NCERT Solutions Class 12 Mathematics Chapter 10

Vector Algebra is one of the important concepts utilised in mathematics and physics, and NCERT solutions for class 12 chapter 10 elaborate on it. We do come across various questions in our daily lives, such as the height of a tree or, to put it in another way, how hard should a ball be hit to score a goal? The only thing that one can say in response to such questions is ‘Magnitude’ and such magnitudes are called scalars. However, if we also need to determine the direction in which a tree is growing or in what direction one must hit the ball to reach the goal, we must use another quantity known as a Vector.

Therefore, a vector is defined as a quantity with both direction and magnitude. We need scalars such as: length, mass, time, distance, speed, area, volume as well as vectors like displacement, velocity, acceleration, force, weight, both for Mathematics and Physical Science. NCERT Solutions for Class 12 Mathematics Chapter 10 demonstrates solutions with the use of vectors.

The problems in the NCERT Solutions for Class 12 Mathematics chapter 10 are based on real-life scenarios that help students understand the topic easily. The lesson begins with an overview of some basic vector concepts and builds on them to thoroughly explain more complex subjects. The best thing about the NCERT solutions for Class 12 Mathematics Chapter 10 vector algebra is that it conveys tough sections in vernacular and easy language so that it could easily be understood by students with different intellect levels.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 10

The topics discussed in NCERT Solutions Class 12 Mathematics Chapter 10 are equally essential because they explain various sections of vectors, such as direction cosines, dot product, cross product, section formula, and associated properties. Furthermore, all topics are interconnected, implying that students cannot move on to the next section without first mastering the prior one. Therefore, studying each and every topic is necessary.

List of NCERT Solutions Class 12 Mathematics Chapter 10 Exercises

The word ‘Vector’ comes from ‘Vectus’, a Latin word, which means “to carry.” Modern vector theory began in 1800 when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) demonstrated how to use a directed line segment in a coordinate plane to give the geometric interpretation of a complex number. Mathematicians conducted additional research, culminating in the topic as we know it today. Below are links to NCERT solutions class 12 Mathematics chapter 10, offering highly engaging facts and recommendations to assist students in studying the subject matter with ease.

There are various lessons in both Mathematics and Physics that need students to have a solid understanding of vectors. As a result, you must regularly review the solutions in these links and take notes on all the procedures and formulas stated in them. NCERT Solutions for Class 12 Mathematics Chapter 10 Vector Algebra for each practice question is provided in the links below.

Class 12 Mathematics Chapter 10 –  Ex 10.1 Solutions – 5 Questions 

Class 12 Mathematics Chapter 10 – Ex 10.2 Solutions – 19 Questions

Class 12 Mathematics Chapter 10 –  Ex 10.3 Solutions – 18 Questions

Class 12 Mathematics Chapter 10 – Ex 10.4 Solutions – 12 Questions

Class 12 Mathematics Chapter 10 Miscellaneous Ex – 19 Questions

Total Questions: There are 63 sums in Chapter 10 of Class 12 Mathematics Vector Algebra, divided into 20 simple computational questions, 37 medium-level problems, and 6 difficult problems.

NCERT Solutions Class 12 Mathematics Chapter 10 Formula List

All the formulas students need to understand this chapter are used in NCERT Solutions for class 12 Mathematics chapter 10 to help execute arithmetic operations on vectors. The same laws that apply to whole numbers or scalars do not apply to vectors, which have their direction. As a result, the computations for vectors alter when this element is taken into consideration. 

Several geometrical implications of vectors are discussed in this chapter, which contains additional formulas and processes some of which are mentioned below.

  • Let S and T be two given vectors; then the dot product is represented by S.T = |S| |T| cos θ.
  • If θ = 0°, meaning both S and T are in the same direction, then T = |S| |T|.
  • If θ = 90°, meaning S and T are orthogonal then T = 0.
  • If we have two vectors S = (S1, S2, S3 ….. Sn) and T = (T1, T2, T3 ….. Tn) then the dot product is given as S.T = (S1T1 + S2T2 + S3T3 ….. SnTn)

Students should also create a chart on important formulae so that they can be accessed easily whenever needed.

NCERT Mathematics Syllabus

Class 12 NCERT Mathematics Syllabus

Term – 1

Unit Name


Chapter Name

Relations and Function

Relations and Functions

Inverse Trigonometric Functions







Continuity and Differentiability

Application of Derivatives 

Linear Programming Linear Programming

 Term – 2

Unit Name Chapter Name




Application of Integrals

Differential Equations

Vectors and Three-Dimensional Geometry  Vector Algebra

Three Dimensional Geometry 

Probability Probability 

Experts at Extramarks create NCERT Solutions to help students understand concepts more easily and precisely. NCERT Solutions provide detailed step-by-step explanations of problems found in textbooks. Solutions are available for various classes mentioned in the below links-

  • 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 Mathematics Exam Pattern

Duration of Marks 3 hours 15 minutes
Marks for Internal 20 marks
Marks for Theory 80 marks
Total Number of Questions 38 Questions
Very short answer question 20 Questions
Short answer questions 7 Questions
Long Answer Questions (4 marks each) 7 Questions
Long Answer Questions (6 marks each) 4 Questions

NCERT Exemplar Class 12 Mathematics 

NCERT Exemplar Class 12 Mathematics are available and can be accessed from NCERT official website. NCERT Exemplar Class 12 Mathematics Chapter 10 consists of problems with their solutions to help students prepare for their final exams. These Exemplar questions are a little more complex, and cover all the important concepts in Class 12 Mathematics Chapter 10. Students will fully understand all the concepts covered in chapter 10 by practising these NCERT Exemplars for Mathematics Class 12.

Introduction to vectors, types of vectors, addition and multiplication of vectors, vector components, section formula, scalar product, projection of a vector on a line, and vector product are all covered here. Each question in these materials is connected to topics covered in the CBSE Class 12 syllabus.

Key Features of NCERT Solutions Class 12 Mathematics Chapter 10

NCERT Solutions for Class 12 Mathematics are not only easy to comprehend, but they also include essential practice questions based on the recent CBSE syllabus.

The following are the important aspects to consider:

  • You would be better positioned to clear your doubts because it provides in-depth knowledge in simple language.
  • It simplifies and categorises every concept in the topic, making it easier to grasp.
  • It offers multiple concepts for approaching the question paper and is based on CBSE pattern.
  • These answers are a result of extensive research and are designed to lay the groundwork for your future studies.


Find |a×b|, if a=i^7j^+7k^ and b=3i^2j^+2k^.


Given vectors are:a=i^7j^+7k^ and b=3i^2j^+2k^Then,   a×b=|i^j^k^177322|      =(14+14)i^+(221)j^+(2+21)k^      =19j^+19k^|a×b|=|19j^+19k^|      =(19)2+(19)2      =219


Find a unit vector perpendicular to each of the vector a+band ab,where a=3i^+2j^+2k^ and b=i^+2j^2k^.


The given vectors are a=3i^+2j^+2k^ andb=i^+2j^2k^.Then,a+b=3i^+2j^+2k^+i^+2j^2k^    =4i^+4j^and   ab=3i^+2j^+2k^i^2j^+2k^    =2i^+4k^(a+b)×(ab)    =|i^j^k^440204|    =(160)i^(160)j^+(08)k^    =16  i^16j^8k^|(a+b)×(ab)|    =|16  i^16j^8k^|    =(16)2+(16)2+(8)2    =24Hence, the unit vector perpendicular to each of the vectors(a+b)  and(ab)is given by the relation,    =±(a+b)×(ab)|(a+b)×(ab)|    =±16  i^16j^8k^24    =±2  i^2j^k^3


If a unit vector a makes angles π3 with i^,π4 withj^ and anacute angle θ with k^, then find θ and hence, the components of a.


