Three Dimensional Geometry Class 12 Maths Important Questions

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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 myCBSEguide 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 and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 11 Maths Extra Questions

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Class 12 Chapter 11 Maths Practice Questions

Chapter 11 Three Dimensional Geometry


  1. Write the vector equation of a line that passes through the given point whose position vector is {tex}\vec a{/tex} and parallel to a given vector {tex}\vec b{/tex} .
    1. {tex}\vec r = \vec a – \lambda \vec b{/tex},{tex} \lambda \in R{/tex}
    2. {tex}\vec r = \vec a + \lambda \vec b{/tex}, {tex} \lambda \in R{/tex}
    3. {tex}\vec r = – \vec a + \lambda \vec b{/tex},{tex} \lambda \in R{/tex}
    4. {tex}\vec r = – \vec a – \lambda \vec b{/tex}, {tex} \lambda \in R{/tex}
  2. If a line has the direction ratios – 18, 12, – 4, then what are its direction cosines ?
    1. {tex}\frac{9}{{11}},\;\frac{6}{{11}},\frac{{ – 2}}{{11}}{/tex}
    2. {tex}\frac{{ – 9}}{{11}},\;\frac{6}{{11}},\frac{{ – 2}}{{11}}{/tex}
    3. {tex}\frac{{ – 9}}{{11}},\;\frac{6}{{11}},\frac{2}{{11}}{/tex}
    4. {tex}\frac{{ – 7}}{{11}},\;\frac{6}{{11}},\frac{{ – 3}}{{11}}{/tex}
  3. 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}and {tex}\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}{/tex}are coplanar if
    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}
    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}
    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}
    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}
  4. Express the Cartesian equation of a line that passes through two points{tex}\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right){/tex} and {tex}\left( {{x_2},{\text{ }}{y_2},{\text{ }}{z_2}} \right){/tex} .
    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}
    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}
    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}
    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}
  5. Two lines {tex}\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \;and\;\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \;{/tex} are coplanar if
    1. {tex}\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right).\left( { – \overrightarrow {{b_1}} \times \overrightarrow { – {b_2}} } \right) = 0{/tex}
    2. {tex}\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right).\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0{/tex}
    3. {tex}\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right).\left( { – \overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0{/tex}
    4. {tex}\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right).\left( {\overrightarrow {{b_1}} \times – \overrightarrow {{b_2}} } \right) = 0{/tex}
  6. Direction ratios of two _________ lines are proportional.
  7. If l, m, n are the direction cosines of a line, then l+ m2 + n2 = ________.
  8. The distance of a point P(a, b, c) from x-axis is ________.
  9. Find the vector equation for the line passing through the points (-1,0,2) and (3,4,6).
  10. 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}
  11. Write the equation of a plane which is at a distance of {tex}5 \sqrt { 3 }{/tex} units from origin and the normal to which is equally inclined to coordinate axes.
  12. 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}.
  13. 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.
  14. 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}.
  15. Write the vector equations of following lines and hence find the distance between them.
    {tex} \frac { x – 1 } { 2 } = \frac { y – 2 } { 3 } = \frac { z + 4 } { 6 },{/tex} {tex} \frac { x – 3 } { 4 } = \frac { y – 3 } { 6 } = \frac { z + 5 } { 12 }{/tex}
  16. 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.
  17. Find the shortest distance between the lines whose vector equations are
    {tex}\overrightarrow r = (1 – t)\hat i + (t – 2)\hat j + (3 – 2t)\hat k{/tex}
    {tex}\overrightarrow r = (s + 1)\hat i + (2s – 1)\hat j – (2s + 1)\hat k{/tex}
  18. 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}and the plane {tex}\vec r.\left( {\hat i – \hat j + \hat k} \right) = 5{/tex}.

