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NCERT Solution for Class 12 Mathematics Chapter 10, “Vector Algebra,” is an important study material designed to help students understand vector algebra’s fundamental concepts and principles. Swastik Classes, a leading coaching institute, has developed comprehensive NCERT solutions that provide step-by-step explanations and solved examples to help students develop a deeper understanding of the subject. The chapter covers the scalar and vector products of two and three-dimensional vectors, triple scalar product, and vector triple product. With the help of Swastik Classes’ NCERT solutions, students can improve their problem-solving skills and gain the confidence to tackle complex vector algebra problems. These solutions are also helpful for students who are preparing for competitive exams like JEE, NEET, and other entrance exams. Overall, Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10 is an essential resource for students who want to excel in mathematics and build a strong foundation in vector algebra.

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Answers of Mathematics NCERT solutions for class 12 Chapter 10 Vector

Exercise 10.1

Question 1:

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

Answer:

Here, vector HSYhD6BnodA1wuvfx1hHcINTrpJmq7WOfM9DNXCzukDSI1hfs0OppKYoOTGCeNLjRtw4ywUX9 PgG8okxe51gwjtXYgqWjrepresents the displacement of 40 km, 30° East of North.

Question 2:

Classify the following measures as scalars and vectors.

  1. 10 kg 
  2. 2 metres north-west 
  3. 40°
  4. 40 Watt 
  5. 10–19 coulomb 
  6. 20 m/s2

Answer: 

(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.

Question 3:

Classify the following as scalar and vector quantities.

  1. time period 
  2. distance
  3. force
  4. velocity
  5. work done 

Answer: 

(i) Time period is a scalar quantity as it involves only magnitude.

(ii) Distance is a scalar quantity as it involves only magnitude.

(iii) Force is a vector quantity as it involves both magnitude and direction.

(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.

(v) Work done is a scalar quantity as it involves only magnitude.

Question 4:

In Figure, identify the following vectors.

OsjeTGpk2bRvaMvI Yu1ApeD5Kn4dASEs8fsL5N4jiK1Zn4QnZF4zvddPyraik2qu6Rg nCZh FEnVwwr8rfyFLAAoeAUH7PxRAuwgN3BUnazSG9B Cm2XDs pMUXWze7VJM57s

(i) Co initial (ii) Equal (iii) Collinear but not equal 

Answer: 

(i) Vectors and are co initial because they have the same initial point

(ii) Vectors and are equal because they have the same magnitude and direction.

(iii) Vectors and are collinear but not equal. This is because although they areparallel, their directions are not the same.

Question 5:

Answer the following as true or false.

(i) and are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

Answer: 

(i) True.

Vectors and are parallel to the same line.

(ii) False.

Collinear vectors are those vectors that are parallel to the same line.

(iii) False.

Two vectors having the same magnitude need not necessarily be  parallel to the same line.

(iv) False

Only if the magnitude and direction of two vectors are the same, regardless of the positions of their initial points the two vector are said to be equal.

Exercise 10.2

Question 1:

Compute the magnitude of the following vectors:
O2Cpqek3zWEULzScGwopft10C1XwCPwwRmG9n8weOHs2ZAHWSwAA04k9CF9O0raojoSkF8Qo0g4wMhHi1Mg6RnH82G y7R1y0dgv nTPz1AHZU2dEU0NVu0OBjyqzODO5 hfguw

Answer:

The given vectors are:Yh2zmhJSaEB1aJ4Naz9YM Al52bGaAr Yme2SkOfx9fSRjk7mYqlZRMa8ORY1rGv8XxAV ffCglsZbNC79ca0w7RTgir3KqG0nZo7Qn oRuQGNj ax9YNAYr5qh VK8ksvr eU

Question 2:

Write two different vectors having same magnitude. 

Answer:9CdqoXuyk96zT3F9t7MmWMYCca xXaeBaTIxke31YsCwy7SqOOgdgee5n ALDxoG64SO9MDbRhxtNpiriXRErEmnFynagg Vjsesw yDfJ7 rp1MTT RqZOVCku3teyvuHtNkPw

Hence, MqwvMS371OTlQ8b3gcQCs2ACxCpjZfCcKHhA7EUmHrdDk4V vgh4CCHd5EvjeiP clI9JjWOGhOEGVP3cZCw are two different vectors having the same magnitude. The vectors are different because they have different directions.

Question 3:

Write two different vectors having same direction. 

Answer: 

C8KKSYLVQyjYrWtiAjgTluWmFWvzyYDnt8MVMPpkv5GHEIdjod2BXt10I5 3c8l5GeFkFBsII7ayknqO8mkncpcELNLFhPJfZs IeC3iUXF

The direction cosines of a4Yn90S35XuHow79jWmyUbF SRXI2n4nELqYtVQl8c6AiSdvRXsd4jyIIXsff8QN16JgPz9fdgaA 5Q9A2f2iTRubWthLR3ppzUiCo 1 Kptaz1rmDZrmiWP8JVAgqLYj71AJ5Y are the same. Hence, the two vectors have the same direction.