Let unit vector a=a1i^+a2j^+a3k^  and  |a|=1.It is given that a makes angles π3  with i^,π4  with j^ and an acute angle θ with k^.Then, we have:cosπ3=a1|a|12=a11a1=12cosπ4=a2|a|12=a21a2=12cosθ=a3|a|cosθ=a31a3=cosθSince,|a|=1        a12+a22+a32=1           a12+a22+a32=1(12)2+(12)2+cos2θ=1  cos2θ=11412  cos2θ=14   cosθ=12=cosπ3So,   θ=π3Thus, θ=π3 and and the components of a  are (12,12,12).


Show that (ab)×(a+b)=2(a×b)


L.H.S.=(ab)×(a+b)=a×a+a×bb×ab×b=0+a×b+a×b0              [b×a=a×b]=2(a×b)=R.H.S.

Q.5 Represent graphically a displacement of 40 km, 30° east of north. 

Here,vector OP represents the displacement of 40 km, 30°East of North.

Q.6 Classify the following measures as scalars and vectors. (i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 watt
(v) 10-19 coulomb
(vi) 20 m/s2


(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40° is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10–19 coulomb is a scalar quantity as it involves only magnitude.
(vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction

Q.7 Classify the following as scalar and vector quantities. (i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done


(i) Time period is a scalar quantity as it has only magnitude.
(ii) Distance is a scalar quantity as it has only magnitude.
(iii) Force is a vector quantity as it has both magnitude and direction.
(iv) Velocity is a vector quantity as it has both magnitude as well as direction.
(v) Work done is a scalar quantity as it has only magnitude.

Q.8 In fig.(a square), identify the following vectors. i) Coinitial
(ii) Equal
(iii) Collinear but not equal


(i) Vectors a  and d are coinitial because they have the same initial point.(ii) Vectors b and d  are equal because they have the same magnitude and direction.(iii) Vectors a  and c are collinear but not equal. This is because although they are parallel, their directions are not the same.


Answer the following as true or false.(i)a  and  aare collinear.(ii)Two collinear vectors are always equal inmagnitude.(iii)Two vectors having same magnitude are collinear.(iv)Two collinear vectors having the same magnitude        are equal.


(i) True. Since, both vectors are parallel to same line.
(ii) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.
(iii) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.
(iv) False.
Collinear vectors are those vectors that are parallel to the same line, irrespective of their magnitudes and directions.


Find λ and μ if (2i^+6j^+27k^)×(i^+λj^+μk^)=0.


(2i^+6j^+27k^)×(i^+λj^+μk^)=0|i^j^k^26271λμ|=0(6μ27λ)i^(2μ27)j^+(2λ6)k^=0i^+0j^+0k^On comparing the corresponding components, we have:6μ27λ=0,2μ27=0 and 2λ6=0μ=272 and λ=3


Compute the magnitude of the following vectors:a=i^+j^+k^;b=2i^7j^3k^;   c=13i^+13j^13k^;


The given vectors are:a=i^+j^+k^;b=2i^7j^3k^;   c=13i^+13j^13k^Magnitude of a=|a|    =(1)2+(1)2+(1)2    =3Magnitude of b=|b|    =(2)2+(7)2+(3)2    =62Magnitude of c=|c|    =(13)2+(13)2+(13)2    =33=1

Q.12 Write two different vectors having same magnitude. 

Two different vectors are:a=i^+2j^+3k^;b=2i^+3j^+k^ThenMagnitudeofa=i^+2j^+3k^=12+22+32=1+4+9=14Magnitude ofb=2i^+3j^+k^=22+32+12=4+9+1=14Hence, aandbaretwodifferentvectorshavingthe same magnitude. The vectors are different because they have different directions.

Q.13 Write two different vectors having same direction.

Two different vectors are:a=2i^+2j^+2k^;b=i^+j^+k^Then,Magnitudeofa=2i^+2j^+2k^=22+22+22=4+4+4=12=23Direction cosinesofaare:223,223,223i.e.,13,13,13Magnitudeof b=i^+j^+k^=12+12+12=1+1+1=3Direction cosinesofbare:13,13,13Direction cosinesofaandbaresame.Hence,these two vectorshavesamedirection.


Find the values of x and y so that the vectors  2i^+3j^ andxi^+yj^ are equal.


Since,  2i^+3j^=xi^+yj^So, on comparing coefficients of i^ and j^, we getx=2 and y=3.

Q.15 Find the scalar and vector components of the vector with initial point (2, 1) and erminal point (– 5, 7).


The vector with the initial point P (2, 1) and terminal point Q (-5, 7) can be given by,PQ=(52)i^+(71)j^       =7i^+6j^Hence, the required scalar components are -7 and 6 while the vector components are  7i^  and  6j^.


Given that a.b=0 and a×b=0.What can you concludeabout the vectors a and b?


(i) a.b=0Either |a|=0 or |b|=0,or ab(if a  and  b are non-zero.)(ii) a×b=0Either |a|=0 or |b|=0, or a    b(if a  and  b are non-zero.)But, a  and b cannot be perpendicular and parallel simultaneously.Hence, ‹|a|=0 or |b|=0.


Find the sum of the vectorsa=i^2j^+k^,b=2i^+4j^+5k^andc=i^6j^7k^.


The given vectors are:a=i^2j^+k^,b=2i^+4j^+5k^  andc=i^6j^7k^Then,a+b  +  c=i^2j^+k^+(2i^+4j^+5k^)+i^6j^7k^=4j^k^


Find the unit vector in the direction of the vector a=i^+j^+2k^.


The unit vector a^  in the direction of vector a=i^+j^+2k^  is given by a^=a|a|=i^+j^+2k^|i^+j^+2k^|=i^+j^+2k^12+12+22=i^+j^+2k^6=i^6+j^6+26k^


Find the unit vector in the direction of the vector PQ, where P and Q are the points (1,2,3) and (4,5,6)respectively.


Here,OP=i^+2j^+3k^ and OQ=4i^+5j^+6k^So, PQ=OQOP=4i^+5j^+6k^i^+2j^+3k^     =3i^+3j^+3k^Unitvectorof PQ     =PQPQ     =3i^+3j^+3k^3i^+3j^+3k^     =3i^+3j^+3k^32+32+32   =3i^+3j^+3k^33   =i^+j^+k^3   =i^3+j^3+k^3


For given vectors, a=2i^j^+2k^ and b=i^+j^k^, find the unit vector in the direction of the vector a+b.


The  given vectors are:        a=2i^j^+2k^andb=i^+j^k^a+b=2i^j^+2k^+(i^+j^k^)   =i^+k^|a+b|=|i^+k^|   =12+12   =2Unit vector of  a+b   =a+b|a+b|=i^2+k^2


Find a vector in the direction of vector 5i^j^+2k^ which has magnitude 8 units.


The unit vector of 5i^j^+2k^=5i^j^+2k^|5i^j^+2k^|=5i^j^+2k^52+(1)2+22=5i^j^+2k^25+1+4=5i^j^+2k^30Thus, the vector in the direction of (5i^j^+2k^) which has magnitude 8 units is:            8a^=8×5i^j^+2k^30=40i^8j^+16k^30=4030i^830j^+1630k^


Show that the vectors 2i^3j^+4k^ and4i^+6j^8k^are collinear.


The given vectors are a=2i^3j^+4k^b=4i^+6j^8k^   =2(2i^3j^+4k^)b=2a, which is in the form of a=λb,where,λ=2Hence, the given vectors are collinear.


Find the direction cosines of the vectori^+2j^+3k^.


The given vectoris i^+2j^+3k^.|i^+2j^+3k^|=12+22+32=1+4+9=14So, the direction cosines of i^+2j^+3k^are  (114,214,314).

Q.24 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, 2, 1) directed A to B.