Chapter 11 Three Dimensional Geometry


Solution

    1. {tex}\vec r = \vec a + \lambda \vec b{/tex}, {tex} \lambda \in R{/tex}
      Explanation: The vector equation of a line that passes through the given point whose position vector is{tex}\vec a{/tex} and parallel to a given vector {tex}\vec b{/tex} is given by : {tex}\vec r = \vec a + \lambda \vec b{/tex}
      {tex}\lambda \in R{/tex}
      Where, {tex}\overrightarrow r = x\hat i + y\hat j + z\hat k{/tex}
      {tex}\overrightarrow a = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k{/tex}
      {tex}\overrightarrow b = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k{/tex}
    1. {tex}\frac{{ – 9}}{{11}},\;\frac{6}{{11}},\frac{{ – 2}}{{11}}{/tex}
      Explanation: If a line has the direction ratios -18, 12, -4, then its direction cosines are given by:
      {tex}l = \frac{{-18}}{{\sqrt {{{( – 18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{/tex}
      {tex}\frac{{ – 18}}{{\sqrt {324 + 144 + 16} }} = \frac{{ – 18}}{{\sqrt {484} }}{/tex}
      {tex} = \frac{{ – 18}}{{22}} = \frac{{ – 9}}{{11}}{/tex}
      {tex}m = \frac{{12}}{{\sqrt {{{( – 18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{/tex}
      {tex}=\frac{{ 12}}{{\sqrt {324 + 144 + 16} }} = \frac{{ 12}}{{\sqrt {484} }}{/tex}
      {tex} = \frac{{ 12}}{{22}} = \frac{{ 6}}{{11}}{/tex}
      {tex}n = \frac{{-4}}{{\sqrt {{{( -18)}^2} + {{(12)}^2} + {{( – 4)}^2}} }}{/tex}
      {tex}=\frac{{- 4}}{{\sqrt {324 + 144 + 16} }} = \frac{{ -4}}{{\sqrt {484} }}{/tex}
      {tex} = \frac{{ -4}}{{22}} = \frac{{-2}}{{11}}{/tex}
    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}.
      Explanation: 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}
      and
      {tex}\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}{/tex}
      are coplanar if
      {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}
    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}
      Explanation: The Cartesian equation of a line that passes through two points {tex}\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right){/tex} and {tex}\left( {{x_2},{\text{ }}{y_2},{\text{ }}{z_2}} \right){/tex} 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}
    1. {tex}\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right).\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0{/tex}
      Explanation: 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}are coplanar if
  1. Parallel
  2. 1
  3. {tex}\sqrt{b^2 + c^2}{/tex}
  4. Let {tex}\overrightarrow a {/tex} and {tex}\overrightarrow b {/tex} be the p.v of the points A (-1,0,2) and B (3, 4, 6)
    {tex}\vec r = \vec a + \lambda \left( {\vec b – \vec a} \right){/tex}
    {tex} = \left( { – \hat i + 2\hat k} \right) + \lambda \left( {4\hat i + 4\hat j + 4\hat k} \right){/tex}
  5. According to the question, The required plane is passing through the point {tex}(a, b, c){/tex} whose position vector is {tex}\vec { p } = a \hat { i } + b \hat { j } + c \hat { k }{/tex} and is parallel to the plane {tex}\vec { r } \cdot ( \hat { i } + \hat { j } + \hat { k } ) = 2{/tex}
    {tex}\therefore{/tex} it is normal to the vector
    {tex}\vec { n } = \hat { i } + \hat { j } + \hat { k }{/tex}
    Required equation of plane is
    {tex}( \vec { r } – \vec { p } ). \vec { n } = 0 \Rightarrow \vec { r } .\vec { n } = \vec { p } .\vec { n }{/tex}