Question 4:

Find the values of x and y so that the vectors WSs9gBJhc are equal 

Answer:

The two vectors WSs9gBJhc will be equal if their corresponding components are equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:

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

Answer:

The vector with the initial point P(2,1) and terminal point Q(–5,7) can be given by,
JoQgxr1em6pPtU06PKNdZ03NDanARNGtheuXkJmK0imHze3kDa eoaacMUMXPI2q 9RBSt ncHP x1dCZlNcpRhjg55ptIyYfvf

Hence, the required scalar components are–7 and 6 while the vector components are and

Question 6:

Find the sum of the vectors PlW2dIKhu6KNxf.

Answer

The given vectors are PlW2dIKhu6KNxf.
vWTlHYPSyO6ESgGSAfvzUWIjiFxWJNUvQhBkecNioHSWQAfBEuea0zGJM8 cuHjE9nhLTyS Bnw kSNu1KVUZjGfOmdl27L aVbr eX8rp6hYd5aQ3BD1jmrFWGBqhLIYEu4uvg

Question 7:

Find the unit vector in the direction of the vector SGijqo7hRKE sJTNz Zy4XZf2XcSLFQYsJoa87up2nfLhpzI5GVdlRy

Answer

The unit vector in the direction of vector is given by

2CEICohGJkFUif9RQAZ2Rq1uxhoVH Hod cvgQo7e69T Fbm3hlm85NznuYOK7jmeNAmgQ1TjtIwv2hxa R 7hogokucaVkIxRbZnugHcuzlj lZVE1mT 8l1ZddVgTDML4NHwQ

Question 8:

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

Answer

The given points are P (1, 2, 3) and Q (4, 5, 6).
VrpYBqIE6wSNxy6LTpe6nfoKAWgZn0aeLCA 0 Q z3UH2 GJzDeuZFC2oLV4xYazQ8unh9qOUEg kHXF7W9Z1qht7U5gvzd3qz2qiZnEW3Vo 7shaoAhQGSHCxpY306WB E0ceE

Hence, the unit vector in the direction ofBDo4lKGm9p1Lz A4ikpZxw6g8VXaPcfu g E3bUgqecku15kNZHqWA17vNt4hzYI2x8UAGZlCZvGct2MfSl qEktwZXiKnisqHj7mghpQuU8HkstkltQ FqFJIlfpvJplOaGf G 5Q0b66BS3U9UQxyfoSHj0ngPUyd

Question 9:

For given vectors, and D1DUymKIW8GCum5OshsCBcIVnPtCSsGDiN9YmX5tWpGi1oMZEpHPrKGbuXfOHAsMMUm0EbiRPC tG7bZPWVJvdK YQgkRQSpvmwHQAW1HlPd9pJhDuHSvKVSlZF9A5B45gdGAIg, find the unit vector in the direction of the vector BnWYUCKF6rktcZ14Lx3wxPsSXa2xXxHS PtIlUqcnW8D2wyIwAxtdaTKvuwZ6b3ctORSA1gNt7pdyKMDJAAekiu1OL0K Y7nOlZz2c sEk 3os oVPl7pp jB3PJQjZwbtWmia0

Answer

The given vectors are and D1DUymKIW8GCum5OshsCBcIVnPtCSsGDiN9YmX5tWpGi1oMZEpHPrKGbuXfOHAsMMUm0EbiRPC tG7bZPWVJvdK YQgkRQSpvmwHQAW1HlPd9pJhDuHSvKVSlZF9A5B45gdGAIg.
liisw0SLqwgPYP5QlKd2vIipm0BANtr4deOeRI4gUxFFsC3gW3tEVg0P3az3N5PAo3 w1FS4Afg9QUbeegWhav kW6KT

Hence, the unit vector in the direction of isrqy7TSBf7IMHoebzxUybPWbdhhocN5hNVNlyNef9jnGmq hNeTqmfRtWDqclBaHCV8i4MLFuGfzznAluzK1CE5 KeE1n 3KOX654qNplMTbWOWFKpNvU3ewSTuD7cndaJNOQzUM

.

Question 10:

Find a vector in the direction of vector g4baKPhb7MdvAQTaH4hCY08 IDWOEADcsrORqzri5kNTwHPB ySV7Z3I1C7cNf9kEJq tMuSv9DF7LThYR100tGVxyRw4oldq5tOpDwKwhich has magnitude 8units. 

Answer
2MtoFhXwozrMvIT2 sEFy9sfJnmNk tzp00Y97qJm6ostJHOwmHIcBaoAYs jMm8qGZhGVLrfmukx s 5Iqi q

Hence, the vector in the direction of vector g4baKPhb7MdvAQTaH4hCY08 IDWOEADcsrORqzri5kNTwHPB ySV7Z3I1C7cNf9kEJq tMuSv9DF7LThYR100tGVxyRw4oldq5tOpDwK which has magnitude 8 units is given by,nAoiBLkY2wK7KbGSW7h3nrrZn0O1RAMt6X2Wr0EMh7mf6qrS8wPEw0YDLHGGvGC06T0Lag5hiDLssm3N2BnT0PqtWfxBRa5vUSfR2KCNDCyblr9mmvm WeJ hKwMbXLAB6ZvLSk

D38ugJNo0U6Ig0JmX1AWbVRWxL

Question 11:

Show that the vectors Ahn99M4h3bphUsfxLDDqphqhiCVXJs1uVCAm are collinear. 