The given points are A(1, 2,3) and B(1,2,1).     AB=(11)i^+(22)j^+(1+3)k^    =2i^4j^+4k^|AB|=(2)2+(4)2+42    =4+16+16    =36    =6Hence, the direction cosines of  AB  are  (26,46,46)=(13,23,23).


Show that the vectori^+j^+k^ is equally inclined to the axes OX, OY and OZ.


The given vector is a=i^+j^+k^,then     |a|=|i^+j^+k^|=12+12+12=3The direction cosines of a are (13,13,13).Let α,β and γ be the angles formed by a with the positive directions of x, y, and z axes.Then, we have cos α=13,cosβ=13, cosγ=13.Hence, the given vector is equally inclined to axes OX, OY, and OZ.


Let the vectors a,b,c be given as a1i^+a2j^+a3k^,b1i^+b2j^+b3k^,  c1i^+c2j^+c3k^. Then show thata×(b+c)=a×b+a×c.


The given vectors are:a1i^+a2j^+a3k^,b1i^+b2j^+b3k^,  c1i^+c2j^+c3k^.L.H.S.=a×(b+c)=(a1i^+a2j^+a3k^)×(b1i^+b2j^+b3k^+c1i^+c2j^+c3k^)=(a1i^+a2j^+a3k^)×{(b1+c1)i^+(b2+c2)j^+(b3+c3)k^}=|i^j^k^a1a2a3(b1+c1)(b2+c2)(b3+c3)|={a2(b3+c3)a3(b2+c2)}i^+{a1(b3+c3)a3(b1+c1)}j^+{a1(b2+c2)a2(b3+c3)}k^={a2b3+a2c3a3b2a3c2}i^+{a1b3+a1c3a3b1a3c1}j^+{a1b2+a1c2a2b3a2c3}k^...(i)a×b=(a1i^+a2j^+a3k^)×(b1i^+b2j^+b3k^)=|i^j^k^a1a2a3b1b2b3|=(a2b3a3b2)i^+(a1b3a3b1)j^+(a1b2a2b1)k^a×c=(a1i^+a2j^+a3k^)×(c1i^+c2j^+c3k^)=|i^j^k^a1a2a3c1c2c3|=(a2b3a3b2)i^+(a1b3a3b1)j^+(a1b2a2b1)k^a×b+a×c=(a2b3a3b2)i^+(a1b3a3b1)j^+(a1b2a2b1)k^+(a2b3a3b2)i^                    +(a1b3a3b1)j^+(a1b2a2b1)k^=(a2b3a3b2+a2b3a3b2)i^+(a1b3a3b1+a1b3a3b1)j^+(a1b2a2b1+a1b2a2b1)k^=(a2b3+a2b3a3b2a3b2)i^+(a1b3+a1b3a3b1a3b1)j^+(a1b2+a1b2a2b1a2b1)k^  ...(ii)From equation(i) and equation(ii), we have  a×(b×c)=a×b+a×c.   Hence, the given result is proved.


Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  i^+2j^k^ and i^+j^+k^ respectively, in the ration 2:1(i) internally (ii) externally


The given vectors are OP=i^+2j^k^ and OQ=i^+j^+k^.The position vector of point R dividing the line segment joining two points P and Q in the ratio 2: 1 is given by:(i)Internally:   OR=2(i^+j^+k^)+1(i^+2j^k^)2+1=i^+4j^+k^3=13i^+43j^+13k^(ii)The position vector of point R which divides the line joining two points P and Q externally in the ratio 2:1 is given by,  OR=2(i^+j^+k^)1(i^+2j^k^)21=3i^+0j^+3k^1=3i^+3k^

Q.28 Find the position vector of the mid-point of the vector joining the points P(2, , 4) and Q(4, 1, –2).


The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1,-2) is given by,  OR=(2i^+3j^+4k^)+(4i^+j^2k^)2=6i^+4j^+2k^2=3i^+2j^+k^


Show that the points A, B and C with position vectors, a=3i^4j^4k^, b=2i^j^+k^ and c=i^3j^5k^ respectively form the vertices of a right angled triangle.


Position vectors of points A, B, and C are respectively given as:a=3i^4j^4k^,b=2i^j^+k^ and c=i^3j^5k^     AB=ba     =(2i^j^+k^)(3i^4j^4k^)     =i^+3j^+5k^        BC=cb     =(i^3j^5k^)(2i^j^+k^)     =i^2j^6k^        CA=ac     =(3i^4j^4k^)(i^3j^5k^)      =2i^j^+k^So, |AB|2=|i^+3j^+5k^|2     =35    |BC|2=|i^2j^6k^|2     =41    |CA|2=|2i^j^+k^|2     =6|AB|2=   |BC|2+|CA|2=36+5=41 Hence, ABC is a right-angled triangle


If either a=0 or b=0, then a×b=0. Is the converse true?Justify your answer with an example.


Let us consider any parallel non-zero vectors so that  a×b=0Leta=4i^+6j^+8k^ and b=2i^+3j^+4k^Then,a×b=|i^j^k^468234|=(2424)i^+(1616)j^+(1212)k^=0i^+0j^+0k^=0And    |a|=|4i^+6j^+8k^|=42+62+82=16+36+64=116|a|0    |b|=|2i^+3j^+4k^|=22+32+42=4+9+16=29|b|0Hence, the converse of the given statement need not be true.

Q.31 Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, ).


The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).The adjacent sides  AB and BC  of ΔABC are given as:   AB=(21)i^+(31)j^+(52)k^=i^+2j^+3k^   BC=(12)i^+(53)j^+(55)k^=i^+2j^AB×BC=12|i^j^k^123120|=12{(06)i^(0+3)j^+(2+2)k^}=12(6i^3j^+4k^)The area ΔABC=12|AB×BC|=12|6i^3j^+4k^|=12(6)2+(3)2+42=1236+9+16=1261 square unitsThus, the area of ΔABC is 1261 square units.


Find the area of the parallelogram whose adjacents sidesare determined by the vectors a=i^j^+3k^ and b=2i^7j^+k^.


The area of the parallelogram whose adjacent sides are a and bis |a×b|.Adjacent sides are in vector form, which are as follows:a=i^j^+3k^ andb=2i^7j^+k^a×b=|i^j^k^113271|    =(1+21)i^+(16)j^+(7+2)k^    =20i^5j^5k^|a×b|=|20i^5j^5k^|    =202+(5)2+(5)2    =400+25+25    =450    =152Thus, the area of the given parallelogram is 152 square units.


Let the vectors a and bbe such that |a|=3 and |b|=32, thena×b is a unit vector, if the angle between a and b is(A)π6       (B)π4       (C)π3       (D)π2


Given:  |a|=3 and |b|=23Since, a×b=|a||b|sinθn^, where  n^ is a unit vector perpendicularto both a and b‹ and θ‹ is the angle between a and b.Since, a×b is a unit vector, so |a×b|=1.       |a×b|=1||a||b|sinθn^|=1     |a||b||sinθ|=1  3×23×sinθ=1         sinθ=12=sinπ4     θ=π4Hence, a×b is a unit vector if the angle between a and b is π4.The correct answer is B.


Area of a rectangle having vertices A, B, C and D withposition vectori^+12j^+4k^,  i^+12j^+4k^,  i^12j^+4k^andi^12j^+4k^respectively is(A)12      (B) 1      (C) 2      D 4


The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:OA=i^+12j^+4k^, OB=i^+12j^+4k^,OC=i^12j^+4k^ and OD=i^12j^+4k^The adjacent sides AB and BC of the given rectangle are given as:AB=(1+1)i^+(1212)j^+(44)k^=2i^BC=(11)i^+(1212)j^+(44)k^=j^AB×BC=|i^j^k^200010|=(20)k^=2k^|AB×BC|=|2k^|=2Now, it is known that the area of a parallelogram whose adjacent sides are a and b  is |a×b|.Hence, the area of the given rectangle is |AB×BC|=2 square units.The correct answer is C.


