NCERT Solutions Class 12 Mathematics Chapter 11

NCERT Solutions Class 12 Mathematics Chapter 11 Three-Dimensional Geometry

Class 12 Mathematics syllabus has a total of 13 essential chapters. Chapter 11 is on Three-Dimensional Geometry, which is a continuation of the chapter: Vector Algebra. The important topics covered are the distance between two points, section formulae and coordinates of a point in space. Geometry guides you in deciding how to design houses and buildings in different geometric shapes in the real world. NCERT Solutions Class 12 Mathematics Chapter 11 helps students find answers to the questions necessary for final exams. This chapter makes the process of learning Three-Dimensional Geometry simpler. This would be helpful for you in your daily practice sessions and while completing homework assignments. The step-by-step walkthrough of the solution is self-explanatory. 

NCERT Solutions Class 12 Mathematics Chapter 11 assists students:

  • In studies.
  • Increases competency levels. 

The solutions provided come in handy for completing the Classwork and preparing for the exams. This focuses on:

  • 3D shapes involving three coordinates, namely X, Y, & Z coordinates.
  • The shapes that absorb space are called 3D shapes. 
  • They are rigid shapes having three dimensions: length, width, and height. 

Three-Dimensional Geometry plays an essential role in exams as multiple questions are included from this chapter in the assessment. NCERT Solutions Class 12 helps students solve complex Three-Dimensional Theories in more straightforward ways. This gives students a solid introduction to Three-Dimensional Mathematics while studying geometry in three different directions. 

Key Topics Covered In NCERT Class 12 Mathematics Chapter 11

This is a practical lesson that provides better suggestions for implementing formulas. In this account, we will explore different formulas, examples, and theories thoroughly to help students build a solid Mathematical foundation. NCERT Solutions Class 12 Mathematics Chapter 11 applies many real-life examples to deliver students a better understanding of the subject. 

Studying Geometry provides many basic skills and builds: 

  • logical thinking.
  • Analytical reasoning.
  • Problem-Solving Skills. 

The main topics covered NCERT Solutions for Class 12 Mathematics Chapter 11 - Three-Dimensional Geometry are 

Serial No  Topic 
1 Introduction 
2 Coordinate Axes and Coordinate Planes in Three-Dimensional Space
3 Coordinates of a Point in Space 
4 Distance Between Two Points 
5 Section Formula 

 

  1. Introduction 

The theory covers the notes prepared as per the CBSE syllabus (2022–2023) and the NCERT curriculum. Prime topics covered in NCERT Solutions Class 12 Mathematics Chapter 11 are Direction Cosine and Direction Ratios of a line joining two points. You can also consider the equation of lines and planes in space under different conditions, the angle between line and plane, space between two lines etc. NCERT Solutions Class 12 Mathematics Chapter 11 will help you practice the formulas thoroughly to know the subject well. This topic has been introduced in Mathematics to find different shapes and figures. In the real world, all the objects are in a three-dimensional form. For example, household objects like tables, kitchen utensils, beds etc., have 3D geometry. You would learn the latest advanced version of 3D geometry in the 12th standard. 

  1. Coordinate Axes and Coordinate Planes in Three-Dimensional Space

The three coordinate axes of a rectangular Cartesian coordinate system are the x-axis, y-axis, and z-axis. These lines are three mutually perpendicular lines. NCERT Solutions Class 12 Mathematics Chapter 11 further explains that the Three-Dimensional Geometry shape has a face, an edge, and a vertex. The size of these forms has been decided by the area they occupy. Space exists in 2D structures, whereas surface space exists in three-dimensional shapes. The opening of all the faces of a three-dimensional form is the surface area, cube, cuboid, cone, and Cylinder shapes that can be seen around us. A coordinate plane is formed by the pair of axes of the three-dimensional coordinate system x, y, and z. NCERT Solutions Class 12 Mathematics Chapter 11 explains this coordination in detail. 

  1. Coordinates of a Point in Space 

NCERT Solutions Class 12 Mathematics Chapter 11 explains a three-dimensional space called 3-space or tri-dimensional space. The point in the space of a geometric setting that contains three values, x, y and z, is required to determine the position of an element. This performs as a three-parameter physical world model where tangible matter exists. These three values are selected from the term's width, height, depth, and length. Just as two-dimensional space consists of many boundless lines, three–dimensional space contains infinitely many planes. These planes are of a particular value. For example, the XY plane consists of the x-axis and y-axis, the YZ plane, which consists of y-axis and z-axis and the XZ plane, which holds the x-axis n z-axis. NCERT Solutions Class 12 Mathematics Chapter 11 provides many examples for a more straightforward understanding of the above-mentioned points. 

  1. Distance Between Two Points

Distance between two points is the length of the line segment that connects the two given points. NCERT Solutions Class 12 Mathematics Chapter 11 preferably uses a formula in Mathematics to calculate the distance between the two points in the coordinate plane. By replacing those points in the formula, we can get the distance between two points. To find out the position of a point in a plane, we require a pair of the coordinate axis. The distance of the point through the x-axis from the Centre is called the x-coordinate. The distance of the point across the y-axis from the starting point is called y-coordinate. The ordered point x and y represents the coordinate of the point. 

  1. Section Formula 

 The section formula has been applied to find out the coordinates of a point. The coordinates of a point in the three–dimensional space are used to locate the point given in the system. This can decide the point that divides a line internally into a specific ratio. Section formula could be divided into internal division formula and external division formula. NCERT Solutions Class 12 Mathematics Chapter 11 also focuses on exceptional cases of section formulas like the midpoint formula and trisection points. Section formulas in three-dimensional can be used to gain more valuable results in 

  • Coordinates of the centroid of a triangle.
  • Collinearity of three points. 

NCERT Solutions Class 12 Mathematics Chapter 11 provides detailed clarity of internal and external section points.

NCERT Solutions Class 12 Mathematics Chapter 11 - Exercises and Answer Solutions.

The exercise and solutions paper are available on the Extramarks website. These questions will clear all doubts, help students understand the concept better, and quickly solve the Mathematical problems. NCERT Solutions Class 12 Mathematics Chapter 11 includes questions according to the latest CBSE syllabus of 2022-23.

Exercise 11.2 Coordinate Axes and Coordinate Planes in Three-Dimensional Space - Questions & Solutions refer to the below link.

Exercise 11.3 Coordinates of a Point in Space - Questions & Solutions click on the below link.

Exercise 11.4 Distance Between Two Points - Questions & Solutions click here to know more.

Exercise 11.5 Section Formula - Questions & Solutions refer to the link below.

Along with Class 12 Mathematics solutions, you can explore NCERT Solutions on our Extramarks website for all primary and secondary Classes. NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11, NCERT Solutions Class 12.

 NCERT Exemplar for Class 12 Mathematics

NCERT Exemplars are practice books that include additional higher-level questions and are meant to aid meticulous learning. They are mainly used for competitive exams. These topics are up to date, consisting of different topics explained in every chapter of the Class 12 Mathematics book. Students can click on the links provided by Extramarks for these Exemplar questions and answers for Class 12 Mathematics chapter wise notes.

Considering the standard and variety, practising Exemplar problems is essential for school boards and competitive examinations. Students can perfectly evaluate themselves by working on these problems and altering their problem-solving skills for future examinations. NCERT Solutions Class 12 Mathematics provides a series of exemplar solutions for Class 12 for detailed clarification to all questions given in the NCERT exemplar Class 12 books. 

 Key Features of NCERT Solutions for Class 12 Mathematics Chapter 11

To score well in competitive exams, one should have a strong command in Mathematics. The NCERT books present you with demanding questions that improveyour analytical skills and give you sufficient exposure to all kinds of questions in any of these exams. Our application is highly beneficial for your preparation for the following reasons. 

  • NCERT Solutions Class 12 Mathematics answers to all numerical problems in the textbook are given with a step-by-step simplification. 
  • Free unlimited access. 
  • Highly detailed solutions to all the theoretical as well as logical reasoning questions from the NCERT Mathematics textbook of Class 12 that are verified by respective experts of the subject. 
  • The reputed teachers' accurate and straightforward answers cover all the complex and specific topics. 
  • The material provided by NCERT helps students to score remarkably in the exams. 

Q.1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

Ans.

Let the direction cosines of the line be  l, m and n. Then,  l= cos90°=0,  m=cos135°=12,n=cos45°=12Therefore, the direction cosines of the line are 0,12 and  12.

 

Q.2 Find the direction cosines of a line which makes equal angles with the coordinate axes.

Ans.

Let the direction cosines of the line, which makes equal angle α with each of the coordinate axes be  l, m and n. Then,  l= cosα,  m=cosα,n=cosαSince,  l2+m2+n2=1cos2α+cos2α+cos2α=13cos2α=1    cos2α=13    cosα=±13Therefore, the direction cosines of the line, which is equally inclined to the coordinate axes, are ±13,±13,±13.

 

Q.3 If a line has the direction ratios –18, 12, –4 then what are its direction cosines?

Ans.

If a line has the direction ratios 18, 12, 4, then directioncosines are:l=18(18)2+(12)2+(4)2  =18324+144+16  =1822  =911m=12(18)2+(12)2+(4)2  =12324+144+16  =1222  =611n=4(18)2+(12)2+(4)2  =4324+144+16  =422  =211Thus, the direction cosines are 911,611,211.

 

Q.4 Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

Ans.

The given points are: A(2, 3, 4), B(–1, –2, 1) and C(5, 8, 7).
Direction ratios of line joining A (2, 3, 4) and B (–1, –2, 1) are:
–1–2, –2 – 3, 1 – 4 i.e., –3, –5, –3.
Direction ratios of line joining B (–1, –2, 1) and C (5, 8, 7) are:
5–(–1), 8 –(–2), 7 –1 i.e., 6, 10, 6.
It can be seen that the direction ratios of BC are –2 times of that of AB.
Therefore, AB is parallel to BC. Since, B is common to both AB and BC, so, points A, B and C are collinear.