Answer

.HSISIBAiWifmmEzGJsfh0NBoR3NdT0Beh0uLaSB5Cb0tfh 4AsHtGHKZ9gfi23xaCVrg8B LSr1KJBMjoUyM emfJz FiHCfDGKqReuMXfgz0CKYJ50ip5l7vRCBthBjhRTDWBA

Hence, the given vectors are collinear.

Question 12:

Find the direction cosines of the vector

Answer5REsVbJZyw0ZECUsWbFkd16 AU8gCNBlV7yFojjM5pOJUUhK2aD19Vs6MMV5PPpxvgUwInHvjh60HIvNJUcbneWSrqzTj4jOR373PHqR5fGr8tid29Hw8kWlbeh3zn gFWqS8Cg

Hence, the direction cosines of are

Question 13:

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

Answer:

The given points are A (1, 2, –3) and B (–1, –2, 1).

q5 CrH6QBims0tj4Dx4oPvVr aw 1 b0NUiA jAq 1ouiaIo4CqJNsrzkP lmZ8BVFJbSQijid5eJcM4PQFzNATnOfyJTHP9DQmkKcdE9taeFhZi7dCxej6 8iGZmNFiW9HuXSc

Hence, the direction cosines of Y8lBPX1N4QZADJ1P9gk4v8qtAC4GFyT4YeRT6NNzc2B9A9bs QoVckkX6Bep4uG4vXOAt7LhkMstuAW4CTe0tHgH0UXMrig TCnG31eWkzr2Ri4u uec D9pGfhZMEdNBjMyg are

Question 14:

Show that the vector is equally inclined to the axes OX, OY, and OZ. 

Answer:

Therefore, the direction cosines of

Now, letα, β, and γ be the angles formed by cRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgo with the positive directions of x, y, and z axes.

Then, we have

Hence, the given vector is equally inclined to axes OX, OY, and OZ.

*Question 15:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ELprZZKDAjXpMHh5ZO PVR2GJQ46477pdNoqy2Q0oq2YrwtHlbsUwNq9utnPVFTnMuwHzI37QZdb95WOAMYsd6b4nkk671zNwZcN7Dz bFBbi4BTvgZI3YrAWs6mbY4ed 8YCiw respectively, in the ration2:1internallyexternally 

Answer:

The position vector of point R dividing the line segment joining two points P and Q in the ratio m: n is given by:

Internally:

S7qAVnGq hOQTOBxMhFWzoR8etLOIho a1yc4H joQukAVApXJwBnSDurR8RmSN9v pnUYwjzsDfDiXWNsbny6J8LCRiC6wQqPji1DXvNmj3oSo7uHVDntBqwccihszmaUKkwTg

Externally:
5QLUIdRu33babzEecF1S64jSFX9s7FHgAQ981LUhHnm pv WaNJdJ4Ntq9OtJFWUUTVwj9pxBIdXHXcb4Wi84Rpg4t2fzBIH50PlDBsDYEP 0ttFfjVtI00oTnkzlkAMhCYT4ng

Position vectors of P and Q are given as:
Bfl2MpgMti0SYz z3FhRDSacet4WLs0vwQoUqJ9MrwE933dc qeLSL7XYnicVKk26nrxdIQbyOL76bqtq92LRmGOxOUxono7TojFELy

ThepositionvectorofpointRwhichdividesthelinejoiningtwopointsPandQ internally in the ratio 2:1 is given by,

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,

*Question 16:

Find the position vector of the mid-point of the vector joining the points 

P (2, 3, 4) and Q (4, 1, – 2).

Answer

The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, –2) is given by,

*Question 17:

Show that the points A, B and C with position vectors, ,

respectively form the vertices of a right-angled triangle. 

Answer:

Position vectors of points A, B, and C are respectively given as:
iOHBR BtPN0esqghtnsoycaUH0N2PTdEGNCultTPTd3Itax wFVKN5 OaCU7pgZ

Hence, ABC is a right-angled triangle.an1QRtayYiSFAIj1Zof dLDIBC4n0l3OkaCPbSHxwoAnkq5xf6nveXTA3s YAqTRypQgBNsrmd2hfnoLr3XMfFukxH1In9 RDp56vAgtByu6A YX y694CnmP5XPNVuMPktYUPo

*Question 18:

In triangle ABC which of the following is not true:SlZ475EMJbmAYcenA3vw7ZePkaloVVwreYReIMjyesM3KcZ0 FLYjiilcUXIo20uXF4FS1wroYAj4UeKM1qGYuQ3udvJ2v5SW70GJZgSF0dn8Vn9qkk7Wz6Vo 6aGsuhtWrF7YM