Find [abc] if a=i^2j^+3k^, b=2i^3j^+k^,c=3i^+j^2k^.


[abc]=|123231312|  =1|3112|+2|2132|+3|2331|  =1(61)+2(43)+3(2+9)  =1(5)+2(7)+3(11)  =514+33  =24[abc]=24


In triangle, which of the following is not true:(A)AB+BC+CA=0(B)AB+BCAC=0(C)AB+BCCA=0(D)ABCB+CA=0


On applying the triangle law of addition in the given triangle, we have:      AB+BC=AC...(i)AB+BC=CAAB+BC+CA=0...(ii)The equation given in alternative A is true.   AB+BC=ACAB+BCAC=0The equation given in alternative B is true.From equation(ii), we have:ABCB+AC=0The equation given in alternative D is true.Now, the equation given in alternative C,AB+BCCA=0      AB+BC=CA  ...(iii)From equation(i) and (iii),‹ we have                 CA=AC        AC=ACAC+AC=0           2AC=0              AC=0, which is not true.Hence, the equation given in alternative C is incorrect.The correct answer is C.


Show that the vectors a=i^2j^+3k^, b=2i^+3j^4k^,c=i^3j^+5k^ are coplanar.


[abc]=|123234135|  =1|3435|+2|2415|+3|2313|  =1(1512)+2(10+4)+3(63)  =1(3)+2(6)+3(3)  =312+9=0Since, [abc]=0. So, given vectors are coplanar.


Findλ if the vectors i^j^+k^,3i^+j^+2k^ andi^+λj^3k^ arecoplanar.


[abc]=|1113121λ3|     [abc]  =1|12λ3|(1)|3213|+1|311λ|  =1(32λ)+1(92)+1(3λ1)  =32λ11+3λ1  =λ15Since, given vectors are coplanar. So,     [abc]=0λ15=0     λ=15


Leta=i^+j^+k^,b=i^,c=c1i^+c2j^+c3k^.Then(a) If c1=1 and c2=2,find c3 which makes a,bandccoplanar.(b) If c1=1 and c3=1,show that no value of c1 canmakea,band  ccoplanar.


(i) Given:a=i^+j^+k^,b=i^andc=c1i^+c2j^+c3k^Putting c1=1 and c2=2, we getc=1i^+2j^+c3k^Since, a,bandcare coplanar. Then,        [abc]=|11110012c3|     [abc]  =1|002c3|1|101c3|+1|1012|  =1(0)1(c30)+1(20)  =0c3+2Since,given vectors are coplanar. So,       [abc]=0c3+2=0     c3=2(ii)Given:a=i^+j^+k^,b=i^andc=c1i^+c2j^+c3k^Putting c2=1 and c3=1, we getc=c1i^1j^+1k^Since, a,bandcare coplanar. Then,        [abc]=|111100c111|     [abc]  =1|0011|1|10c11|+1|10c11|  =1(0)1(10)+1(10)  =011=2Since,       [abc]0So, there will be no effect on the value of [abc] by any value of c1.Thus, no value of c1 can make a,b and  c coplanar.


Show that the four points with position vectors4i^+8j^+12k^,2i^+4j^+6k^,3i^+5j^+4k^ and 5i^+8j^+5k^are coplanar.


Let the four points with position vectors beOA=4i^+8j^+12k^, OB=2i^+4j^+6k^,OC=3i^+5j^+4k^and OD=5i^+8j^+5k^, then   AB=OBOA       =(2i^+4j^+6k^)(4i^+8j^+12k^)       =2i^4j^6k^            AC=OCOA       =(3i^+5j^+4k^)(4i^+8j^+12k^)       =i^3j^8k^            AD=ODOA       =(5i^+8j^+5k^)(4i^+8j^+12k^)       =i^+0j^7k^Thus,[ABACAD]=|246138107|       =2|3807|(4)|18  17|+(6)|13  1  0|       =2(210)+4(7+8)6(0+3)       =42+6018       =0Thus, the given four points with position vectors are coplanar.


If a bare two collinear vectors, then which of the following are incorrect:(A)b=λa, for some scalar λ(B)a=±b(C) the respective components of a and b are proportional(D) both the vectors a and b have same direction, but      different magnitudes.


If a  and b are two collinear vectors, then they are parallel.Therefore, we have:  b=λa(λ is a scalar quantities.)If λ=±1, then  b=±1a.If a=a1i^+a2j^+a3k^ and b=b1i^+b2j^+b3k^, then  b=λab1i^+b2j^+b3k^=λ(a1i^+a2j^+a3k^) b1i^+b2j^+b3k^=(λa1)i^+(λa2)j^+(λa3)k^b1=λa1,b2=λa2,b3=λa3b1a1=b2a2=b3a3=λThus, the respective components of a  and b  are proportional.However, vectors a  and b  can have different directions.Hence, the statement given in D is incorrect.The correct answer is D.


Find x such that the four points A(3,2,1), B(4,x,5),C(4,2,2) and D(6,5,1) are coplanar.


Let the four points with position vectors beOA=3i^+2j^+k^,OB=4i^+xj^+5k^,OC=4i^+2j^2k^andOD=6i^+5j^1k^, then   AB=OBOA       =(4i^+xj^+5k^)(3i^+2j^+k^)       =i^+(x2)j^+4k^            AC=OCOA       =(4i^+2j^2k^)(4i^+8j^+12k^)       =0i^6j^14k^            AD=ODOA       =(6i^+5j^1k^)(4i^+8j^+12k^)       =2i^3j^13k^Since, given vectors are coplanar. So,[ABACAD]=0       [ABACAD]=|1x2406142313|    =1|614313|(x2)|014213|+4|0623|    =1(7842)(x2)(0+28)+4(0+12)    =3628x+56+48    =14028x       14028x=0x=14028=5Thus, the value of x is 5.


Show that the vectorsa,b,care coplanar if a+b,b+cand c+aare coplanar.


[a+b,b+c,c+a]=(a+b).((b+c)×(c+a))  =(a+b).(b×c+b×a+c×c+c×a)  =(a+b).(b×c+b×a+0+c×a)  =a.(b×c)+a.(b×a)+a(c×a)+b.(b×c)+b.(b×a)+b.(c×a)  =[a,b,c]+[a,b,a]+[a,c,a]+[b,b,c]     +[b,b,a]+[b,c,a]  =[a,b,c]+0+0+0+0+[a,b,c][[a,b,a]=[a,c,a]=[b,b,c]=[b,b,a]=0]  =2[a,b,c]Since, a+b,b+cand c+aare coplanar. So,[a+b,b+c,c+a]=0  2[a,b,c]=0     [a,b,c]=0So, the vectorsa,b,care coplanar.

Q.44 Write down a unit vector in XY-plane, making an angle of 30° with the positive irection in X-axis.


If is r a unit vector in the XY-plane, then r=cosθi^+sinθj^.Here, θ is the angle made by the unit vector with the positive direction of the x-axis.Therefore, for θ = 30°:r=cos30°  i^+sin30° j^=32i^+12j^Thus, the required unit vector is 32i^+12j^.

Q.45 Find the scalar components and magnitude of the vector joining the points P(x, y1, z1) and Q(x2, y2, z2).


The vector joining the points P(x1,  y1,  z1) and Q(x2,  y2,  z2) is:   PQ=(x2x1)i^+(y2y1)j^+(z2z1)k^|PQ|=|(x2x1)i^+(y2y1)j^+(z2z1)k^|        =(x2x1)2+(y2y1)2+(z2z1)2Hence, the scalar components and the magnitude of the vectorjoining the given points are respectively{(x2x1),  (y2y1),  (z2z1)} and (x2x1)2+(y2y1)2+(z2z1)2.