 

Q.5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).

Ans.

LettheverticesoftriangleABCbeA3,5,4, B1,1,2andC5,5,2. ThedirectionratiosofsideAB​ are:13,  15,  24  i.e.,4,4,6.So,thedirectioncosinesofABare:  442+42+62,442+42+62,642+42+62i.e.,  4217,4217,6217217,217,317ThedirectionratiosofsideBCare:51,  51,22  i.e.,4,6,4.So,thedirectioncosinesofABare:  442+42+62,642+42+62,442+42+62i.e.,  4217,6217,4217217,317,217ThedirectionratiosofsideACare:53,  55,24  i.e.,8,  10,  2.So,thedirectioncosinesofABare:  882+102+22,1082+102+22,282+102+22i.e.,  8242,10242,2242442,542,142

 

Q.6

Show that the three lines with direction cosines1213,313,413; 413,1213,313; 313,413,1213 are mutually perpendicular.

Ans.

Two lines with direction cosines l 1 ,m 1 ,n 1 and l 2 ,m 2 ,n 2 are perpendicular to each other if l 1 l 2 + m 1 m 2 + n 1 n 2 =0 ( i ) For the lines with direction cosines, 12 13 , 3 13 , 4 13 and 4 13 , 12 13 , 3 13 , we have l 1 l 2 + m 1 m 2 + n 1 n 2 = 12 13 . 4 13 + 3 13 . 12 13 + 4 13 . 3 13 = 48 169 36 169 12 169 =0 Thus, lines are perpendicular. ( ii )For the lines with direction cosines 4 13 , 12 13 , 3 13 and 3 13 , 4 13 , 12 13 , we have l 1 l 2 + m 1 m 2 + n 1 n 2 = 4 13 . 3 13 + 12 13 .( 4 13 )+ 3 13 . 12 13 = 12 169 48 169 + 36 169 =0 Thus, lines are perpendicular. ( iii )For the lines with direction cosines 3 13 , 4 13 , 12 13 and 12 13 , 3 13 , 4 13 , we have l 1 l 2 + m 1 m 2 + n 1 n 2 = 3 13 . 12 13 + 4 13 .( 3 13 )+ 12 13 .( 4 13 ) = 36 169 + 12 169 48 169 =0 Thus, lines are perpendicular. Therefore, all lines are mutually perpendicular. 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Q.7 Show 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).

Ans.

Let line AB has end points A(1,–1, 2) and B(3, 4, –2). And line CD has end points C (0, 3, 2) and D(3, 5, 6).

The direction ratios a1, a2 and a3 of AB are (3 – 1), {4 –(–1)} and (–2–2) i.e., 2, 5, – 4.The direction ratios b1, b2 and b3 of CD are (3 – 0), (5 –3) and (6 – 2) i.e., 3, 2, 4.AB and CD will be perpendicular to each other, if a1b1+ a2b2 + a3b3 = 0.Then,
a1b1+ a2b2 + a3b3 = (2)(3) + (5)(2) + (– 4)(4)

= 6 + 10 – 16
= 0

Therefore, AB and CD are perpendicular to each other.

 

Q.8 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

Ans.

Let line AB has end points A (4, 7, 8) and B (2, 3, 4). And line CD has end points C (–1, –2, 1) and D (1, 2, 5).

The direction ratios a1, a2 and a3 of AB are (2 – 4), (3 –7) and (4 – 8) i.e., –2, – 4, – 4.The direction ratios b1, b2 and b3 of CD are {1–(–1)}, {2 –(–2)} and (5 – 1) i.e., 2, 4, 4.AB will be parallel to CD, if(a1/b1) = (a2/b2) = (a3/b3). Then,
(a1 / b1) = (–2/2) = –1

(a2 / b2) = (– 4/4) = – 1

(a3 / b3) = (– 4/4) = – 1

So, (a1/b1) = (a2/b2) = (a3/b3)

Therefore, AB is parallel to CD.

 

Q.9 Find the equation of the line which passes through the point ( 1 , 2 , 3 ) and is parallel to the vector 3 i ^ + 2 j ^ 2 k ^ .

Ans.

The position vector of point A1,2,3 is a=i^+2j^+3k^ and it isparallelt to vector b=3i^+2j^2k^.Since, the line passes through the point A1,2,3 and parallel tovector b is given byr=a+λb, where λ is a constant.    r=i^+2j^+3k^+λ3i^+2j^2k^  = 1+3λi^+2+2λj^+3-2λk^This is the required equation of the line.

 

Q.10 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 i ^ j ^ + 4 k ^ and is in the direction i ^ + 2 j ^ k ^ .

Ans.

Let a =2 i ^ j ^ +4 k ^ b = i ^ +2 j ^ k ^ Since, the line passes through the point with position vector a and parallel to vector b is given by r = a b , where λ is a constant. r = 2 i ^ j ^ +4 k ^ i ^ +2 j ^ k ^ i This is the required equation of the line in vector form. Putting r =x i ^ +y j ^ +z k ^ in equation i , we get x i ^ +y j ^ +z k ^ = 2 i ^ j ^ +4 k ^ i ^ +2 j ^ k ^ = 2+λ i ^ 1 j ^ + 4λ k ^ x=2+λ,y=2λ-1 and z=4-λ λ=x-2,λ= y+1 2 and λ=4-z x2 1 = y+1 2 = z4 1 This is the required equation of the given line in cartesian form.

 

Q.11 Findthecartesianequation of the line which passes through the point 2, 4, 5 and parallel to the line given by x+3 3 = y4 5 = z+8 6 .

Ans.

The direction ratios of x+3 3 = y4 5 = z+8 6 are 3,5 and 6. Let the direction ratios of required line parallel to x+3 3 = y4 5 = z+8 6 are 3k, 5k and 6k. Since, equation of the line through the point x 1 , y 1 , z 1 and with the direction ratios a, b and c is x x 1 a = y y 1 b = z z 1 c Therefore, the equation of required linethrough the point 2,4,5 is x+2 3k = y4 5k = z+5 6k or x+2 3 = y4 5 = z+5 6 which is the required equation of line.

 

Q.12

Thecartesianequationofalineisx53=y+47=z62.Writeitsvectorform.

Ans.

Thecartesianequation of the line isx53=y+47=z62...iThe given lines passes through the point 5,-4,6.The positionvector of this point is a = 5i^4j^+6k^.The direction ratios of the given line are 3, 7 and 2.The direction vector of given line is b = 3i^+7j^+2k^.So, the equation of line through the point a and in the directionof the vector b is given by the equation,r =a+λb, where λ is a contant.So, r =5 i^4j^+6k^+λ3 i^+7j^+2k^This is the required equation of the given line in the vector form.

 

Q.13 Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2 , 3).

Ans.

The required line passes through origin.Therefore its position vector is a =0. The direction ratios of line through origin and the point 5,2,3 are: 5-0 , 20 , 30 i.e., 5,-2,3 The line is parallel to the position vector b =5 i ^ -2 j ^ +3 k ^ . So, the equation of line passes through origin and parallel to b is r = a b ,λR r =0+λ 5 i ^ 2 j ^ +3 k ^ r 5 i ^ 2 j ^ +3 k ^ The equation of line through the point x 1 ,y 1 ,z 1 and direction ratios a,b,c is given by x x 1 a = y y 1 b = z z 1 c Therefore, equation of the required line in the cartesian form is x-0 5 = y-0 2 = z-0 3 x 5 = y 2 = z 3

 

Q.14 Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

Ans.

Let the given points are A(3,2,5) and B(3,2, 6)throughwhich line AB passes.So, position vector of point (3,2,5)(a)=3i^2j^5k^The direction ratios of AB are given as(33),(2+2),(6+5) i.e., 0,0,11The equation of vector in the direction of AB isb=0i^+0j^+11k^=11k^The equation of AB in the vector form is r=(3i^2j^5k^)+λ(11k^)The equation of AB in the cartesian form isx30=y+20=z+511

 

Q.15

Findtheanglebetweenthefollowingpairs​​oflines:ir=2i^5j^+k^+λ3i^+2j^+6k^and    r=7i^6k^+μi^+2j^+2k^iir=3i^+j^2k^+λi^j^2k^​ and      r=2i^j^56k^+μ3i^5j^4k^

Ans.

iHere,b1=3i^+2j^+6k^andb2=i^+2j^+2k^Thenangleθbetweenthetwolinesisgivenbycosθ=b1.b2b1b1=3i^+2j^+6k^.i^+2j^+2k^3i^+2j^+6k^i^+2j^+2k^=3+4+1232+22+6212+22+22=199+4+361+4+4=197×3=1921  θ=cos11921iiHere, b1=i^j^2k^andb2=3i^5j^4k^Then angle θ between the two lines is given bycosθ=b1.b2b1b1=i^j^2k^.3i^5j^4k^i^j^2k^3i^5j^4k^=3+5+812+12+2232+52+42=161+1+49+25+16=166×50=16103  θ=cos1853

 

Q.16

Findtheanglebetweenthefollowingpairoflines:i  x22=y15=z+33andx+21=y48=z54ii  x2=y2=z1andx54=y21=z38

Ans.

(i)  x22=y15=z+33andx+21=y48=z54The direction ratios of the first line are 2,5,3 and the directionratios of the second line are 1,8,4. If θ is the angle betweenthem, thencosθ=|a1a2+b1b2+c1c2a12+b12+c12.a22+b22+c22|=|2×1+5×8+(3)×4(2)2+(5)2+(3)2.(1)2+82+42|=|2+40124+25+9.1+64+16|=|2+401238.81|    θ=cos1(26938)(ii)  x2=y2=z1andx54=y21=z38The direction ratios of the first line are 2,2,1 and the directionratios of the second line are 4, 1, 8. If θ is the angle betweenthem, thencosθ=|a1a2+b1b2+c1c2a12+b12+c12.a22+b22+c22|=|2×4+2×1+1×8(2)2+(2)2+(1)2.(4)2+12+82|=|8+2+84+4+1.16+1+64|=|189.81|=|183×9|    θ=cos1(23)

 

Q.17

Findthevaluesofpsothatthelines1x3=7y142p=z32and  77x3p=y51=6z5areatrightangles.