A. PO9YC3lbUHRvXpjKJNFehXCLJINu3PZ6io ppwY8KdEFh y2R5mHtRDCOIaXGe9azXBPJ2c Drn9BceB3qM0u52wAkI23yPKSbXazdhbJeVj0SdpWst7oCBFJzy2IqjPtzaTZYA

B. 3NaV4pBiDufTX0iQbTRsMHicXWRTaIA6y4z4C8IKeHWhWC2efjCfu6m22J8UEW4QpQVjHSp5mInLjyabQ59cVbjKzwtBeDnC46ED2oOl2K

C. j4a 7s5GqTThoK6CccbsJP4nIrULFlMJXhZdVc4 QQr1ZNLCte3GsDhISrNjQBYY xlpTewVfOJA fxtvyOWU2x5ymaAT4T

D. 65JtILV02iNsMlEnDF79rFP8qGNK02dqIeRUyXpVoFTll5Nkbg8pDTG67Mbvh7etWqvZc4 yPNqgi6szPsfozVDjlupF6Gp1zvNr d7Ae597x0rYFWqihxaKV5RlnnTom E 08

Answer:SlZ475EMJbmAYcenA3vw7ZePkaloVVwreYReIMjyesM3KcZ0 FLYjiilcUXIo20uXF4FS1wroYAj4UeKM1qGYuQ3udvJ2v5SW70GJZgSF0dn8Vn9qkk7Wz6Vo 6aGsuhtWrF7YM

On applying the triangle law of addition in the given triangle, we have:
JBXKs0EnImnoEqBjGFnSkwGYNt4wNQSxxIkiKsWyWbQKatTxkau5DoAWEUXGfiqJKCeiuKufuiYFsSBhG1qtv5El UNcKrdEkOhalOGFS3N XZq0bl867AaKsUWm6UiyTKAca2o

From equations (1) and (3), we have:
Z p5nNLZYsiU9M81ke khn30cdgEM3DBIcg9HduxBLBr4L7Wzi h1QpLcrweYCHGo6gtaPUWw

Hence, the equation given in alternative C is incorrect. The correct answer is C.

*Question 19:

If  KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP are two collinear vectors, then which of the following are incorrect:

A.  IHkkoxoHO7nPYuOaao STWhATpzXZ ZzoxJUDWJcLbeTuP0F3wf1QDffX3sGcXC7v7uaZbrrzGSoGRB f6 dAAXRrTQMB aCTbExWf F46YztEe 9lzy5S8Vu0uHR075j7BuyX0, for some scalar λ

B. f 4hBSNAWv5pYaaNpiXvzawBI 9SPFXKY gUCZ2RrRLB0va8ibVAkwWLdAss95yoezoZRsoPOObtNwlIW4A2KKGGNIDVjDbnm23 EnfBqxF3uOaf3d0R3zY eo7BL2YBYy8RpK4

C. the respective components of KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP are proportional

D. both the vectors KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRPhave same direction, but different magnitudes 

Answer:

If KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP are two collinear vectors, then they are parallel. Therefore, we have:

IHkkoxoHO7nPYuOaao STWhATpzXZ ZzoxJUDWJcLbeTuP0F3wf1QDffX3sGcXC7v7uaZbrrzGSoGRB f6 dAAXRrTQMB aCTbExWf F46YztEe 9lzy5S8Vu0uHR075j7BuyX0(For some scalar λ)

If λ = ±1, thenf 4hBSNAWv5pYaaNpiXvzawBI 9SPFXKY gUCZ2RrRLB0va8ibVAkwWLdAss95yoezoZRsoPOObtNwlIW4A2KKGGNIDVjDbnm23 EnfBqxF3uOaf3d0R3zY eo7BL2YBYy8RpK4.
cn0SSyckBWkg1eHcAF7FeW4MnwZNTbRM4qT0ZdjwjXPu0Xw vtzxp74NJ3D169ZhsXTpva4DAQuqQq

Thus, the respective components of KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP are proportional. However, vectors KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP can have different directions.

Hence, the statement given in D is incorrect.

The correct answer is D.

Exercise 10.3

Question 1:

Find the angle between two vectors and XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o with magnitudes and 2, respectively having qB5Fjfj48nfAgC98SLl1.cRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgo

Answer:

It is given that,
fFwKu94f4gUe1IzH2JL8HP8gPbuJqMsxjVVT AlRxhxPJ42kYKTwlRZK5bnkv6aaG0Hh eQVy rKW24yv6IWcXjoNPl AGao2bii7fot41tuGCrYAnlYZ3Ug2075AUNwciDV8d5 zFOa Kz7uFjSgOXUBLG65FrXcVn3cRH4ov6uD0mYp5CXHBR3OLNKfFUVKzwSkC8Gtyt7Snzw4BPTbDC

Hence, the angel between the given vectors and is

Question 2:

Find the angle between the vectors

Answer:

The given vectors are and

t ZGW3OyFl5sFYJGbZLOLzEXGZOLTjozyPjoRkDPz9jVdYBy13waZu 75BIYRX2 oylYRO4ZgTQttOdU5FADhpN1 62jW27 8os1PGTz8FGMWp lGAhT YqX PW63aISFwFcqIg

Also, we know that

ZP9kPIt485dScsBGVaP 8yWC FU4c4KPLabS3uQ S9T7wgIcRdlmSLH5BISpkJ5Y4l 7np9EWGmPt0mVdeASyhkESY2M4RDgDlJLD7qXwCetClJR7iZ3Lau mBwUN4F SRXYGkE

Question 3:

Find the projection of the vector on the vector

Answer:                 

Let hOG76kJ8OyB XDshE33WeJ5wEUthb8eCWIOvSMGJkHw676Q MaVDadubmkntN6iKX4LJskvHoB5uMU9l 5uUit16sgdulC5RMFGSdUxgWzux WyoM550QebsVJhP F1VGsNWQlo and .