    {tex}\Rightarrow{/tex} {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}
    {tex}\therefore \quad \vec { r }. ( \hat { i } + \hat { j } + \hat { k } ) = a + b + c{/tex}
  6. 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} units
    {tex}\therefore{/tex} the direction cosines are {tex}\frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } \text { and } \frac { 1 } { \sqrt { 3 } }{/tex}
    The required equation of plane is
    {tex}\frac { 1 } { \sqrt { 3 } } \cdot x + \frac { 1 } { \sqrt { 3 } } \cdot y + \frac { 1 } { \sqrt { 3 } } \cdot z = 5 \sqrt { 3 }{/tex}
    {tex}\Rightarrow x+y+z=5\times 3{/tex}
    {tex}\Rightarrow \quad x + y + z = 15{/tex}
    [{tex}\because{/tex} 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}]
  7. {tex}\frac{{x – 0}}{2} = \frac{{y – 0}}{2} = \frac{{z – 0}}{1}{/tex}
    {tex}\frac{{x – 5}}{4} = \frac{{y – 2}}{1} = \frac{{z – 3}}{8}{/tex}
    a1 = 2, b1 = 2, c1 = 1
    a2 = 4, b2 = 1, c2 = 8
    {tex}\cos \theta = \frac{{\left| {{{\vec b}_1}.{{\vec b}_2}} \right|}}{{\left| {{{\vec b}_1}} \right|\left| {{{\vec b}_2}} \right|}}{/tex}
    {tex} = \left| {\frac{{2(4) + 2(1) + 1(8)}}{{\sqrt {{2^2} + {2^2} + 1} \sqrt {{4^2} + {1^2} + {8^2}} }}} \right|{/tex}
    {tex} = \left| {\frac{{8 + 2 + 8}}{{\sqrt 9 \sqrt {81} }}} \right|{/tex}
    {tex} = \frac{{18}}{{27}}{/tex}
    {tex} = \frac{2}{3}{/tex}
    {tex}\theta = {\cos ^{ – 1}}\left( {\frac{2}{3}} \right){/tex}
  8. Let the point P divide QR in the ratio {tex}\lambda :1{/tex}, then the co-ordinate of P are
    {tex}\left( {\frac{{5\lambda + 2}}{{\lambda + 1}},\frac{{\lambda + 2}}{{\lambda + 1}},\frac{{ – 2\lambda + 1}}{{\lambda + 1}}} \right){/tex}
    But x – coordinate of P is 4. Therefore,
    {tex}\frac{{5\lambda + 2}}{{\lambda + 1}} = 4 \Rightarrow \lambda = 2{/tex}
    Hence, the z – coordinate of P is {tex}\frac{{ – 2\lambda + 1}}{{\lambda + 1}} = – 1{/tex}.
  9. {tex}\vec a = 5\hat i + 2\hat j – 4\hat k,\vec b = 3\hat i + 2\hat j – 8\hat k{/tex}
    Vector equation of line is
    {tex}\vec r = \vec a + \lambda \vec b{/tex}
    {tex} = 5\hat i + 2\hat j – 4\hat k + \lambda (3\hat i + 2\hat j – 8\hat k){/tex}
    Cartesian equation is
    {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}
    {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}
    {tex}\Rightarrow x = 5 + 3\lambda ,y = 2 + 2\lambda ,z = – 4 – 8\lambda {/tex}
    {tex}\Rightarrow\frac{{x – 5}}{3} = \frac{{y – 2}}{2} = \frac{{z + 4}}{{ – 8}}{/tex}{tex}=\lambda{/tex}
    Therefore, required equation is,
    {tex}\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}{/tex}
  10. The given equations of lines are
    {tex} \frac { x – 1 } { 2 } = \frac { y – 2 } { 3 } = \frac { z + 4 } { 6 }{/tex}
    and {tex} \frac { x – 3 } { 4 } = \frac { y – 3 } { 6 } = \frac { z + 5 } { 12 }{/tex}
    Now, the vector equation of given lines are
    {tex} \vec { r } = ( \hat { i } + 2 \hat { j } – 4 \hat { k } ) + \lambda ( 2\hat { i } + 3 \hat { j } + 6 \hat { k } ){/tex}……(i)
    [{tex}\because{/tex} vector form of equation of line is {tex} \vec { r } = \vec { a } + \lambda \vec { b } ]{/tex}
    and {tex} \vec { r } = ( 3 i + 3 \hat { j} – 5 \hat { k } ) + \mu ( 4 \hat { i } + 6 \hat { j } + 12 \hat { k } ){/tex}……………….(ii)