Ans.

The eqution of given lines are      1x3=7y142p=z32and77x3p=y51=6z5x13=y2(2p7)=z32andx1(3p7)=y51=z65The direction ratios of the first line are 3,2p7, 2 and the directionratios of the second line are 3p7, 1,5. Since, both lines are perpendicular, so              a1a2+b1b2+c1c2=0(3).(3p7)+2p7.1+2.(5)=0          9p7+2p710=0                  11p70=0  p=7011Thus, the value of p is 7011.

 

Q.18

Showthatthelinesx57=x+25=z1andx1=y2=z3areperpendiculartoeachother.

Ans.

The eqution of given lines are      x57=x+25=z1andx1=y2=z3The direction ratios of the first line are 7, 5, 1 and the directionratios of the second line are 1, 2,3. So,a1a2+b1b2+c1c2=7.1+5.2+1.3    =710+3    =0Thus, both lines are perpendicular to each other.

 

Q.19

Findtheshortestdistancebetweenthelinesr=i^+2j^+k^+λi^j^+k^andr=2i^j^+k^+μ2i^+j^+2k^

Ans.

Theequationsofthegivenlinesare:r=i^+2j^+k^+λi^j^+k^    ...ir=2i^j^k^+μ2i^+j^+2k^...iiComparing equatini with r=a1+λb1 and equationii with  r=a2+μb2, we geta1=i^+2j^+k^, b1=i^j^+k^and  a2=2i^j^k^, b2=2i^+j^+2k^The shortest distance between the lines        d=b1×b2.a2a1b1×b2...iiiSo,b1×b2=i^j^k^111212=21i^22j^+1+2k^=3i^0.j^+3k^    b1×b2=3i^0.j^+3k^=32+32=9+9=32    a2a1=2i^j^k^i^+2j^+k^=i^3j^2k^From equationi, we get            d=3i^+3k^.i^3j^2k^32      =3632      =32×22      =322Thus, the distance between the given two lines is 322 units.

 

Q.20

Findtheshortestdistancebetweenthelinesx+17=y+16=z11andx31=y52=z71

Ans.

Theequationsofthegivenlinesare:X+17=Y+16=Z11andX31=Y52=Z71Comparingboth  the  equations  withXX1a1=YY1b1=ZZ1C1  and  XX2a2=Y5b2=Z7C2  respectively.X1=1,  X2=3Y1=1,  Y2=5Z1=1,  Z2=7a1=7,  a2=1b1=6,  b2=2c1=1,  c2=1Distance  between  two  lines  is​ given  by    d=x2x1y2y1z2z1a1b1c1a2b2c2b1c2b2c12+c1a2c2a12+a1b2a2b12=3151717611216×12.12+1×11×72+7×21×62      d=46876112142+62+82=46+2671+814+616+36+64=163664116=116229=229So,thedistancebetweentwogivenlinesis 229units.

 

Q.21

Findtheshortestdistancebetweenthelineswhosevectorequationsarer=i^+2j^+3k^+λi^j^+2k^andr=4i^+j^+6k^+μ2i^+3j^+k^

Ans.

Theequationsofthegivenlinesare:r=i^+2j^+3k^+λi^3j^+2k^...ir=4i^+5j^+6k^+μ2i^+3j^+k^  ...iiComparingequatiniwithr=a1+λb1andequationii with  r=a2+μb2, we get  a1=i^+2j^+3k^, b1=i^3j^+2k^and  a2=4i^+5j^+6k^, b2=2i^+3j^+k^Theshortestdistancebetweenthelines        d=b1×b2.a2a1b1×b2...iiiSo,b1×b2=i^j^k^132231=36i^14j^+3+6k^=9i^+3j^+9k^    b1×b2=9i^+3j^+9k^=92+32+92=81+9+81=171=319    a2a1=4i^+5j^+6k^i^+2j^+3k^=3i^+3j^+3k^Fromequationi,weget            d=9i^+3j^+9k^.3i^+3j^+3k^319      =33i^+j^+3k^.i^+j^+k^19      =33+1+319      =319×1919      =31919So,thedistancebetweentwogivenlinesis31919units.

 

Q.22

Findtheshortestdistancebetweenthelineswhosevectorequationsarer=1ti^+t2j^+32tk^andr=s+1i^+2s1j^2s+1k^

Ans.

Theequationsofthegivenlinesare:r=1ti^+t2j^+32tk^r=i^2j^+3k^+ti^+j^2k^...ir=s+1i^+2s1j^2s+1k^r=i^j^k^+si^+2j^2k^  ...iiComparingequatiniwithr=a1+λb1andequationii with  r=a2+μb2, we get  a1=i^2j^+3k^, b1=i^+j^2k^and  a2=i^j^k^, b2=i^+2j^2k^Theshortestdistancebetweenthelines        d=b1×b2.a2a1b1×b2...iiiSo,b1×b2=i^j^k^112122=2+4i^2+2j^+21k^=2i^4j^3k^  b1×b2=2i^4j^3k^=22+42+32=4+16+9=29    a2a1=i^j^k^i^2j^+3k^=j^4k^Fromequationi, weget            d=2i^4j^3k^.j^4k^29=4+1229=829×2929=82929Therefore,theshortestdistancebetweentwolinesis82929  units.

 

Q.23 In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2 (b) x + y + z = 1
(c) 2x + 3y – z = 5 (d) 5y + 8 = 0

Ans.

a The equatin of the plane is z=20.x+0.y+z=2 i The direction ratios of normal are 0, 0 and 1. 0 2 +0 2 +1 2 =1 So, according to rule, dividing equation i by 1, we get 0.x+0.y+z=2, which is in the form of lx+my+nz=d, where direction cosines, l=0,m=0 and n=1 and the distance of the plane from origin is 2 units. b x+y+z=1 i The direction ratios of normal are 1, 1 and 1. 1 2 +1 2 +1 2 = 3 Dividing equation i by 3 , we get x 3 + y 3 + z 3 = 1 3 ii Comparing equation ii with the equation lx+my+nz=d,we get Direction cosines: 1 3 , 1 3 , 1 3 and the distance of plane from the origin is 1 3 units. c 2x+3y-z=5 i The direction ratios of normal are 2, 3 and-1. 2 2 +3 2 + -1 2 = 14 Dividing equation i by 14 , we get 2x 14 + 3y 14 z 14 = 5 14 ii Comparing equation ii with the equation lx+my+nz=d,we get Direction cosines: 2 14 , 3 14 , -1 14 and the distance of plane from the origin is 5 14 units. d 5y+8=0 -0x-5y-0z=8 i The direction ratios of normal are 0, -5 and0. 0 2 + -5 2 +0 2 =5 Dividing equation i by 5, we get 0x 5 5y 5 0z 5 = 8 5 ii Comparing equation ii with the equation lx+my+nz=d,we get Direction cosines: 0,-1,0 and the distance of plane from the origin is 8 5 units.

 

Q.24 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

3 i ^ +5 j ^ 6 k ^ . MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWFZaGaaGPaVpaaHaaabaGaa8xAaaGaayPadaGaa83kaiaa=vdacaaMc8+aaecaaeaacaWFQbaacaGLcmaacqGHsislcaWF2aGaaGPaVpaaHaaabaGaae4AaaGaayPadaGaaiOlaaaa@472D@

Ans.

Here,n=3i^+5j^6k^  n^=nn      =3i^+5j^6k^3i^+5j^6k^      =3i^+5j^6k^32+52+62      =3i^+5j^6k^9+25+36      =3i^+5j^6k^70Theequationoftheplanewithpositionvectorrisgivenby        r.n^=dr3i^+5j^6k^70=7Thisisthevectorequtionof​ theplane.

 

Q.25

Findthecartesianequationofthefollowing​ planes:a  r.i^+j^k^=2br.2i^+3j^4k^=1cr.s2ti^+3tj^+2s+tk^=15

Ans.

aThegivenequationoftheplaneisri^+j^k^=2...iForanyarbitrarypointPx,y,zontheplane,positionvectorrisgivenby,r=xi^+yj^+zk^.Substitutinvalueofr  inequationi,wegetxi^+yj^+zk^i^+j^k^=2                          x+yz=2Thisisthecartesianequationoftheplane.bThegivenequationoftheplaneis      r2i^+3j^4k^=1...iForanyarbitrarypointPx,y,zontheplane,positionvectorrisgivenby,r=xi^+yj^+zk^Substitutinvalueofr  inequationi,wegetxi^+yj^+zk^2i^+3j^4k^=1              2x+3y4z=1Thisisthecartesianequationoftheplane.cThegivenequationoftheplaneisr.s2ti^+3tj^+2s+tk^=15...iForanyarbitrarypointPx,y,zontheplane,positionvectorrisgivenby,r=xi^+yj^+zk^Substitutinvalueofr  inequationi,wegetxi^+yj^+zk^.s2ti^+3tj^+2s+tk^=1              s2tx+3ty+2s+tz=1Thisisthecartesianequationoftheplane.