Q.46 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of orth and stops. Determine the girl’s displacement from her initial point of departure.


Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:

Now, we have:  OA=4i^   AB=i^|AB|cos60°+j^|AB| sin60°=i^(3)12+j^(3) 32=32i^+332j^By the triangle law of vector addition, we have:   OB=OA+AB=4i^+32i^+332j^=(4+32)i^+332j^=52i^+332j^Hence, the girl’s displacement from her initial point of departure is  52i^+332j^.


If a=b+c, then is it true that |a|=|b|+|c|? Justify youranswer.


In ΔABC,letCB=a,CA=b and AB=c.Now, by the triangle law of vector addition, we have   a=b+c.Since, sum of any two sides of a triangle is always greater than third side.|a|<|b|+|c|Hence, it is not true that |a|=|b|+|c|.


Find the value of x for which x(i^+j^+k^) is a unit vector.


x(i^+j^+k^) is a unit vector, so  |x(i^+j^+k^)|=1x|(i^+j^+k^)|=1x12+12+12=1x3=1     x=±13Thus, the required value of x is ±13.


Find a vector of magnitude 5units, and parallel to theresultant of the vectors a=2i^+3j^k^ and b=i^2j^+k^.


The given vectorsarea=2i^+3j^k^ and b=i^2j^+k^Let c be the resultant vectors of a and b.Then,      c=a+b          =2i^+3j^k^+i^2j^+k^=3i^+j^|c|=|3i^+j^|=32+12=10So,c^=3i^+j^10Thus, the vector of magnitude 5 units and parallel to the resultant of vectors a and bis±5.c^=±5.3i^+j^10=±3102i^±102j^


Find the angle between two vectors a and b with magnitude 3 and  2, respectively having a.b=6.


It is given that,|a|=3,|b|=2 and a.b=6Now, we know that       a.b=|a||b|cosθ6=3.2cosθcosθ=63.2       =12 cosθ=cosπ4   θ=π4Hence, the angle between the given vectors a and b is π4.


Find the angle between the vectors i^2j^+3k^ and 3i^2j^+k^.


The given vectors are a=i^2j^+3k^ and b=3i^2j^+k^.     |a|=|i^2j^+3k^|=12+(2)2+32=1+4+9=14     |b|=|3i^2j^+k^|=32+(2)2+12=9+4+1=14Now,    a.b=(i^2j^+3k^).(3i^2j^+k^)=3+4+3=10 Also, we know thata.b=|a|.|b|cosθ10=14.14cosθcosθ=1014=57     θ=cos1(57)


Find the projection of the vector i^j^ on the vector i^+j^.


Let a=i^j^ and b=i^+j^Now, projection of vector a  on b  is given by,1|b|(a.b)=1|i^+j^|(i^+j^).(i^j^)     =112+12(11)=0Hence, the projection of vector a on  b is 0.


Find the projection of the vector i^+3j^+7k^ on the vector7i^j^+8k^.


Let a=i^+3j^+7k^ and b=7i^j^+8k^Now, projection of vector a  on b  is given by,1|b|(a.b)=1|7i^j^+8k^|(i^+3j^+7k^).(7i^j^+8k^)     =172+(1)2+82(73+56)     =149+1+64(60)=1114(60)Hence, the projection of vector a on  b is 60114.


Show that each of the given three vectors is a unit vector:17(2i^+3j^+6k^),  17(3i^6j^+2k^),17(6i^+2j^3k^)Also, show that they are mutually perpendicular to each other.


Let  a=17(2i^+3j^+6k^)        b=17(3i^6j^+2k^)        c=17(6i^+2j^3k^)|a|=|17(2i^+3j^+6k^)|   =1722+32+62  =174+9+36  =77=1    |b|=|17(3i^6j^+2k^)|  =1732+(6)2+22  =179+36+4  =77=1and    |c|=|17(6i^+2j^3k^)|  =1762+22+(3)2  =1736+4+9  =77=1Thus, each of the given three vectors is a unit vector.Now,  a.b=17(2i^+3j^+6k^).17(3i^6j^+2k^)=149(618+12)=0 b.c=17(3i^6j^+2k^).17(6i^+2j^3k^)=149(18126)=0c.a=17(6i^+2j^3k^).17(2i^+3j^+6k^)=149(12+618)=0Hence, the given three vectors are mutually perpendicular to each other.


Find a and b, if a+b.ab=8 and a=8b.


We have,(a+b).(ab)=8 and |a|=8|b|So,               (a+b).(ab)=8a.aa.b+b.ab.b=8|a|2|b|2=8[a.b=b.a]    (8|b|)2|b|2=8     64|b|2|b|2=8 | b |= 8 63 = 2 2 3 7 | a |=8| b | =8× 2 2 3 7 = 16 2 3 7


Evaluate the product(3a5b).(2a+7b).


(3a5b).(2a+7b)=6a.a+21a.b10b.a35b.b         =6|a|2+21a.b10b.a35|b|2         =6|a|2+11a.b35|b|2   [a.b=b.a]


Find the magnitude of the vectors a and b, having the same magnitude and such that the angle between them is 60° and their scalar product is12.


Let θ be the angle between the vectors a and b. Then,a . b=|a||b|cosθ   12=|a||a|cos60°[|a|=|b|]    12=|a|2×12    1=|a|2  |a|=1So,     |b|=|a|=1


Find |x|, if for a unit vector a,(xa).(x+a)=12.


Since,(xa).(x+a)=12         x.x+x.aa.xa.a=12  |x|2|a|2=12[x.a=a.x]  |x|212=12[a is a unit vector so,|a|=1]          |x|2=12+1          |x|=13


If a=i^+j^+k^,  b=2i^j^+3k^ and c=i^2j^+k^, find a unitvector parallel to the vector 2ab+3c.


The given vectors are:a=i^+j^+k^,  b=2i^j^+3k^ and c=i^2j^+k^So,  2ab+3c=2(i^+j^+k^)(2i^j^+3k^)+3(i^2j^+k^)    =2i^+2j^+2k^2i^+j^3k^+3i^6j^+3k^    =3i^3j^+2k^|2ab+3c|=|3i^3j^+2k^|    =32+(3)2+22    =9+9+4    =22Thus, the unit vector along (2ab+3c)is2ab+3c|2ab+3c|=3i^3j^+2k^22    =322i^322j^+222k^


If a=2i^+2j^+3k^, b=i^+2j^+k^ and c=3i^+j^ are such that a+λb is perpendicular to c, then find the value ofλ.


The given vectors are:a=2i^+2j^+3k^,b=i^+2j^+k^ and c=3i^+j^      a+λb=(2i^+2j^+3k^)+λ(i^+2j^+k^)           =(2λ)i^+(2+2λ)j^+(3+λ)k^Since, (a+λb) is perpendicular to c, so(a+λb).c=0{(2λ)i^+(2+2λ)j^+(3+λ)k^}.(3i^+j^)=03(2λ)+(2+2λ)=0         63λ+2+2λ=0            8λ=0                   λ=8Therefore, the required value of λ is 8.

Q.61 Show that the points A(1,–2,–8), B(5, 0, –2) and C(11, 3, 7) are collinear, and ind the ratio in which B divides AC.