    Here, {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}
    and {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}
    Now, {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}
    {tex} = 2 \hat { i } + \hat { j } – \hat { k }{/tex}……………..(iii)
    and {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}
    {tex} = \hat { i } ( 36 – 36 ) – \hat { j } ( 24 – 24 ) + \hat { k } ( 12 – 12 ){/tex}
    {tex} = 0 \hat { i} – \hat { 0 } \hat { j } + 0 \hat { k } = \vec { 0 }{/tex}
    {tex}\Rightarrow \quad \vec { b } _ { 1 } \times \vec { b } _ { 2 } = \vec { 0 },{/tex}
    i.e. Vector b1 is parallel to {tex}\vec { b } _ { 2 }{/tex}
    {tex}[ \because \text { if } \vec { a } \times \vec { b } = \vec { 0 } , \text { then } \vec { a } \| \vec { b } ]{/tex}
    Thus, two lines are parallel.
    {tex} \therefore \quad \vec { b } = ( 2 \hat { i } + 3 \hat { j } + 6 \hat { k } ){/tex}……………….(iv)
    [since, DR’s of given lines are proportional] Since, the two lines are parallel, we use the formula for shortest distance between two parallel lines
    {tex} d = \left| \frac { \vec { b } \times \left( \vec { a } _ { 2 } – \vec { a _ { 1 } } \right) } { | \vec { b } | } \right|{/tex}
    {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}…………..(v)
    [from Eqs. (iii) and (iv) ] Now, {tex}( \hat { 2 } i + 3 \hat { j } + 6 \hat { k } ) \times ( 2 \hat { i } + \hat { j } – \hat { k } ){/tex}
    {tex}= \left| \begin{array} { c c c } { \hat { i } } & { \hat { j } } & { \hat { k } } \\ { 2 } & { 3 } & { 6 } \\ { 2 } & { 1 } & { – 1 } \end{array} \right|{/tex}
    {tex}= \hat { i } ( – 3 – 6 ) – \hat { j } ( – 2 – 12 ) + \hat { k } ( 2 -6 ){/tex}
    {tex}= -9 \hat { i } + 14 \hat { j } – 4 \hat { k }{/tex}
    From Eq, (v), we get
    {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}
    {tex} \therefore d = \frac { \sqrt { 81 + 196 + 16 } } { 7 } = \frac { \sqrt { 293 } } { 7 }{/tex}units
  11. 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} Also, the diagonals of a parallelogram intersect each other at mid-point.
    Given points are A (4,5,10), B (2, 3,4) and C(1,2,-1).
    Three Dimensional Geometry Class 12 Maths Important Questions
    We know that, two point vector form of line is
    given by
    {tex} \vec { r } = \vec { a } + \lambda { ( \vec b } – \vec { a } ){/tex}…………………….. ……(i)
    where, {tex} \vec { a }{/tex} and {tex} \vec { b }{/tex} are the position vector of points through which the line is passing through. Here, for line AB, position vectors are
    {tex} \vec { a } = \vec { O A } = 4 \hat { i } + 5 \hat { j } + 10 \hat { k }{/tex}
    and {tex} \vec { b } = \vec { O B } = 2 \hat { i } + 3 \hat { j } + 4 \hat { k }{/tex}
    Using Equation. (i), the required equation of line AB is
    {tex}\vec { r } = ( 4 \hat { i } + 5 \hat { j } + 10 \hat { k } ) + \lambda [ ( 2 \hat { i} + 3 \hat { j } + 4 \hat { k } ){/tex}{tex}- ( 4\hat{ i }+ 5 \hat { j } + 10 \hat { k } ) ]{/tex}
    {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}
    Similarly, vector equation of line BC, where B(2,3,4) and C (1, 2, -1) is
    {tex}\vec { r } = ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) + \mu (\hat { i } + 2 \hat { j } – \hat { k } ){/tex}{tex}- ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) ]{/tex}