 

Q.26 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12= 0 (b) 3y + 4z – 6 = 0
(c) x + y + z = 1 (d) 5y + 8 = 0

Ans.

aThe given equation of the plane is 2x+3y+4z-12=0    2x+3y+4z=12...iLet the coordinates of the foot of perpendicular from origin tothe plane is x1, y1, z1. Then pointx1, y1, z1will satisfyequation i, so      2x1+3y1+4z1=12Then, the direction ratios of the perpendicular are  2,  3,  4.Dividing equation  i by 22+32+42=29, we get  229x+329y+429z=1229This equation is in the form of lx+my+nz=d, where l, m and nare the direction cosines of normal to the plane and d is the distance of the normal from the origin.The coordinates of the foot of the perpendicular are given byld,md,nd.Therefore, the coordiantes of the foot of the perpendicular are229.1229,329.1229,429.1229=2429,3629,4829.bThe given equation of the plane is             3y+4z-6=0    0.x+3y+4z=12...iLet the coordinates of the foot of perpendicular from origin tothe plane is x1, y1, z1. Then pointx1, y1, z1will satisfyequation i, so     0.x1+3y1+4z1=6Then, the direction ratios of the perpendicular are  0,  3,  4.Dividing equation  i by 02+32+42=5, we get            05x+35y+45z=65This equation is in the form of lx+my+nz=d, where l, m and nare the direction cosines of normal to the plane and d is thedistance of the normal from the origin.The coordinates of the foot of the perpendicular are given byld,md,nd.Therefore, the coordiantes of the foot of the perpendicular are05.65,35.65,45.65=0,1825,2425.cThe given equation of the plane is                 x+y+z=1...iLet the coordinates of the foot of perpendicular from origin tothe plane is x1, y1, z1. Then pointx1, y1, z1will satisfyequation i, so    x+y+z=1Then, the direction ratios of the perpendicular are  1,1, 1.Dividing equation  i by 12+12+12=3, we get      13x+13y+13z=13This equation is in the form of lx+my+nz=d, where l, m and nare the direction cosines of normal to the plane and d is thedistance of the normal from the origin.The coordinates of the foot of the perpendicular are given byld,md,nd.Therefore, the coordiantes of the foot of the perpendicular are13.13,13.13,13.13=13,13,13.dThe given equation of the plane is                         5y+8=0                   0.x-5y+0.z= 8...iLet the coordinates of the foot of perpendicular from origin tothe plane is x1, y1, z1. Then pointx1, y1, z1will satisfyequation i, so         0.x15y1+0.z1= 8Then, the direction ratios of the perpendicular are  0,5, 0.Dividing equation  i by 02+52+02=5, we get            05x55y+05z=85This equation is in the form of lx+my+nz=d, where l, m and nare the direction cosines of normal to the plane and d is thedistance of the normal from the origin.The coordinates of the foot of the perpendicular are given byld,md,nd.Therefore, the coordiantes of the foot of the perpendicular are0×85,55×85,0×85=0,-85,0.

 

Q.27

Findthevectorandcartesianequationsoftheplanesa thatpassesthroughthepoint1,0,2andthenormaltotheplaneisi^+j^k^.b thatpassesthroughthepoint1,4,6andthenormalvectortotheplaneisi^2j^+k^.

Ans.

aThepositionvectorofpoint1,0,2,  a=i^+0.j^2k^andthenormalvectorNperpendiculartotheplaneasN=i^+j^k^Therefore,thevectorequationoftheplaneisgivenby                  ra.N=0orri^+0.j^2k^.i^+j^k^=0...iristhepositionvectorofanypointPx,y,zintheplane.  r=xi^+yj^+zk^Therefore,equationibecomes  xi^+yj^+zk^i^+0.j^2k^.i^+j^k^=0      x1i^+yj^+z+2k^.i^+j^k^=0                                    x1+yz+2=0                                      x+yz=3Thisisthecartesianequationoftherequiredplane.bThepositionvectorofpoint​ 1,4,6,  a=i^+4j^+6k^andthenormalvectorNperpendiculartotheplaneasN=i^2j^+k^Therefore,thevectorequationoftheplaneisgivenby                  ra.N=0orri^+4j^+6k^.i^2j^+k^=0...iristhepositionvectorofanypointPx,y,zintheplane.  r=xi^+yj^+zk^Therefore,equationibecomes          xi^+yj^+zk^i^+4j^+6k^.i^2j^+k^=0    x1i^+y4j^+z6k^.i^2j^+k^=0                                x12y4+z6=0                                            x2y+z+1=0Thisisthecartesianequationoftherequiredplane.

 

Q.28 Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, –5), (–4, –2, 3)

(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)

Ans.

( a )Let vector forms of the given three points are: a = i ^ + j ^ k ^ , b =6 i ^ +4 j ^ 5 k ^ and c =4 i ^ 2 j ^ +3 k ^ Then, the vector equation of the plane passing through a , b and c is given by ( r a ).[ ( b a )×( c a ) ]=0 i.e., [ r ( i ^ + j ^ k ^ ) ].[ ( 5 i ^ +3 j ^ 4 k ^ )×( 5 i ^ 3 j ^ +4 k ^ ) ]=0 [ r ( i ^ + j ^ k ^ ) ].| i ^ j ^ k ^ 5 3 4 5 3 4 |=0 [ r ( i ^ + j ^ k ^ ) ].( 0. i ^ 0 .j ^ +0. k ^ )=0 0=0 Since, the given three points are collinear, so there will be infininte numbers of planes passing through the given points. ( b )Let vector forms of the given three points are: a = i ^ + j ^ +0 k ^ , b = i ^ +2 j ^ + k ^ and c =2 i ^ +2 j ^ k ^ Then, the vector equation of the plane passing through a , b and c is given by ( r a ).[ ( b a )×( c a ) ]=0 i.e., [ r ( i ^ + j ^ ) ].[ ( 0 i ^ + j ^ + k ^ )×( 3 i ^ + j ^ k ^ ) ]=0 [ r ( i ^ + j ^ ) ].| i ^ j ^ k ^ 0 1 1 3 1 1 |=0 [ x i ^ +y j ^ +z k ^ ( i ^ + j ^ ) ].( 2 i ^ 3 j ^ +3 k ^ )=0 [ ( x1 ) i ^ +( y1 ) j ^ +z k ^ ].( 2 i ^ 3 j ^ +3 k ^ )=0 2( x1 )3( y1 )+3z=0 2x+23y+3+3z=0 2x+3y3z=5 This is the cartesian equation of the required plane. 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Q.29 Find the intercepts cut off by the plane
2x + y – z = 5

Ans.

Thegivenequationoftheplaneis:2x+yz=5...iDividingbothsidesofequationiby5,weget2x5+y5z5=55  x52+y5+z5=1Comparingthisequationoftheplanewithxa+yb+zc=1,we getinterceptscutbyplaneatx, yandzaxes:a=52,b=5andc=5Thus,theinterceptscutoffbytheplaneare52,5and5.

 

Q.30 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Ans.

The equation of ZOX plane is y=0

Equation of any plane parallel to ZOX plane is

y = a

Since, y-intercept of the plane is 3, so

a = 3

Thus, the equation of the required plane is y = 3.

 

Q.31 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 with (2, 2, 1).

Ans.

Theequationsofgivenplanesare: 3xy+2z4=0 and x+y+z2=0 Theequationofplanethroughintersectionofgivenplanesis 3xy+2z4 +λ x+y+z2 =0whereλR i Theplanepassesthroughthepoint 2,2,1 ,sofromequation i 3×22+2×14 +λ 2+2+12 =0 2+λ 3 =0λ= 2 3 Puttingvalueofλinequation i ,weget 3xy+2z4 2 3 x+y+z2 =0 3 3xy+2z4 2 x+y+z2 =0 9x3y+6z122x2y2z+4=0 7x5y+4z8=0 Thisistherequiredequationoftheplane.

 

Q.32n

Findthevectorequationof​​theplanepassing​ throughtheintersectionofthe​ planes  r.2i^+2j^3k^=7,r.2i^+5j^+3k^=9andthroughthepoint2,1,3.

Ans.

The givenen equations of the plane are:r.(2i^+2j^3k^)=7 or r.(2i^+2j^3k^)7=0...(i)r.(2i^+5j^+3k^)=9 or r.(2i^+5j^+3k^)9=0...(ii)The equation of any plane throught the intersection of theplanes given in equation (i) and (ii) is given by{r.(2i^+2j^3k^)7}+λ{r.(2i^+5j^+3k^)9}=0,λR          r.[(2i^+2j^3k^)+λ(2i^+5j^+3k^)]=9λ+7        r.[(2+2λ)i^+(2+5λ)j^(33λ)k^]=9λ+7  ...(iii)The plane passes through the point (2,1,3). Therefore,itsposition vector is given by,r=2i^+j^+3k^Substituting value of rin equation(iii), we get(2i^+j^+3k^).[(2+2λ)i^+(2+5λ)j^(33λ)k^]=9λ+72(2+2λ)+(2+5λ)3(33λ)=9λ+7    4+4λ+2+5λ9+9λ=9λ+79λ3=7λ=109Putting λ=109 in equation(iii), we getr.[(2+2×109)i^+(2+5×109)j^(33×109)k^]=9×109+7r.[38i^+68j^+3k^]=153This is the vector equation of the required plane.

 

Q.33 Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Ans.

Theequationsofgivenplanesare: x+y+z=1 and2x+3y+4z=5 Theequationofplanethroughintersectionofgivenplanesis x+y+z1 +λ 2x+3y+4z5 =0whereλR 1+2λ x+ 1+3λ y+ 1+4λ z=5λ+1 i The direction ratios of the plane are 1+2λ , 1+3λ and 1+4λ . Plane i is perpendicular to xy+z=0, whose direction ratios are 1,1,1. Since, plane i and ii are perpendicular, so a 1 a 2 +b 1 b 2 +c 1 c 2 =0 1+2λ 1 + 1+3λ 1 + 1+4λ 1 =0 1+2λ13λ+1+4λ=0 3λ=1 λ= 1 3 Substituting λ= 1 3 in equation i , we get 1+2× 1 3 x+ 1+3× 1 3 y+ 1+4× 1 3 z=5× 1 3 +1 1 3 x+ 0 y 1 3 z= 2 3 xz=2 xz+2=0 This is the required equation of the plane. The direction ratios of the plane are 1+2λ , 1+3λ and 1+4λ . Plane i is perpendicular to xy+z=0, whose direction ratios are 1,1,1. Since, plane i and ii are perpendicular, so a 1 a 2 +b 1 b 2 +c 1 c 2 =0 1+2λ 1 + 1+3λ 1 + 1+4λ 1 =0 1+2λ13λ+1+4λ=0 3λ=1 λ= 1 3 Substituting λ= 1 3 in equation i , we get 1+2× 1 3 x+ 1+3× 1 3 y+ 1+4× 1 3 z=5× 1 3 +1 1 3 x+ 0 y 1 3 z= 2 3 xz=2 xz+2=0 This is the required equation of the plane.