The given points are A (1,2,8), B (5, 0,2) and C(11, 3, 7).AB=(51)i^+(0+2)j^+(2+8)k^=4i^+2j^+6k^   BC=(115)i^+(30)j^+(7+2)k^=6i^+3j^+9k^   AC=(111)i^+(3+2)j^+(7+8)k^=10i^+5j^+15k^|AB|=|4i^+2j^+6k^|=42+22+62=16+4+36=56=214|BC|=|6i^+3j^+9k^|=62+32+92=36+9+81=126=314|CA|=|10i^+5j^+15k^|=102+52+152=100+25+225=350=514|CA|=|AB|+|BC|Thus, the given points A, B, and C are collinear.Now, let point B divide AC in the ratio λ:1. Then, we have:       OB=λOC+1OAλ+15i^2k^=λ(11i^+3j^+7k^)+(i^2j^8k^)λ+1(λ+1)(5i^2k^)=λ(11i^+3j^+7k^)+(i^2j^8k^)5(λ+1)i^2(λ+1)k^=(11λ+1)i^+(3λ2)j^+(7λ8)7k^     5(λ+1)=(11λ+1)             5λ+5=11λ+1                       4=6λ        λ=46=23Thus, point B divides AC in the ratio 2:3.


Show that | a | b +| b | a is perpendicular to | a | b | b | a , for any two nonzero vectors a and b . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabofacaqGObGaae4BaiaabEhacaqGGaGaaeiDaiaabIga caqGHbGaaeiDaiaabccadaabdaqaamaaFiaabaacbeGaa8xyaaGaay 51GaaacaGLhWUaayjcSdWaa8HaaeaacaWFIbaacaGLxdcacaWFRaWa aqWaaeaadaWhcaqaaiaa=jgaaiaawEniaaGaay5bSlaawIa7amaaFi aabaGaa8xyaaGaay51GaGaa8hiaiaabMgacaqGZbGaaeiiaiaabcha caqGLbGaaeOCaiaabchacaqGLbGaaeOBaiaabsgacaqGPbGaae4yai aabwhacaqGSbGaaeyyaiaabkhacaqGGaGaaeiDaiaab+gacaWFGaWa aqWaaeaadaWhcaqaaiaa=fgaaiaawEniaaGaay5bSlaawIa7amaaFi aabaGaa8NyaaGaay51GaGaeyOeI0YaaqWaaeaadaWhcaqaaiaa=jga aiaawEniaaGaay5bSlaawIa7amaaFiaabaGaa8xyaaGaay51GaGaa8 hlaiaa=bcacaqGMbGaae4BaiaabkhacaqGGaGaaeyyaiaab6gacaqG 5bGaaeiiaaqaaiaabshacaqG3bGaae4BaiaabccacaqGUbGaae4Bai aab6gacaqG6bGaaeyzaiaabkhacaqGVbGaaeiiaiaabAhacaqGLbGa ae4yaiaabshacaqGVbGaaeOCaiaabohacaWFGaWaa8HaaeaacaWFHb aacaGLxdcacaWFGaGaaeyyaiaab6gacaqGKbGaa8hiamaaFiaabaGa a8NyaaGaay51GaGaa8Nlaaaaaa@9B27@


( | a | b +| b | a ).( | a | b | b | a )= | a | 2 | b | 2 | a | b | b | a +| b | a | a | b | b | 2 | a | 2 =| a || b | b a +| b || a | a b =0 [ b a = a b ] So,( | a | b +| b | a )and ( | a | b | b | a ) are perpendicular to each other. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaamaabmaabaWaaqWaaeaadaWhcaqaaiaadggaaiaawEniaaGa ay5bSlaawIa7amaaFiaabaGaamOyaaGaay51GaGaey4kaSYaaqWaae aadaWhcaqaaiaadkgaaiaawEniaaGaay5bSlaawIa7amaaFiaabaGa amyyaaGaay51GaaacaGLOaGaayzkaaGaaiOlamaabmaabaWaaqWaae aadaWhcaqaaiaadggaaiaawEniaaGaay5bSlaawIa7amaaFiaabaGa amOyaaGaay51GaGaeyOeI0YaaqWaaeaadaWhcaqaaiaadkgaaiaawE niaaGaay5bSlaawIa7amaaFiaabaGaamyyaaGaay51GaaacaGLOaGa ayzkaaGaeyypa0ZaaqWaaeaadaWhcaqaaiaadggaaiaawEniaaGaay 5bSlaawIa7amaaCaaaleqabaGaaGOmaaaakmaaemaabaWaa8Haaeaa caWGIbaacaGLxdcaaiaawEa7caGLiWoadaahaaWcbeqaaiaaikdaaa GccqGHsisldaabdaqaamaaFiaabaGaamyyaaGaay51GaaacaGLhWUa ayjcSdWaa8HaaeaacaWGIbaacaGLxdcadaabdaqaamaaFiaabaGaam OyaaGaay51GaaacaGLhWUaayjcSdWaa8HaaeaacaWGHbaacaGLxdca cqGHRaWkdaabdaqaamaaFiaabaGaamOyaaGaay51GaaacaGLhWUaay jcSdWaa8HaaeaacaWGHbaacaGLxdcadaabdaqaamaaFiaabaGaamyy aaGaay51GaaacaGLhWUaayjcSdWaa8HaaeaacaWGIbaacaGLxdcacq GHsisldaabdaqaamaaFiaabaGaamOyaaGaay51GaaacaGLhWUaayjc SdWaaWbaaSqabeaacaaIYaaaaOWaaqWaaeaadaWhcaqaaiaadggaai aawEniaaGaay5bSlaawIa7amaaCaaaleqabaGaaGOmaaaaaOqaaiaa xMaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlabg2da9iabgkHiTmaaemaabaWaa8Ha aeaacaWGHbaacaGLxdcaaiaawEa7caGLiWoadaabdaqaamaaFiaaba GaamOyaaGaay51GaaacaGLhWUaayjcSdWaa8HaaeaacaWGIbaacaGL xdcadaWhcaqaaiaadggaaiaawEniaiabgUcaRmaaemaabaWaa8Haae aacaWGIbaacaGLxdcaaiaawEa7caGLiWoadaabdaqaamaaFiaabaGa amyyaaGaay51GaaacaGLhWUaayjcSdWaa8HaaeaacaWGHbaacaGLxd cadaWhcaqaaiaadkgaaiaawEniaaqaaiaaxMaacaWLjaGaaCzcaiaa xMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlabg2da9iaaicdacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcamaa dmaabaGaeSynIe1aa8HaaeaacaWGIbaacaGLxdcadaWhcaqaaiaadg gaaiaawEniaiabg2da9maaFiaabaGaamyyaaGaay51GaWaa8Haaeaa caWGIbaacaGLxdcaaiaawUfacaGLDbaaaeaacaqGtbGaae4BaiaabY cacaaMc8+aaeWaaeaadaabdaqaamaaFiaabaGaamyyaaGaay51Gaaa caGLhWUaayjcSdWaa8HaaeaacaWGIbaacaGLxdcacqGHRaWkdaabda qaamaaFiaabaGaamOyaaGaay51GaaacaGLhWUaayjcSdWaa8Haaeaa caWGHbaacaGLxdcaaiaawIcacaGLPaaacaqGHbGaaeOBaiaabsgaca qGGaWaaeWaaeaadaabdaqaamaaFiaabaGaamyyaaGaay51GaaacaGL hWUaayjcSdWaa8HaaeaacaWGIbaacaGLxdcacqGHsisldaabdaqaam aaFiaabaGaamOyaaGaay51GaaacaGLhWUaayjcSdWaa8HaaeaacaWG HbaacaGLxdcaaiaawIcacaGLPaaacaqGGaGaaeyyaiaabkhacaqGLb GaaeiiaiaabchacaqGLbGaaeOCaiaabchacaqGLbGaaeOBaiaabsga caqGPbGaae4yaiaabwhacaqGSbGaaeyyaiaabkhacaqGGaGaaeiDai aab+gacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGVbGa aeiDaiaabIgacaqGLbGaaeOCaiaab6caaaaa@3FEA@


Find the position vector of a point R which divides the linejoining two points P and Q whose position vectors are(2a+b)and(a3b) externally in the ratio1:2. Also, showthat P is the mid point of the line segment RQ.


