aaaad21519","name":"myCBSEguide"}]}},"_links":{"self":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/27937","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/comments?post=27937"}],"version-history":[{"count":2,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/27937\/revisions"}],"predecessor-version":[{"id":27965,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/27937\/revisions\/27965"}],"wp:attachment":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media?parent=27937"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/categories?post=27937"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/tags?post=27937"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}