 

Q.34

Findtheanglebetweentheplaneswhosevectorequationsarer2i^+2j^k^=5and  r3i^3j^+5k^=

Ans.

Theplaneswhosevectorequationsarer(2i^+2j^3k^)=5andr(3i^3j^+5k^)= 3comparing with r.n1=d1 and r.n2=d2, thenn1=(2i^+2j^3k^) and n2=(3i^3j^+5k^)n1.n2=(2i^+2j^3k^).(3i^3j^+5k^)    =6615    =15      |n1|=|2i^+2j^3k^|    =22+22+(3)2    =4+4+9=17      |n2|=|3i^3j^+5k^|    =32+(3)2+52    =9+9+25=43Let angle between two planes be θ, then  cosθ=n1.n2|n1||n2|    =151743    =15731        θ=cos1(15731)

 

Q.35 In the following cases, determine whether
the given planes are parallel or perpendicular and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x –2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y +6 z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y +3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Ans.

a 7x+5y+6z+30=0 and 3xy10z+4=0 Direction ratios of the given planes are: a 1 = 7, b 1 = 5, c 1 = 6 a 2 = 3, b 2 = 1, c 2 =10 a 1 a 2 + b 1 b 2 + c 1 c 2 =7×3+5×1+6×10 =21560=440 Therefore, the given planes are not perpendicular to each other. a 1 a 2 = 7 3 , b 1 b 2 = 5 1 and c 1 c 2 = 6 10 = 3 5 a 1 a 2 b 1 b 2 c 1 c 2 Therefore, the given planes are not parallel to each other. The angle between two planes is θ= cos 1 a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 1 2 a 2 2 + b 2 2 + c 2 2 = cos 1 7×3+5×1+6×10 7 2 + 5 2 + 6 2 3 2 + 1 2 + 10 2 = cos 1 44 110 110 = cos 1 44 110 = cos 1 2 5 b 2x+y+3z2=0 and x2y+5=0 Direction ratios of the given planes are: a 1 = 2, b 1 = 1, c 1 =3 a 2 = 1, b 2 = 2, c 2 =0 a 1 a 2 + b 1 b 2 + c 1 c 2 =2×1+1×2+3×0 =22+0=0 Therefore, the given planes are perpendicular to each other. c 2x2y+4z+5=0 and 3x3y+6 z1=0 Direction ratios of the given planes are: a 1 = 2, b 1 = 2, c 1 =4 a 2 = 3, b 2 = 3, c 2 =6 a 1 a 2 + b 1 b 2 + c 1 c 2 =2×3+2×3+4×6 =6+6+24=360 Therefore, the given planes are not perpendicular to each other. a 1 a 2 = 2 3 , b 1 b 2 = 2 3 = 2 3 and c 1 c 2 = 4 6 = 2 3 a 1 a 2 = b 1 b 2 = c 1 c 2 Therefore, the given planes are parallel to each other. d 2xy+3z1=0 and 2xy+3z+3=0 Direction ratios of the given planes are: a 1 = 2, b 1 = 1, c 1 =3 a 2 = 2, b 2 = 1, c 2 =3 a 1 a 2 + b 1 b 2 + c 1 c 2 =2×2+1×1+3×3 =4+1+9=140 Therefore, the given planes are not perpendicular to each other. a 1 a 2 = 2 2 =1, b 1 b 2 = 1 1 =1 and c 1 c 2 = 3 3 =1 a 1 a 2 = b 1 b 2 = c 1 c 2 Therefore, the given planes are parallel to each other. e 4x+8y+z8=0 and y+z4=0 Direction ratios of the given planes are: a 1 = 4, b 1 = 8, c 1 =1 a 2 = 0, b 2 = 1, c 2 =1 a 1 a 2 + b 1 b 2 + c 1 c 2 =4×0+8×1+1×1 =0+8+1=90 Therefore, the given planes are not perpendicular to each other. a 1 a 2 = 4 0 , b 1 b 2 = 8 1 =8 and c 1 c 2 = 1 1 =1 a 1 a 2 b 1 b 2 c 1 c 2 Therefore, the given planes are notparallel to each other. The angle between two planes is θ= cos 1 a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 1 2 a 2 2 + b 2 2 + c 2 2 = cos 1 4×0+8×1+1×1 4 2 + 8 2 + 1 2 0 2 + 1 2 + 1 2 = cos 1 9 16+64+1 0+1+1 = cos 1 9 81 2 = cos 1 1 2 = π 4 Thus, the angle between two planes is π 4 .

 

Q.36 In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0) 3x – 4y + 12 z = 3

(b) (3, –2, 1) 2x – y + 2z + 3 = 0

(c) (2, 3, –5) x – 2y + 2z = 9

(d) (–6, 0, 0) 2x – 3y + 6z – 2 =0

Ans.

a The given equation of plane is 3x4y+12 z=3 and point is 0,0,0 . The distance of plane from given point d = 3 0 4 0 +12 0 3 3 2 + 4 2 + 12 2 = 3 9+16+144 = 3 169 = 3 13 b The given equation of plane is 2xy+2z+3=0 and point is 3,2,1 . The distance of plane from given point d = 2 3 2 +2 1 +3 2 2 + 1 2 + 2 2 = 13 4+1+4 = 13 9 = 13 3 c The given equation of plane is x2y+2z9=0 and point is 2,3,5 . The distance of plane from given point d = 2 2 3 +2 5 9 1 2 + 2 2 + 2 2 = 23 1+4+4 = 23 9 = 23 3 d Thegivenequationofplaneis2x3y+6z2=0andpoint is 6,0,0 . Thedistanceof planefromgivenpoint d = 2 6 3 0 +6 0 2 2 2 + 3 2 + 6 2 = 14 4+9+36 = 14 49 = 14 7 =2

 

Q.37 Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, –1), (4, 3, –1).

Ans.

Let OA be the line joining the origin, O (0, 0, 0) and the point, A (2, 1, 1). Then, the direction ratios of OA are: (2 – 0), (1 – 0), (1 – 0) i.e., 2, 1, 1.
Let BC be the line joining the origin, B (3, 5, –1) and the point, C (4, 3,–1).Then, the direction ratios of BC are: (4 – 3), (3 – 5), (–1+1) i.e., 1, –2, 0.

OA is perpendicular to BC, if a1a2+b1b2+c1c2= 0

a1a2+b1b2+c1c2 = (2)(1)+(1)(–2)+(1)(0)
= 2 – 2+0 = 0
Thus, OA is perpendicular to BC.

 

Q.38 If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 – m2n1, n1l2 – n2l1, l1m2l2 m1.

Ans.

Since l 1 , m 1 , n 1 and l 2 , m 2 , n 2 are the direction cosines of two mutually perpendicular lines. Therefore, l 1 l 2 + m 1 m 2 + n 1 n 2 =0 ( i ) l 1 2 + m 1 2 + n 1 2 =1 ( ii ) l 2 2 + m 2 2 + n 2 2 =1 ( iii ) Let l,m,n be the direction cosines of a line which is perpendicular to each given line with direction cosines l 1 , m 1 , n 1 and l 2 , m 2 , n 2 . l 1 l+ m 1 m+ n 1 n=0 l l 2 +m m 2 +n n 2 =0 l m 1 n 2 m 2 n 1 = m n 1 l 2 n 2 l 1 = n l 1 m 2 m 1 l 2 l 2 ( m 1 n 2 m 2 n 1 ) 2 = m 2 ( n 1 l 2 n 2 l 1 ) 2 = n 2 ( l 1 m 2 m 1 l 2 ) 2 = l 2 + m 2 + n 2 ( m 1 n 2 m 2 n 1 ) 2 + ( n 1 l 2 n 2 l 1 ) 2 + ( l 1 m 2 m 1 l 2 ) 2 ( iv ) Since, l,m,n are the direction cosines of line. l 2 + m 2 + n 2 =1 ( v ) It is known that ( l 1 2 + m 1 2 + n 1 2 )( l 2 2 + m 2 2 + n 2 2 ) ( l 1 l 2 + m 1 m 2 + n 1 n 2 ) 2 = ( m 1 n 2 m 2 n 1 ) 2 + ( n 1 l 2 n 2 l 1 ) 2 + ( l 1 m 2 m 1 l 2 ) 2 From equations( i ),( ii ) and ( iii ), we get 1.10= ( m 1 n 2 m 2 n 1 ) 2 + ( n 1 l 2 n 2 l 1 ) 2 + ( l 1 m 2 m 1 l 2 ) 2 ( m 1 n 2 m 2 n 1 ) 2 + ( n 1 l 2 n 2 l 1 ) 2 + ( l 1 m 2 m 1 l 2 ) 2 =1 ( vi ) Substituting the values from equations ( v ) and ( vi ) in equation ( iv ), we get l 2 ( m 1 n 2 m 2 n 1 ) 2 = m 2 ( n 1 l 2 n 2 l 1 ) 2 = n 2 ( l 1 m 2 m 1 l 2 ) 2 = 1 1 l 2 ( m 1 n 2 m 2 n 1 ) 2 = m 2 ( n 1 l 2 n 2 l 1 ) 2 = n 2 ( l 1 m 2 m 1 l 2 ) 2 =1 l= m 1 n 2 m 2 n 1 , m= n 1 l 2 n 2 l 1 and n= l 1 m 2 m 1 l 2 Thus, the direction cosines of the required line are m 1 n 2 m 2 n 1 , n 1 l 2 n 2 l 1 and l 1 m 2 m 1 l 2 . 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Q.39 Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Ans.