Given that OP=(2a+b) and OQ=(a3b)Since, point R divides a line segment joining two points P and Qexternally in the ratio 1: 2. Then, by using the section formula,we get:  OR=2(2a+b)(a3b)21=4a+2ba+3b1=3a+5b Therefore, the position vector of point R is 3a+5b.Position vector of the mid-point of RQ=OQ+OR2    =a3b+3a+5b2    =4a+2b2    =2a+b=OPHence, P is the mid-point of the line segment RQ.


Ifa.a=0 and a.b=0, then what can be concluded about the vectorb?


Since, it is given that     a.a=0|a|2=0   |a|=0a is a zero vector.Since, vector b satisfying a.b=0 can be any vector.


If a,  b,care unit vectors such thata+b+c=0, find the value of a.b+b.c+c.a.


Given: a,  b,c are unit vectors such that a+b+c=0Then,         [a+b+c]2=0|a|2+|b|2+|c|2+2[a.b+b.c+c.a]=0  1+1+1+2[a.b+b.c+c.a]=0a.b+b.c+c.a=32


The two adjacent sides of a parallelogram are2i^4j^+5k ^and i^2j^3k^.Find the unit vector parallel to its diagonal. Also, find its area.


Adjacent sides of a parallelogram are:a=2i^4j^+5k^ and b=i^2j^3k^Then, the diagonal of a parallelogram=a+b     =2i^4j^+5k^+i^2j^3k^     =3i^6j^+2k^The unit vector parallel to the diagonal=3i^6j^+2k^|3i^6j^+2k^|       =3i^6j^+2k^32+(6)2+22       =3i^6j^+2k^9+36+4       =3i^6j^+2k^49       =37i^67j^+27k^          Now, Area of parallelogram ABCD=|a×b|     a×b=|i^j^k^245123|         =(12+10)i^(65)j^+(4+4)k^       =22i^+11j^           |a×b|=11|2i^+j^|       =1122+12       =115Thus, the area of the parallelogram is 115 square units.


Show that the direction cosines of a vector equally inclined to the axesOX,OY and OZ are13,13,13.


Let a vector be equally inclined to axes OX, OY, and OZ at angle α. Then, the direction cosines of the vector are cos α,cos α, and cos α.Since, cos2α+cos2β+cos2γ=1So,       cos2α+cos2α+cos2α=1    3 cos2α=1       cos2α=13        cosα=13Hence, the direction cosines of the vector which are equally inclined to the axes are 13,13,13.


If either vector a=0 or b=0, then a.b=0. But the converse need not be true. Justify your answer with an example.


Let a=3i^+4j^+3k^ and b=2i^+3j^6k^Then,  a.b=(3i^+4j^+3k^)(2i^+3j^6k^) =6+1218=0And    |a|=|3i^+4j^+3k^|=32+42+32=9+16+9=340    |b|=|2i^+3j^6k^|=22+32+(6)2=4+9+36=49=70Hence, the converse of the given statement need not be true.


Leta=i^+4j^+2k^,  b=3i^2j^+7k^ and c=2i^j^+4k^. Find avector dwhich is perpendicular to both aa nd b,c.d=15.


The given vectors area=i^+4j^+2k^,  b=3i^2j^+7k^ andc=2i^j^+4k^. Let d=d1i^+d2j^+d3k^Since, d  is perpendicular to a  and b, so        a.d=0  (i^+4j^+2k^).(d1i^+d2j^+d3k^)=0     d1+4d2+2d3=0  ...(i)      b.d=0(3i^2j^+7k^).(d1i^+d2j^+d3k^)=0    3d12d2+7d3=0...(ii)But,           c.d=15    (2i^j^+4k^).(d1i^+d2j^+d3k^)=15         2d1d2+4d3=15  ...(iii)On solving (i), (ii), and (iii), we get:d1=1603,   d2=53,d3=703d=1603i^53j^703k^=13(160i^5j^70k^)Hence, the required vector is 13(160i^5j^70k^).


The scalar product of the vector i^+j^+k^ with a unit vectoralong the sum of vectors 2i^+4j^5k^ and λi^+2j^+3k^ is equalto one. Find the value of λ.


Sum of vectors 2i^+4j^5k ^and λi^+2j^+3k^=2i^+4j^5k^+λi^+2j^+3k^        =(2+λ)i^+6j^2k^Unit vector of (2+λ)i^+6j^2k^=(2+λ)i^+6j^2k^|(2+λ)i^+6j^2k^|        =(2+λ)i^+6j^2k^(2+λ)2+62+(2)2        =(2+λ)i^+6j^2k^4+4λ+λ2+36+4        =(2+λ)i^+6j^2k^44+4λ+λ2Scalar product of unit vector and (i^+j^+k^)=1[Given](i^+j^+k^).(2+λ)i^+6j^2k^44+4λ+λ2=1(2+λ)+62=44+4λ+λ2   (6+λ)2=44+4λ+λ2  36+12λ+λ2=44+4λ+λ2    8λ=8       λ=1Thus, the value ofλ is 1.


If a,b,c are mutually perpendicular vector of equal magnitudes,show that the vector a+b+c is equally inclined to a,b and c.


Since, a,b,c are  mutually perpendicular vectors,soa.b=b.c=c.a=0Magnitudes‹ of a,b,c are equal, so|a|=|b|=|c|Leta,b,c are making angles θ1, θ2 and θ3 with (a+b+c)respectively.Then, we havecosθ1=(a+b+c).a|a+b+c||a|=|a|2|a+b+c||a|=|a||a+b+c|cosθ2=(a+b+c).b|a+b+c||b|=|b|2|a+b+c||b|=|b||a+b+c|cosθ3=(a+b+c).c|a+b+c||c|=|c|2|a+b+c||c|=|c||a+b+c|Since,|a|=|b|=|c|, so cosθ1=cosθ2=cosθ3    θ1=θ2=θ3Hence, the vector (a+b+c)  is equally inclined to  a,b and c.


Prove that(a+b).(a+b)=|a|2+|b|2,if and only ifa,bare perpendicular,given a0, b0.


L.H.S.=(a+b).(a+b)=a.a+a.b+b.a+b.b=|a|2+2a.b+|b|2                       [a.b=b.a]=|a|2+2(0)+|b|2                        [a.b=0 as a and b are perpendicular to each other.]=|a|2+|b|2=R.H.S.


If θ is the angle between two vectors a and b,then a.b0only when(A)0<θ<π2          (B)0θπ2(C)0<θ<π           (D)0θπ


If θ is the angle between two vectors a and b,then      a.b0|a|.|b|cosθ0  cosθ0[|a|0,|b|0]  0θπ2So,  a.b0 if   0θπ2.Thus, option B is correct.