Now, projection of vector cRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgo on XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o is given by,
QegRrsYwb0EuuU2cOBTXvmuOA1dh791h8hsgo Ule8Dl7GBVMa9wDOBzHkTaAjzcgCAqYInPa5d6Byi8Sh0bNqZCnXq J1rw VeV7AUUjAiCJRwOhGF6agReCiojLLuYSa5FdOs

Hence, the projection of vector cRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgo on XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o is 0.

Question 4:

Find the projection of the vector Dp XdlKVEkAniRdk LuCys 7yxKtcHatK9JkeWjM SCjCj0zGPVLQ4yO IuH6Sa7z3uI3QSGlOBLvfCwpFPqkqllSxqV EGKVnSlPwiuNUgkfgLOSmNw5D8f DD T5Gju8wr4aU on the vector

Answer:

Let  pWYlTFFGD6ACxJ6kg21NILAXF K2ubYOGodosjpmZNx5RnLeRsbk9jMPk3m7phs wB5zT3PrnLzv6LGv GH7rz8ue2WAqVkdDHHta16T 65G8J5eqiKaztWuDiwbSfM1d2Vb0E and GBcX5c2Bw7ArLdMaGqcJn8MF8XkHryikzAN1XSb3ne QGxn0SNDCpBu6FUR6WXnEPgqDWIGue1FtL.

Now, projection of vector cRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgo on XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o is given by,

HQV4gi90omGXBB7DtjVW7u4cT59DJJggnPnLiox4iLQWaJBIDFzySwsdKazRzXc2L0I3D4UImiXPd0XjyYEpNCasxN7oJ9FCxxAxVwi53CGYLsV3ro1jiRIDrUV

Question 5:

Show that each of the given three vectors is a unit vector:
zgHNZlYer90DQgmQwR099JSLLsRU8EuW0tfECT q6uAPPwswSvA7YKqCKWL7byLWa6PevYbCZeCwonPmaShyFJP

Also, show that they are mutually perpendicular to each other. 

Answer:

Kpba0b0wf9gW YnQxZO1fEGosTHkd6nMIHsqYeJnfEWRd9eYvhLBIOekLYjgXxscKPiBLz80 E83HVnZwhuDcm2rzdw

aZGLcPUl1VqYvUNXsk6veIQG1eyzW23lyC50sV 0P9kF jr6G1DenQtuYNlWDBXvljvkeZWtThOaUCGCr77P c5w5surn1cFPmfXSM8qWL62R4NRBHlHHzWGF44yj8akpypLFEI

Thus, each of the given three vectors is a unit vector.yqi26JWkkIPctNCR5j2lbjjwJw3v FOrO4SGNKWov4cMqkH

Hence, the given three vectors are mutually perpendicular to each other.

Question 6:

Find and  , if .

Answer

4
tD4WJw7CmNAZIXNDzmXy QO6AD1lQ2Fi83AiW4PJzgdQuPDECZ6isOUKDCMJQypxdLPK1eeaAZq5b9YRHQYce Z0iGI6 vwv4J

Question 7:

Evaluate the product

Answer
FSnw79gbd7lgA11Gav4chlEZXuipQVtNtQTe9FccGpIPGVSvyuKM0R8Uct5hLqLJBphP0ctirDIJ n2Ld 0aF3zYND0X6qcP2wgwQh7Oz Y5DZxHAatOJKpAAlC0F4sY6wdC rg

Question 8:

Find the magnitude of two vectors KCYpKyvFR riYPy7fH Lfw3bOn0w7lvTiSwaUpT0O6KNiBvwZR4cYJX6thcsWRP, having the same magnitude and such that the angle between them is 60° and their scalar product is

Answer

Let θ be the angle between the vectors FEdQQN2MrrdhtZyUBajYpRri CT3NdNExGdD4sYGQqQC0GuUwKtb4a1HxMJ1znLVwrakG5pZTJ5cPKwCvuHAiLMP42BhEbhD2q0VF 3KSUrAYe0j1SRxKCoKr TG471pYMR3EQ It is given that
TXG Zbv mdZvr9yjBewlD8SWcwyqUM6LGirKSG8Magg0VSzpIlvh7vnUj35owgn1JojEdDCGdA6k5HpKmkz