Let the angle between the lines whose direction ratios are a, b, c and bc, ca, ab be θ, then cosθ=| a( bc )+b( ca )+c( ab ) a 2 + b 2 + c 2 ( bc ) 2 + ( ca ) 2 + ( ab ) 2 | =| abac+bcba+cabc a 2 + b 2 + c 2 ( bc ) 2 + ( ca ) 2 + ( ab ) 2 | =0 cosθ=cos90°θ=90° Thus, the angle between the lines is 90°. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2BBB@

 

Q.40 Find the equation of a line parallel to x-axis and passing through the origin.

Ans.

Since, the line parallel to x-axis and passing through the origin is x-axis itself. Let A( a,0,0 ) be a point on x-axis, where aR. So, direction cosines of OA are ( a0 ),( 00 ),( 00 )i.e., a,0,0. The equation of OA is given as x0 a = y0 0 = z0 0 x 1 = y 0 = z 0 =a Thus, the equation of line parallel to x-axis and passing through origin is x 1 = y 0 = z 0 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadofacaWGPbGaamOBaiaadogacaWGLbGaaiilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiBaiaabMgacaqGUbGaaeyzaiaabccacaqGWbGaaeyyaiaabkhacaqGHbGaaeiBaiaabYgacaqGLbGaaeiBaiaabccacaqG0bGaae4BaiaabccacaqG4bGaaeylaiaabggacaqG4bGaaeyAaiaabohacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabchacaqGHbGaae4CaiaabohacaqGPbGaaeOBaiaabEgacaqGGaGaaeiDaiaabIgacaqGYbGaae4BaiaabwhacaqGNbGaaeiAaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4BaiaabkhacaqGPbGaae4zaiaabMgacaqGUbGaaeiiaaqaaiaabMgacaqGZbGaaeiiaiaabIhacaqGTaGaaeyyaiaabIhacaqGPbGaae4CaiaabccacaqGPbGaaeiDaiaabohacaqGLbGaaeiBaiaabAgacaqGUaGaaeiiaiaabYeacaqGLbGaaeiDaiaabccacaqGbbWaaeWaaeaacaWGHbGaaiilaiaaicdacaGGSaGaaGimaaGaayjkaiaawMcaaiaabccacaqGIbGaaeyzaiaabccacaqGHbGaaeiiaiaabchacaqGVbGaaeyAaiaab6gacaqG0bGaaeiiaiaab+gacaqGUbGaaeiiaiaabIhacaqGTaGaaeyyaiaabIhacaqGPbGaae4CaiaabYcacaqGGaGaae4DaiaabIgacaqGLbGaaeOCaiaabwgacaqGGaGaaeyyaiabgIGiolaabkfacaqGUaaabaGaae4uaiaab+gacaqGSaGaaeiiaiaabsgacaqGPbGaaeOCaiaabwgacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGJbGaae4BaiaabohacaqGPbGaaeOBaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaab+eacaqGbbGaaeiiaiaabggacaqGYbGaaeyzaiaabccadaqadaqaaiaadggacqGHsislcaaIWaaacaGLOaGaayzkaaGaaiilamaabmaabaGaaGimaiabgkHiTiaaicdaaiaawIcacaGLPaaacaGGSaWaaeWaaeaacaaIWaGaeyOeI0IaaGimaaGaayjkaiaawMcaaiaadMgacaGGUaGaamyzaiaac6cacaGGSaGaaeiiaiaabggacaqGSaGaaeimaiaabYcacaqGWaGaaeOlaaqaaiaabsfacaqGObGaaeyzaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqGpbGaaeyqaiaabccacaqGPbGaae4CaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGHbGaae4Caaqaamacyc4caaqaiGjGcGaMaoiEaiadycOHsislcGaMaIimaaqaiGjGcGaMaoyyaaaacqGH9aqpdaWcaaqaaiaadMhacqGHsislcaaIWaaabaGaaGimaaaacqGH9aqpdaWcaaqaaiaadQhacqGHsislcaaIWaaabaGaaGimaaaacqGHshI3daWcaaqaaiaadIhaaeaacaaIXaaaaiabg2da9maalaaabaGaamyEaaqaaiaaicdaaaGaeyypa0ZaaSaaaeaacaWG6baabaGaaGimaaaacqGH9aqpcaWGHbaabaGaamivaiaadIgacaWG1bGaam4CaiaacYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaabYgacaqGPbGaaeOBaiaabwgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgacaqGSbGaaeyzaiaabYgacaqGGaGaaeiDaiaab+gacaqGGaGaaeiEaiaab2cacaqGHbGaaeiEaiaabMgacaqGZbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGWbGaaeyyaiaabohacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabshacaqGObGaaeOCaiaab+gacaqG1bGaae4zaiaabIgaaeaacaqGVbGaaeOCaiaabMgacaqGNbGaaeyAaiaab6gacaqGGaGaaeyAaiaabohacaqGGaWaaSaaaeaacaWG4baabaGaaGymaaaacqGH9aqpdaWcaaqaaiaadMhaaeaacaaIWaaaaiabg2da9maalaaabaGaamOEaaqaaiaaicdaaaGaaiOlaaaaaa@5D9A@

 

Q.41 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Ans.

The coordinates of the points A, B, C, D be are ( 1, 2, 3 ), ( 4, 5, 7 ), ( 4,3,6 ) and ( 2, 9, 2 ) respectivley. The direction ratios of AB are: ( 41 ),( 52 ),( 73 )i.e, 3, 3, 4. The direction ratios of CD are: ( 2+4 ),( 93 ),( 2+6 )i.e, 6, 6, 8. Then, a 1 a 2 = 3 6 = 1 2 b 1 b 2 = 3 6 = 1 2 c 1 c 2 = 4 8 = 1 2 So, a 1 a 2 = b 1 b 2 = c 1 c 2 Thus, AB is parallel to CD. Therefore, the angle between AB and CD is either 0° or 180°. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9300@

 

Q.42

Ifthelinesx13=y22k=z32andx13k=y11=z65areperpendicular,findthevalueofk.

Ans.

Thegivenlinesarex13=y22k=z32andx13k=y11=z65.The direction ratios of given lines are:a1=3,b1=2k,  c1=2 and a2=3k,b2=1,  c2=5Then,  a1a2+b1b2+c1c2=03×3k+2k×1+2×5=09k+2k10=07k13=0    k=137Therefore, given lines are perpendicular for k=137.

 

Q.43

Findthevectorequationofthelinepassingthrough(1,2,3)andperpendiculartotheplaneri^+2j^5k^+9=0.

Ans.

The position vector of point (1,2,3)=i^+2j^+3k^The perpendicular plane is r.(i^+2j^5k^)+9=0 thendirection ratios to the plane are 1,2,5 and normal vector isN=i^+2j^5k^The equation of line passing through a point and perpendicularto the given plane isl=r+λN,  λRl=(i^+2j^+3k^)+λ(i^+2j^5k^)

 

Q.44

Findtheequationoftheplanepassingthrough(a,b,c)andparalleltotheplaneri^+j^+k^=2.

Ans.

Theequationoftheplanepassingthrough(a,b,c)andparalleltotheplaner(i^+j^+k^)=2.Equation of the plane parallel to the plane r(i^+j^+k^)=2is:r(i^+j^+k^)=λ  ...(i)The plane passes through the point (a,b,c). Therefore, theposition vector r of this point is r=ai^+bj^+ck^Substituting the value of r in equation(i), we get(ai^+bj^+ck^)(i^+j^+k^)=λa+b+c=λSo,from equation (i), the equation of required plane isr(i^+j^+k^)=a+b+c...(ii)This is the vector equation of required plane.Substituting r=xi^+yj^+zk^ in equation(ii), we get    (xi^+yj^+zk^)(i^+j^+k^)=a+b+c  x+y+z=a+b+c

 

Q.45

Findtheshortestdistancebetweenlinesr=6i^+2j^+2k^+λi^2j^+2k^andr=4i^k^+μ3i^2j^2k^.

Ans.

The equations of given lines are:r=6i^+2j^+2k^+λ(i^2j^+2k^)...(i)r=4i^k^+μ(3i^2j^2k^)...(ii)Comparing equation(i) and equation(ii) with r=a1+λb1 andr=a2+λb2, we geta1=6i^+2j^+2k^,b1=i^2j^+2k^a2=4i^k^b2=3i^2j^2k^The shortest distance between two lines(d)=|(b1×b2).(a2a1)|b1×b2||  b1×b2=|i^j^k^122322|      =(4+4)i^(26)j^+(2+6)k^      =8i^+8j^+4k^        |b1×b2|=82+82+42      =12a2a1=(4i^k^)(6i^+2j^+2k^)      =10i^2j^3k^  d=|(8i^+8j^+4k^).(10i^2j^3k^)12|      =|80161212|      =10812=9Therefore, the shortest distance between two lines is 9 units.

 

Q.46 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

Ans.

The equation of the line passing through the points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is x x 1 x 2 x 1 = y y 1 y 2 y 1 = z z 1 z 2 z 1 The equation of the line passing through the points ( 5,1,6 ) and ( 3,4,1 ) is given as x5 35 = y1 41 = z6 16 x5 2 = y1 3 = z6 5 =k( let ) x=52k,y=1+3k and z=65k Any point on the line is of the form ( 52k,1+3k,65k ). The equation of YZ-plane is x=0 Since, the line passes through YZ-plane, 52k=0k= 5 2 So,y=1+3× 5 2 = 17 2 and z=65× 5 2 = 13 2 Therefore, the required point is ( 0, 17 2 , 13 2 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@273D@

 

Q.47 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Ans.