If the vertices A, B, C of a triangle ABC are ( 1,2,3 ),( -1,0,0 ), ( 0,1,2 ) respectively, then find ABC. [ ABC is the angle between the vectors BA and BC ]. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMeacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqG2bGaaeyzaiaabkhacaqG0bGaaeyAaiaabogacaqGLbGaae4Cai aabccacaqGbbGaaeilaiaabccacaqGcbGaaeilaiaabccacaqGdbGa aeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeiDaiaabkhaca qGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaaeyqaiaa bkeacaqGdbGaaeiiaiaabggacaqGYbGaaeyzaGqabiaa=bcadaqada qaaiaabgdacaqGSaGaaeOmaiaabYcacaqGZaaacaGLOaGaayzkaaGa aeilamaabmaabaGaaeylaiaabgdacaqGSaGaaeimaiaabYcacaqGWa aacaGLOaGaayzkaaGaaeilaaqaamaabmaabaGaaeimaiaabYcacaqG XaGaaeilaiaabkdaaiaawIcacaGLPaaacaqGGaGaaeOCaiaabwgaca qGZbGaaeiCaiaabwgacaqGJbGaaeiDaiaabMgacaqG2bGaaeyzaiaa bYgacaqG5bGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGUbGaae iiaiaabAgacaqGPbGaaeOBaiaabsgacaqGGaGaeyiiIaTaaeyqaiaa bkeacaqGdbGaaeOlaiaa=bcaaeaadaWadaqaaiabgcIiqlaabgeaca qGcbGaae4qaiaabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaae OyaiaabwgacaqG0bGaae4DaiaabwgacaqGLbGaaeOBaiaabccacaqG 0bGaaeiAaiaabwgacaqGGaGaaeODaiaabwgacaqGJbGaaeiDaiaab+ gacaqGYbGaae4CaiaabccadaWhcaqaaiaabkeacaqGbbaacaGLxdca caqGGaGaaeyyaiaab6gacaqGKbGaaeiiamaaFiaabaGaaeOqaiaabo eaaiaawEniaaGaay5waiaaw2faaiaa=5caaaaa@B9F0@


The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2). MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqG2bGaaeyzaiaabkha caqG0bGaaeyAaiaabogacaqGLbGaae4CaiaabccacaqGVbGaaeOzai aabccacaqGuoGaaeyqaiaabkeacaqGdbGaaeiiaiaabggacaqGYbGa aeyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccaca qGHbGaae4CaiaabccacaqGbbGaaeiiaiaabIcacaqGXaGaaeilaiaa bccacaqGYaGaaeilaiaabccacaqGZaGaaeykaiaabYcacaqGGaGaae OqaiaabccacaqGOaGaae4eGiaabgdacaqGSaGaaeiiaiaabcdacaqG SaGaaeiiaiaabcdacaqGPaGaaeilaiaabccaaeaacaqGHbGaaeOBai aabsgacaqGGaGaae4qaiaabccacaqGOaGaaeimaiaabYcacaqGGaGa aeymaiaabYcacaqGGaGaaeOmaiaabMcacaqGUaaaaaa@7427@ Also, it is given that ABC is the angle between the vectors BA and BC . BA = ( 1 + 1 ) i ^ + ( 2 – 0 ) j ^ + ( 3 – 0 ) k ^ = 2 i ^ + 2 j ^ + 3 k ^ BC = ( 0 + 1 ) i ^ + ( 1 – 0 ) j ^ + ( 2 – 0 ) k ^ = i ^ + j ^ + 2 k ^ BA BC = ( 2 i ^ + 2 j ^ + 3 k ^ )( i ^ + j ^ + 2 k ^ ) = 2 + 2 + 6 = 10 | BA | = | 2 i ^ + 2 j ^ + 3 k ^ | = 2 2 + 2 2 + 3 2 = 4 + 4 + 9 = 17 | BC | = | i ^ + j ^ + 2 k ^ | = 1 2 + 1 2 + 2 2 = 1 + 1 + 4 = 6 Now, it is known that: BA BC = | BA || BC |cos( ABC ) 10 = 17 6 cos( ABC ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabgeacaqGSbGaae4Caiaab+gacaqGSaGaaeiiaiaabMga caqG0bGaaeiiaiaabMgacaqGZbGaaeiiaiaabEgacaqGPbGaaeODai aabwgacaqGUbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqGGaGa eyiiIaTaaeyqaiaabkeacaqGdbGaaeiiaiaabMgacaqGZbGaaeiiai aabshacaqGObGaaeyzaiaabccacaqGHbGaaeOBaiaabEgacaqGSbGa 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If the vertices A, B, C of a triangle ABC are (1,2,3),(-1,0,0),(0,1,2) respectively, then find ABC.[ABC is the angle between the vectors BA and BC].


The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2). Also, it is given that ABC is the angle between the vectors BAand  BC.BA = (1 + 1)i^ + (2 – 0)j^ + (3 – 0)k^ = 2i^ + 2j^ + 3 k^  BC = (0 + 1)i^ + (1 – 0)j^ + (2 – 0)k^ = i^ + j^ + 2k^BABC = (2i^ + 2j^ + 3 k^)(i^ + j^ + 2k^) = 2 + 2 + 6 = 10|BA| = |2i^ + 2j^ + 3 k^| = 22 + 22 + 32 = 4 + 4 + 9 = 17|BC| = |i^ + j^ + 2k^| = 12 + 12 + 22 = 1 + 1 + 4 = 6Now, it is known that: BABC = |BA||BC|cos(ABC)    10 = 176 cos (ABC) cos(ABC) = 10102ABC = cos–1(10102) |AC| = |2i^ + 8j^ – 8k^|= 22 + 82 + (–8)2= 4 + 64 + 64= 132 = 233|AC|=|AB|+|BC|Hence, the given points A, B, and C are collinear.


Show that the vectors 2i^j^+k^,i^3j^5k^ and 3i^4j^4k^form the vertices of a right angled triangle.


Let vectors 2i^j^+k^,i^3j^5k^ and 3i^4j^4k^ be position vectorsof points A, B and C respectively.i.e., OA=2i^j^+k^,  OB=i^3j^5k^,  OC=3i^4j^4k^AB=(12)i^+(3+1)j^+(51)k^=i^2j^6k^   BC=(31)i^+(4+3)j^+(4+5)k^=2i^j^+k^   CA=(23)i^+(1+4)j^+(1+4)k^=i^+3j^+5k^|AB|=|i^2j^6k^|         =(1)2+(2)2+(6)2         =1+4+36         =41|BC|=|2i^j^+k^|          =22+(1)2+12         =4+1+1         =6|CA|=|i^+3j^+5k^|         =(1)2+32+52         =1+9+25         =35|BC|2+|CA|2=(6)2+(35)2 =6+35=41=|AB|2Hence, ΔABC is a right-angled triangle.


If a is a non zero vector of magnitude ‘a’ and λ a non zero scalar, then λ a is unit vector if(A)  λ = 1 (B)  λ = –1 (C) a = |λ| (D) a = 1|λ|


Vector λa  is a unit vector if |λa| = 1.So,  |λa| = 1    |λ||a| = 1       |a| = 1|λ|[λ 0]      a = 1|λ|[|a| = a]Hence, λa vector is a unit vector if a = 1|λ|.The correct answer is D.


Let ‹‹and ‹‹‹be‹two ‹unit ‹vectors ‹and‹θ‹is‹the ‹angle ‹betweenthem.‹Then a‹‹+b‹‹is ‹a ‹unit ‹vector‹ifAθ=π4Bθ=π3Cθ=π2Dθ=2π3


Let a and b  be two unit vectors i.e., |a|=|b|=1and θ is the angle between them.Thena+bis a unit vector if |a+b|=1.              |a+b|=1|a+b|2=1(a+b).(a+b)=1a.a+a.b+b.a+b.b=1        |a|2+2a.b+|b|2=1                1+2a.b+1=1[|a|=|b|=1]                                a.b=12                  |a|.|b|cosθ=12                       1.1cosθ=12[|a|=|b|=1]                              cosθ=cos2π3                              θ=2π3Thus, (a+b) is unit vector if θ=2π3.Then, the correct option is D.




i^.(j^×k^)+j^.(i^×k^)+k^.(i^×j^)=i^.i^+j^.(j^)+k^.k^j^×k^=i^,i^×k^=j^i^×j^=k^   =11+1[i^.i^=j^.j^=k^.k^=1]   =1Thus, the correct option is C.


Ifθ istheanglebetweenanytwovectorsaandb,thena.b=a×bwhenθisequaltoA0Bπ4Cπ2Dπ


Let θ be the angle between two vectors a and b and |a|0,|b|0.Given:        |a.b|=|a×b||a|.|b|cosθ=|a|.|b|sinθ      cosθ=sinθ                 [|a|and|b| are positive.]      tanθ=1       =tanπ4   θ=π4So, the correct option is B.

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