    {tex}\Rightarrow \quad \vec { r } = ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) + \mu ( – \hat { i} – \hat { j } – 5 \hat { k } ){/tex}
    We know that, mid-point of diagonal BD
    = Mid-point of diagonal AC
    [{tex}\therefore{/tex} diagonal of a parallelogram bisect each other] {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}
    Therefore, on comparing corresponding coordinates, we get
    {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}
    {tex}\Rightarrow \quad x = 3 , y = 4 \text { and } z = 5{/tex}
    Therefore, coordinates of point D (x, y, z) is (3,4,5) and vector equations of sides AB and BC are
    {tex}\vec { r } = ( 4 \hat { i } + 5 \hat { j} + 10 \hat { k } ) – \lambda ( 2 \hat { i } + 2 \hat { j } + 6 \hat { k } ){/tex} and
    {tex}\vec { r } = ({ 2 } \hat {i} + 3 \hat { j} + 4 \hat { k } ) – \mu \hat { ( i } + \hat { j } + \ { 5 \hat { k } } ),{/tex} respectively.
    {tex}\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+t(-\hat{i}+\hat{j}-2 \hat{k}){/tex}
    {tex}\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2 \hat{j}-2 \hat{k}){/tex}
    {tex}\overrightarrow{a_{1}}=\hat{i}-2 \hat{j}+3 \hat{k}{/tex}
    {tex}\overrightarrow{b_{1}}=-\hat{i}+\hat{j}-2 \hat{k}{/tex}
    {tex}\vec{a}_{2}=\hat{i}-\hat{j}-\hat{k}{/tex}
    {tex}\vec{b}_{2}=\hat{i}+2 \hat{j}-2 \hat{k}{/tex}
    {tex}\vec{a}_{2}-\overrightarrow{a_{1}}=\hat{j}-4 \hat{k}{/tex}
    {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}
  12. {tex}2 \hat{i}-4 \hat{j}-3 \hat{k}{/tex}
    {tex}\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right){/tex}{tex}=(0 \vec{i}+\vec{j}-4 \vec{k}) \cdot(2 \vec{i}-4 \vec{j}-3 \vec{k}){/tex}{tex}=0-4+12=8{/tex}
    {tex}\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}}{/tex}
    {tex}=\sqrt{29}{/tex}
    {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}
  13. {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}
    {tex}\Rightarrow\frac{{x – 2}}{3} = \frac{{y + 1}}{4} = \frac{{z – 2}}{2} = \lambda {/tex} …(1)
    Any point on line (1) is,
    {tex}P(3\lambda + 2,4\lambda – 1,2\lambda + 2){/tex}
    Now, {tex}\vec r.\left( {\hat i – \hat j + \hat k} \right) = 5{/tex}
    {tex}(x\hat i + y\hat j + z\hat k).(\hat i – \hat j + \hat k) = 5{/tex}
    {tex}x – y + z = 5{/tex} …(2)
    Since point P lies on (2), therefore, from (2), we have,
    {tex}(3\lambda+2)-(4\lambda-1)+(2\lambda+2)=5{/tex}
    {tex}\Rightarrow \lambda+5=5{/tex}
    {tex}\Rightarrow\lambda = 0{/tex}
    We get (2, -1, 2)
    as the coordinate of the point of intersection of the given line and the plane
    Now distance between the points (-1, -5, -10) and (2, -1, 2)
    req. distance {tex} = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( { – 1 + 5} \right)}^2} + {{\left( {2 + 10} \right)}^2}} {/tex}
    = {tex}\sqrt{9+16+144}{/tex}=13

Chapter Wise Important Questions Class 12 Maths Part I and Part II

  1. Relations and Functions
  2. Inverse Trigonometric Functions
  3. Matrices
  4. Determinants
  5. Continuity and Differentiability
  6. Application of Derivatives
  7. Integrals
  8. Application of Integrals
  9. Differential Equations
  10. Vector Algebra
  11. Three Dimensional Geometry
  12. Linear Programming
  13. Probability

http://mycbseguide.com/examin8/




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