The equation of the line passing through the points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is x x 1 x 2 x 1 = y y 1 y 2 y 1 = z z 1 z 2 z 1 The equation of the line passing through the points ( 5,1,6 ) and ( 3,4,1 ) is given as x5 35 = y1 41 = z6 16 x5 2 = y1 3 = z6 5 =k( let ) x=52k,y=1+3k and z=65k Any point on the line is of the form ( 52k,1+3k,65k ). The equation of ZX-plane is y=0 Since, the line passes through ZX-plane, 1+3k=0k= 1 3 So,x=52× 1 3 = 17 3 and z=65× 1 3 = 23 3 Therefore, the required point is ( 17 3 ,0, 23 3 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2825@

 

Q.48 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

Ans.

The equation of the line passing through the points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is x x 1 x 2 x 1 = y y 1 y 2 y 1 = z z 1 z 2 z 1 The equation of the line passing through the points ( 3,4,5 ) and ( 2,3,1 ) is given as x3 23 = y+4 3+4 = z+5 1+5 x3 1 = y+4 1 = z+5 6 =k( let ) x=3k,y=k4 and z=6k5 Any point on the line is of the form ( 3k,k4,6k5 ). Since, point lies on the plane, 2x+y+z=7 2( 3k )+k4+6k5=7 62k+k4+6k5=7 5k3=7 k= 10 5 =2 So,x=32=1, y=24=2 and z=6×25=7 Therefore, the required point is ( 1,2,7 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@73C6@

 

Q.49 Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

Ans.

The equation of the plane passing through the point ( 1,3,2 ) is a( x+1 )+b( y3 )+c( z2 )=0 ( i ) where a, b, c are the direction ratios of the plane. Since, two planes are perpendicular to each other, if a 1 a 2 + b 1 b 2 + c 1 c 2 =0 Plane( i ) is perpendicular to the plane, x+2y+3z=5 a.1+b.2+c.3=0 a+2b+3c=0 ( ii ) Plane( i ) is perpendicular to the plane, 3x+3y+z=0 a.3+b.3+c.1=0 3a+3b+c=0 ( iii ) From equation( ii ) and equation( iii ), we have a 29 = b 19 = c 36 a 7 = b 8 = c 3 =k( let ) a=7k, b=8k and c=3k Substituting the values of a, b and c in equation( i ), we get 7k( x+1 )+8k( y3 )3k( z2 )=0 7x+78y+24+3z6=0 7x8y+3z+25=0 This is the required equation of the plane. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadsfacaWGObGaamyzaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiCaiaabYgacaqGHbGaaeOBaiaabwgacaqGGaGaaeiCaiaabggacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiAaiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccadaqadaqaaiabgkHiTiaaigdacaGGSaGaaG4maiaacYcacaaIYaaacaGLOaGaayzkaaGaaeiiaiaabMgacaqGZbaabaGaaeyyamaabmaabaGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPaaacqGHRaWkcaWGIbWaaeWaaeaacaWG5bGaeyOeI0IaaG4maaGaayjkaiaawMcaaiabgUcaRiaadogadaqadaqaaiaadQhacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyypa0JaaGimaiaaxMaacaWLjaGaaiOlaiaac6cacaGGUaWaaeWaaeaacaWGPbaacaGLOaGaayzkaaaabaGaam4DaiaadIgacaWGLbGaamOCaiaadwgacaqGGaGaaeyyaiaabYcacaqGGaGaaeOyaiaabYcacaqGGaGaae4yaiaabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabsgacaqGPbGaaeOCaiaabwgacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGYbGaaeyyaiaabshacaqGPbGaae4BaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabchacaqGSbGaaeyyaiaab6gacaqGLbGaaeOlaaqaaiaabofacaqGPbGaaeOBaiaabogacaqGLbGaaeilaiaabccacaqG0bGaae4Daiaab+gacaqGGaGaaeiCaiaabYgacaqGHbGaaeOBaiaabwgacaqGZbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGWbGaaeyzaiaabkhacaqGWbGaaeyzaiaab6gacaqGKbGaaeyAaiaabogacaqG1bGaaeiBaiaabggacaqGYbGaaeiiaiaabshacaqGVbGaaeiiaiaabwgacaqGHbGaae4yaiaabIgacaqGGaGaae4BaiaabshacaqGObGaaeyzaiaabkhacaqGSaGaaeiiaiaabMgacaqGMbaabaGaiGjGdggadGaMaUbaaSqaiGjGcGaMaIymaaqajGjGaOGaiGjGdggadGaMaUbaaSqaiGjGcGaMaIOmaaqajGjGaOGamGjGgUcaRiaadkgadaWgaaWcbaGaaGymaaqabaGccaWGIbWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIaam4yamaaBaaaleaacaaIXaaabeaakiaadogadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaqGWaaabaGaaeiuaiaabYgacaqGHbGaaeOBaiaabwgadaqadaqaaiaadMgaaiaawIcacaGLPaaacaqGGaGaaeyAaiaabohacaqGGaGaaeiCaiaabwgacaqGYbGaaeiCaiaabwgacaqGUbGaaeizaiaabMgacaqGJbGaaeyDaiaabYgacaqGHbGaaeOCaiaabccacaqG0bGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiCaiaabYgacaqGHbGaaeOBaiaabwgacaqGSaGaaeiiaiaabIhacqGHRaWkcaaIYaGaamyEaiabgUcaRiaaiodacaWG6bGaeyypa0JaaGynaaqaaiabgsJiCjaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGHbGaaiOlaiaaigdacWaMaA4kaSIaamOyaiaac6cacaaIYaGaey4kaSIaam4yaiaac6cacaaIZaGaeyypa0JaaeimaaqaaiabgkDiElaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGHbGaey4kaSIaaGOmaiaadkgacqGHRaWkcaaIZaGaam4yaiabg2da9iaaicdacaWLjaGaaCzcaiaaxMaacaWLjaGaaiOlaiaac6cacaGGUaWaaeWaaeaacaWGPbGaamyAaaGaayjkaiaawMcaaaqaaiaabcfacaqGSbGaaeyyaiaab6gacaqGLbWaaeWaaeaacaWGPbaacaGLOaGaayzkaaGaaeiiaiaabMgacaqGZbGaaeiiaiaabchacaqGLbGaaeOCaiaabchacaqGLbGaaeOBaiaabsgacaqGPbGaae4yaiaabwhacaqGSbGaaeyyaiaabkhacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabchacaqGSbGaaeyyaiaab6gacaqGLbGaaeilaiaabccacaqGZaGaaeiEaiabgUcaRiaaiodacaWG5bGaey4kaSIaamOEaiabg2da9iaaicdaaeaacqGH0icxcaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaamyyaiaac6cacaaIZaGamGjGgUcaRiaadkgacaGGUaGaaG4maiabgUcaRiaadogacaGGUaGaaGymaiabg2da9iaabcdaaeaacqGHshI3caWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaIZaGaamyyaiabgUcaRiaaiodacaaMc8UaamOyaiabgUcaRiaadogacqGH9aqpcaaIWaGaaCzcaiaaxMaacaWLjaGaaCzcaiaac6cacaGGUaGaaiOlamaabmaabaGaamyAaiaadMgacaWGPbaacaGLOaGaayzkaaaabaGaamOraiaadkhacaWGVbGaamyBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBamaabmaabaGaamyAaiaadMgaaiaawIcacaGLPaaacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbWaaeWaaeaacaWGPbGaamyAaiaadMgaaiaawIcacaGLPaaacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabIgacaqGHbGaaeODaiaabwgaaeaacaWLjaWaaSaaaeaacaWGHbaabaGaaGOmaiabgkHiTiaaiMdaaaGaeyypa0ZaaSaaaeaacqGHsislcaWGIbaabaGaaGymaiabgkHiTiaaiMdaaaGaeyypa0ZaaSaaaeaacaWGJbaabaGaaG4maiabgkHiTiaaiAdaaaaabaGaeyO0H4TaaCzcamaalaaabaGaamyyaaqaaiabgkHiTiaaiEdaaaGaeyypa0ZaaSaaaeaacaWGIbaabaGaaGioaaaacqGH9aqpdaWcaaqaaiaadogaaeaacqGHsislcaaIZaaaaiabg2da9iaadUgadaqadaqaaiaadYgacaWGLbGaamiDaaGaayjkaiaawMcaaaqaaiabgkDiElaadggacqGH9aqpcqGHsislcaaI3aGaam4AaiaacYcacaqGGaGaaeOyaiabg2da9iaaykW7caaI4aGaam4AaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4yaiabg2da9iabgkHiTiaaiodacaWGRbaabaGaam4uaiaadwhacaWGIbGaam4CaiaadshacaWGPbGaamiDaiaadwhacaWG0bGaamyAaiaad6gacaWGNbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyyaiaabYcacaqGGaGaaeOyaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4yaiaabccacaqGPbGaaeOBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBamaabmaabaGaamyAaaGaayjkaiaawMcaaiaacYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaeyOeI0IaaG4naiaadUgadaqadaqaaiaadIhacqGHRaWkcaaIXaaacaGLOaGaayzkaaGaey4kaSIaaGPaVlaaiIdacaWGRbWaaeWaaeaacaWG5bGaeyOeI0IaaG4maaGaayjkaiaawMcaaiabgkHiTiaaiodacaWGRbWaaeWaaeaacaWG6bGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da9iaaicdaaeaacqGHshI3caWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaI3aGaamiEaiabgUcaRiaaiEdacqGHsislcaaI4aGaamyEaiabgUcaRiaaikda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Q.50

Ifthepoints1,1,p​ and3,0,1beequidistantfromtheplaner.3i^+4j^12k^+13=0,thenfindthevalueofp.

Ans.