We know thatfFwKu94f4gUe1IzH2JL8HP8gPbuJqMsxjVVT AlRxhxPJ42kYKTwlRZK5bnkv6
jnjtvobLxt5mJUmp4BAoNcL70R9gqjfpKQjAaSPhXPblBABfFVrLQBTTesTM2Pdsl krmJ7XCRIclLXqfF8zzZOTng1xT1IjNN8hl6MsT COnUWe3Rk5GyzRPtOFkHYD4W2Jh6U

Question 9:

Find , if for a unit vector

Answer3HpH2Hch 8EV9lJ2MqHI5VPFERos5 2vkNxipUGNdeqney4gezF Ljea6JzCg0vINqqQsZKtBthC1izFI18hOir2284ERLPdoJ0OpSWBssS d1XcbQHbQH7 0fYtkb2YBOtHJ Q

Question 10:

If R AcXZubk0iwfMpaNWO2g5lJYO3vXBZ9Hq1CIuwSsE WQYL4EBdUV5eQNiehl7fKoUXd9I8are such that KXRE57Ylr1pQbYV2Y2HWCANR0mcV5RTFjtAyIn8m6GfoOv7KOwDDRm3LQrjkRnncHAtqKdwXtr2OGsv is perpendicular to , then find the value of λ.

Answer:lH6so 9BprAavlBdi0NZ3xH8IbU7tw7tjVFpyXHRZGCekQJqUl8QJ1LQhLf7qj7F5nob rGxHywZ3ObvPUwk70CZkrzPi8PwMgOpXQV5UJClx7g9cXehoYcMuSN uHd941EgPI

Hence, the required value of λ is 8.

Question 11:

Show that is perpendicular to  , for any two nonzero vectors

Answer:
DOne6T2I5jHPtYZyXBkhiIGD1CxhDEWBIaEcLcJ2nWsk3uOfn9RHu82enacNVoJ1FAamDx1Sazb2HDPR2dQNn9gahvEr1WbA7AOOrdmNp2GTiWt5Pdwz4 h02ms1IuaIDHZ32X8

Hence,  and are perpendicular to each other.

Question 12:

IfS PV7kJ0MKcpSqukl8 vH9fYuoXFBRmlyY wyfEMvUYBR6QaeI ecU1yLdWRB9CwQHwODQqxXsM2YVvNMlIxLnjzmFs3FXfyM2pbvBpqQkMcAXm336m2LGshiss OL laVfdsl0, then what can be concluded about the vector XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o?

Answer: 

It is given thatS PV7kJ0MKcpSqukl8 vH9fYuoXFBRmlyY wyfEMvUYBR6QaeI ecU1yLdWRB9CwQHwODQqxXsM2YVvNMlIxLnjzmFs3FXfyM2pbvBpqQkMcAXm336m2LGshiss OL laVfdsl0.
jPhW1pwi WGNCKKEq9BdRU00CPsb3bA5a868E56aOsnvWi t 3cGYkWQc8SwyagCBt68kPasIfApIxj8OHiDKu jR6PsSXRYFu9rpwFXN52rlUdwACy9sgEpN3bJ6eKIn yVDBg

Hence, vector XPYs4ox5m4Aag1qL5tqy3hdRbZv8n6WOqAtONhelFuFZJfr9HKs7D2yRTzgLmaLtb x8NEvN1uMsgJjMbcCwTQZvLyO2k hkSIQVBJcmXUHHPYUA QkkfU TVd LvWKsGIBaH5o satisfying yNlUuZ12rDWQwldMjuzgL5JrZ YurlAybKTREyfqhOu3444o3SWOe4JgR 1wDBJF9bj8AX ivCqWhfdbCSN 6v 4gxby2mi5Pc9n1DMZMalFikZUox50SIqoRamt3pDR0 2bKys can be any vector.

*Question 13:

If and are unit vectors such that find the value of

Answer: 

Consider the given vectors,

Given  

So,

om9UPFHc8E gRwmaXI4kUCgbIOSMCEWXSFh2mRFdZCorQOl4DWS29tcZ7ARrAsDktkkbMpLYAwuu hb 0upwR31c45ilw

[Distributivity of scalar product over addition]

…..(1)

Next

H16LrSmnib7UjgVmSOyOa6lSib99PobP43 VqCxTc6QvPv4TChp5Hu9Y4rek 6fI20PE
95JeSUXDSEQj2Kzea7HLzv5NhSfLkx 7ZNIyD 1iFqB ur YqGfJoXCG9aWDKlPXIHj4cBonfcLqXfJ5W7HRB8aO

….(2)

And,

6DW9Z3nke6 yLBzTp7tEKgXgAvThW37id30rCBSdg3ucBf0i2r5eSfmZ0gkTMDWnUsymSc1GnD1n634xdj G4uZ4UCzcGiGv7kCOHYgjRn nz90QfSMQmCJsM4QMeEmw0 L7cw
V6sl8GJkTI5L3oeqxCiOJ xyR2K0E8E0gykFJOi48diF8p1zqBx h6g15b4PgZJKnG5TlbW6 8NGkAjXXpc AF4m6I yaP366tpN0Mp79n96zTmbCcSqrtGzgYdFSC7gC61Tors

…..(3)

From (1), (2) and (3),

aR5 4UBeH44yWu7heHa9UBsOGosaDTV VpevG O05QTFM2NSgefsCRmhw8LrX rMlWaBYldMhm5XZunzRZ 4S9kdhsT78LRnUrPW9u12HLCB4Sj9wr55tedXa77Mi93mfmPrtc

[Scalar product is commutative]

fmlA8dYbxz7RPcrzQyB9V9vhg9e RDJCQKQaa05fYW McmLZiM1 ve1 G pvdwWVIoK0
XpakpM sc9qUwYRk4m 92V0uerQ1h0nO zn2mSS1ifi6ODpqesj4 h5hp8GSbpgJPt1iPcwo2ULtDe1 J8NSNLDTbKUmO PTLJsPo2D2 cHVp gHp sbXK44t1u3PojWpTWWD4

Hence the value is

*Question 14:

If either vector , then yNlUuZ12rDWQwldMjuzgL5JrZ YurlAybKTREyfqhOu3444o3SWOe4JgR 1wDBJF9bj8AX ivCqWhfdbCSN 6v 4gxby2mi5Pc9n1DMZMalFikZUox50SIqoRamt3pDR0 2bKys. But the converse need not be true. Justify your answer with an example.

Answer:
fEDdi8w4KEgbCckIm35zenrTLto3ijIBktksygJJSriaFHfUn1EVRUhuiGrTmlr

We now observe that:

SdrjqKh IqSEGQFf7xW9WkYg7UfFXvHyUdpyGkW1kmb8ciZMPZvxlVIe0kVqWqnXanJgykEEvjzXxBpSZL4KRIHDgqQcjsoaQciabAKGd2DWGZzcCjWCl N9vDPzxpNm7ZL DYE

Hence, the converse of the given statement need not be true.

*Question 15:

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 GOlZPNAeHeiomGfdL8JEq6dDDhIjF2WsGl VslyFokiSzzrhnx4zG1qEx9KQ0uKscCcRy EvdiXwCcQPeYvI57okhdKqQ3F CwRlHvb80YM0Cu1RfJs4yK0UbiIQvnATfIOfMyE and

Answer:

The vertices of OABC 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 y upft3LHHOt0VgrSEl VgBy52EqICiD8g6KeH7i5AnAFtJZbAL8Pegr SGLAL5wHgsjSaJs9osQaK6b1ot8JySj72QM and nyeFnnNwWIWIxKN3W o2KCqxVf2sCUY3AvI0Nc8THVKbKU6aZf SpR.

VSEOQjNhmy4O6zOghZhHp8w7DbiGAl1fQ 0BTfWw tg17iW sMfZ6eSRHmFu9msAtLMhiJJUjHn4OjgOS f01XfLhIk2mCLmNXHgOcLSNsbK4eIreGjhCGM9BQgRnT 5wTXviAQ

Now, it is known that:

t3gFH3h2J6fu22Eu28odHXRqt Af5f InWAezAwNC s6 RgbtGnKaDPTk0PlIXhDwxnUmRwoF 7RZ322dYwkISJJhdr58eXSLgImK94lCZ B2GSdr078gUAQ4TGy7tCCk8zdmvg.
4jPH ggNi4c2NYMPDJG Wle738ncI pTQKexd MusMlLynss5XUO8tg0a17mcMy5BN9A4IgWArXx7NMbDQiSpkci5rUBz1eJX0m77bT Dd9zmU2nnfOLVT NDzpRWBMKuTaKR98

*Question 16:

Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear. 

Answer:

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).
14KXiqnluFRjNbvf ddNe4nr29OawGPAmpMRaMSucbs3a3FmPNDKTv b46cEkwimDPnJZztpVxtZl9LfXnXFJsqUJmsYMJh4D2i 5UEhH72RLGOGpuZN08kndcbUI7ewdlpHJLw

Hence, the given points A, B, and C are collinear.

*Question 17:

Show that the vectors Kgabwh3zEt7sbz ObhB2Q2qfWkljFhR Hk cpC1t46Oc QKe CcO0Td7DyqYdXP lRuyI9UIHDDOz9Ncn juo8TDn9HP6pME8YrTnJsRRFSyWylyo7AGyPauS8teeUsGQFl KyY form the vertices of a right-angled triangle.

Answer:

Let vectors Kgabwh3zEt7sbz ObhB2Q2qfWkljFhR Hk cpC1t46Oc QKe CcO0Td7DyqYdXP lRuyI9UIHDDOz9Ncn juo8TDn9HP6pME8YrTnJsRRFSyWylyo7AGyPauS8teeUsGQFl KyY be position vectors of points A, B, and C respectively.TfpCXecIZexvC4RTux36Uof Wyf8 b4tiFXTtB1e3LMUOrAOQZABdHH5vY Fd7e5igs7hUJNWvbdPvNfJZPjbhLtff12QDSS wq841PlL3q hmZclMqO7PhstJfw4LWyY6gUjH8

Now, vectors XvFrDtnyadQ9aUnDOjqP3azTc7xB0OxdRMBX2Myb9Gcbb7CuAQOp3T0b38302JKws1Yi05hqC2iJxiYwgSyxHXWN q represent the sides of OABC.

xdAtb2CxH UbmtRw ermghE JqzGva2wr6L qkWN0vzb

Hence, OABC is a right-angled triangle.