The position vector through the point (1,1,p) is a1=i^+j^+pk^Similarly,the position vector through the point (3,0,1) is      a2=3i^+k^The equation of the given plane is r.(3i^+4j^12k^)+13=0Since, the perpendicular distance of a point and the plane isd=|a.Nd|N||Here,N=3i^+4j^12k^ and d=13Therefore, the distance between point (1,1,p) and plane r.(3i^+4j^12k^)+13=0 isd1=|(i^+j^+pk^).(3i^+4j^12k^)+13|3i^+4j^12k^||=|3+412p+1332+42+(12)2|  d1=|2012p|13Similarly, the distance between point (3,0,1) and plane r.(3i^+4j^12k^)+13=0 isd2=|(3i^+k^).(3i^+4j^12k^)+13|3i^+4j^12k^||=|912+1332+42+(12)2|  d2=|8|13=813Since, distance of points (1,1,p)and (3,0,1) from plane r.(3i^+4j^12k^)+13=0 is equal.  d1=d2|2012p|13=813Either (2012p)=8 or (2012p)=8p=1 or 73

 

Q.51

Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanesr.i^+j^+k^=1andr.2i^+3j^k^+4=0andparalleltoxaxis.

Ans.

Theequationoftheplanepassingthroughthelineof intersectionoftheplanesr.(i^+j^+k^)=1andr.(2i^+3j^k^)+4=0is  [r.(i^+j^+k^)1]+λ[r.(2i^+3j^k^)+4]=0      r[(i^+j^+k^)+λ(2i^+3j^k^)]+(4λ1)=0      r[(1+2λ)i^+(1+3λ)j^+(1λ)k^]+(4λ1)=0...(i)The direction cosines of plane are (1+2λ),(1+3λ),(1λ).The required plane is parallel to xaxis. Therefore, its normalis perpendicular to xaxis.The direction ratios of xaxis are 1, 0, 0.a1a2+b1b2+c1c2=0 [For perpendicular line on plane.](1+2λ).1+(1+3λ).0+(1λ).0=0        1+2λ=0              λ=12Substituting λ=12 in equation (i), we getr[(1+2×12)i^+(1+3×12)j^+(1+12)k^]+(4×121)=0  r[(11)i^+(132)j^+32k^]+(21)=0  r[12j^+32k^]3=0        r[j^+3k^]6=0Putting r=xi^+yj^+zk^, we get    (xi^+yj^+zk^)[j^+3k^]6=0y+3z6=0 or y3z+6=0This is the equation of the required plane.

 

Q.52 If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

Ans.

The coordinates of the origin, O and point P are (0, 0, 0) and (1, 2, –3) respectively.

So, the direction ratios of OP are: 1–0, 2 – 0, –3 – 0, i.e., 1, 2, –3.

The equation of plane passing through (1, 2, –3) and perpendicular to OP having direction ratios 1, 2, –3 is

1(x – 1) + 2(y – 2) – 3 (z + 3) = 0
x – 1 + 2y – 4 –3z –9 = 0
x + 2y – 3z – 14 = 0

This is the required equation of the plane.

 

Q.53

Findtheequationoftheplanewhichcontainsthelineofintersectionoftheplanesr.i^+2j^+3k^4=0,r.2i^+j^k^+5=0andwhichisperpendiculartotheplaner.5i^+3j^6k^+8=0.

Ans.

The equation of the planes are:r.(i^+2j^+3k^)4=0...(i)r.(2i^+j^k^)+5=0...(ii)The equation of the plane passing through the intersection of the planes  (i) and (ii) is[r.(i^+2j^+3k^)4]+λ[r.(2i^+j^k^)+5]=0  r[(1+2λ)i^+(2+λ)j^+(3λ)k^]+5λ4=0    ...(iii)Plane (iii) is perpendicular to the plane r.(5i^+3j^6k^)+8=0,  (1+2λ).5+(2+λ).3+(3λ).(6)=0  5+10λ+6+3λ18+6λ=0                  19λ=7                      λ=719Substituting λ=719 in equation (iii), we get  r[(1+2×719)i^+(2+719)j^+(3719)k^]+5×7194=0      r[3319i^+4519j^+5019k^]4119=0          r(33i^+45j^+50k^)41=0(iv)This is the vector equation of the  required plane.Substituting r=xi^+yj^+zk^ in equation  (iv), we get      (xi^+yj^+zk^)(33i^+45j^+50k^)41=0    33x+45y+50z41=0This is the cartesian equation of the  required plane.

 

Q.54

Findthedistanceofthepoint1,5,10fromthepointofintersectionoftheliner=2i^j^+2k^+λ3i^+4j^+2k^andtheplaner.i^j^+k^=5.

Ans.

The equation of line is:r=2i^j^+2k^+λ(3i^+4j^+2k^)...(i)Equation of plane is:r.(i^j^+k^)=5 ...(ii)Putting r in equation(ii), we get[2i^j^+2k^+λ(3i^+4j^+2k^)].(i^j^+k^)=5    2+1+2+λ(34+2)=5        λ=0Putting λ=0 in equation(i), we get      r=2i^j^+2k^+0.(3i^+4j^+2k^)=2i^j^+2k^Then, the position vector of the point of intersection of the lineand the plane is r=2i^j^+2k^.The coordinates of the point of intersection of line and the planeis(2,1,2).So, the distance of the point (1,5,10) from the point of intersection i.e.,(2,1,2)=(2+1)2+(1+5)2+(2+10)2    =9+16+144    =169  =13 units.Thus, the required distance is 13 units.

 

Q.55

Findthevectorequationofthelinepassingthrough1,2,3andparalleltotheplanesr.i^j^+2k^=5andr.3i^+j^+k^=6.

Ans.

Here, a line through (1,2,3) is parallel to the two planesr.(i^j^+2k^)=5andr.(3i^+j^+k^)=6, so given line will be perpendicular to the normals on the given planes.Then,Position vector of the point (1,2,3),  a=i^+2j^+3k^Let vector parallel to required line beb=b1i^+b2j^+b3k^Comparing given planes with rN1d=0 and rN2d=0, we getN1=i^j^+2k^ and N2=3i^+j^+k^      b.N1=0 and b.N2=0(b1i^+b2j^+b3k^).(i^j^+2k^)=0 b1b2+2b3=0...(i)    (b1i^+b2j^+b3k^).(3i^+j^+k^)=0    3b1+b2+b3=0  ...(ii)On solving equation (i) and (ii), we getb11×12×1=b21×12×3=b31×1+1×3    b13=b25=b34Therefore, the direction ratios of b are 3,5 and 4.b=3i^5j^+4k^So, the eqution of the line passing through (1,2,3) and parallelto vector b is given as:r=(i^+2j^+3k^)+λ(3i^5j^+4k^)

 

Q.56

Findthevectorequationofthelinepassingthroughthepoint1,2,4andperpendiculartothetwolines:x83=y+1916=z107andx153=y298=z55.

Ans.

The equations of lines arex83=y+1916=z107...(i)x153=y298=z55...(ii)Let vector b=b1i^+b2j^+b3k^, be parallel to required line.The equation of line passing through point (1,2,4) and parallelto vector b is:r=i^+2j^4k^+λ(b1i^+b2j^+b3k^)...(iii)Since, b is perpendicular to line(i),then3b116b2+7b3=0...(iv)And, b is perpendicular to line(ii),then3b1+8b25b3=0...(v)From equation (iv) and (v), we haveb18056=b21521=b324+48b124=b236=b372b12=b23=b36Direction ratios of b are 2,3 and 6.So,    b=2i^3j^+6k^From equation(iii), we haver=i^+2j^4k^+λ(2i^3j^+6k^)This is the equation of the required line.

 

Q.57

Provethatifaplanehastheinterceptsa,b,candisatadistanceofpunitsfromtheorigin,then1a2+1b2+1c2=1p2.

Ans.

The equation of a plane having intercepts a, b, c on x, y and zaxis respectively is given below:xa+yb+zc=1...(i)The distance(p) of the plane from origin is:p=|0a+0b+0c1(1a)2+(1b)2+(1c)2|p=1(1a)2+(1b)2+(1c)2Squaring both sides,we getp2=11a2+1b2+1c2    1p2=1a2+1b2+1c2

 

Q.58 Distance between the two planes:

2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units
(C) 8 units (D)

2 29 units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaGabaiaa=jdaaeaadaGcaaqaaiaa=jdacaWF5aaaleqaaaaakiaa=bcacaWF1bGaa8NBaiaa=LgacaWF0bGaa83Caaaa@4086@ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaGqabiaa=jdaaeaadaGcaaqaaiaa=jdacaWF5aaaleqaaaaakiaa=bcacaWF1bGaa8NBaiaa=LgacaWF0bGaa83Caaaa@40C4@

Ans.

The equations of two planes: 2x+3y+4z=4 ( i ) 4x+6y+8z=12 2x+3y+4z=6 ( ii ) We can see that both the planes are parallel, so distance between parallel planes D=| d 2 d 1 a 2 + b 2 + c 2 | =| 64 2 2 + 3 2 + 4 2 | = 2 4+9+16 = 2 29 Thus, the distance between two parallel lines is 2 29 . So, the correct option is D. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8CD1@

 

Q.59 The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular

(B) Parallel
(C) intersect y-axis
(D) passes through

( 0,0, 5 4 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaqadaqaaGqabiaa=bdacaWFSaGaa8hmaiaa=XcadaWcaaqaaiaa=vdaaeaacaWF0aaaaaGaayjkaiaawMcaaaaa@3ED6@

Ans.

The equations of the planes are: 2xy+4z=5 ( i ) 5x2.5y+10z=6 ( ii ) So, a 1 a 2 = 2 5 , b 1 b 2 = 1 2.5 = 10 25 = 2 5 and c 1 c 2 = 4 10 = 2 5 a 1 a 2 = b 1 b 2 = c 1 c 2 Therefore, the planes are parallel. Thus, the correct option is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1DB9@

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