*Question 18:

If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λcRnhovuOLqsOfMOh7SVKJANArE6Us9j2XnkTaCRNjEbTiE1L1ofywVqwRfM6r lliFia1ZfSVnCOi udOvDyg8znbQLCG5bpL7Ddc1FMGC8RWalubkBShto5TNhMpdcGBp37lgois unit vector if 

(A) λ = 1

(B) λ = –1

(C)     

(D)

Answer:

Vector is a unit vector if

lRO6ObRqgNA0AoyLwPn8F4iknuwn6fw3fyD1HULbW8dMZUyIUA4PLLLF1lwYJh8OXosuCui017ly3 IRcPvABu Zyv1uI8QZR7Rd2IF5t9r1CbQE0k5 52OE0NWN9Rw6uQH0vAg
SvjTLMdQOKjufSntHnbJA CPL6D CvwOkSpFiXDrL6tOpHYQo8MNPCVpsUipv3F2eZo0gdP8gosPCAtEAvlRuAa 8mgy8liTPArwh0dxjp0v FcJDmKjJodQULuebzZozQFo7s0

Hence, vector is a unit vector if The correct answer is D.

Exercise 10.4

Question 1:

Find if and

Answer:

We have,

and
gVbwS7gicShd8ZhMoaBWIAiGKaukdf mHjt uwiAkgEP2 fJl2PAWf0R5sUM4nQrtTM2 Kq6hOJAnqEl85S1fyBOh

Question 2:

Find a unit vector perpendicular to each of the vector and , where and .

Answer:

We have,

and

VrWOhN65CHtTZMtDcHMswFBAD5ncIATsSzKZtsTW7Rkulkmf6qaD84iigth0 rZ5GDuib

Hence, the unit vector perpendicular to each of the vectors and is given by the relation,KXWsw0XAP38yeg SC1i2zbRhEa1Rs7m 90fal9CKP42 g3CqDvHunbafs9UoewB 3WCnGwvizicfdGY6UI7WOVDSAuWQmgAwOVw8C16NFrg0yySaOGTTHYP9LrR6eEEai4ddAfs

Question 3:

If a unit vector makes an angles with with and an acute angle θ with then find θ and hence, the compounds of

Answer:

Let unit vector have (a1, a2, a3) components.
pREIenNU1Y2nW98DlmzBdSZpm3gDLay5 5q6xge9l7gDf48RcPiXlqa2wCHVqG0hToZ2 TUyGkwo2gapWh83cJu4fDULJ7v7ZNhDiiq sU6BG4pfFvPpQp3JCA Uyi9GwmdUA0

Since  is a unit vector, .

Also, it is given that makes an angles with with and an acute angle θ with then we have:

6fJFgtmf6jS xKA K8qj

ztttmaay6h8BF5NtHeFG
XczqpvDjwzo1AKEed987MxmVQ1TIs1IgrCSHiQvmR6xF76v2bW3URz4rWQt9p5r6MCxUMI2c5oj7EnbuX9BpYclJp69LZ Bupvl5txvxw dDQDC0r pXPGtGollFIqM4eQRdtDQ

Hence, and the components of are

Question 4:

Show that
x nBQvTsGyTMr0smhhQNviVsnKMA44G1BBJhRGAsQJl7gdjGemSfDA5qf5RYyddcy2hk3jnT700KdWC p0s1mkb4lD

Answer:
ptJFeT fGuvUGjGQiG LyEzNJcdW1IAoIQeMz57bNZqHbM

Question 5:

Find λ and µ if

5m5BItRD2cMElRLsZtUX1qkkZR2gb5U Ag7WSqpm5suL lR3AyDnwynFuthhfkvuU58zJgoUPI0Hbg8zmT3jfTcnqKxvP9rrlNVRFJicdFAn UGWbaDAE65RqYI1vXcSjX78Iac

Answer:8Tdh8QNDpt0MJCU Dmujx6fdPn5hCZYOq4j7BC3HhJxQFhpwLnk7LMjjO8XF16dO6iPSsE31F3OBPLVHc0LdbsVF2Q5gUBn7AzPwtNU2UqIa omuuLp9deEeICW4PWJXgpQdLVc

On comparing the corresponding components, we have:
yLiLXWxb6TxNbCVeaVN6ih96NXm7SEBDWL4t2wfV CcfqHxzkqnQDNcbd3KSkP9v4QxS6IT0KNcrfBbrPqndjoKDjCjQfShKmVFquHv3xKzUV8wp1bTxM8culgKZAFJ29WzvGvKzoqClNk8zC1G

Hence,

Question 6:

Given that HGyKVMtuPlnQsUX1LZRItsDnbzU2lIUr9rdsKUGxsrxx4j PDAy0mjKlWfS and  0zGaO3ZhC hXYgG79Hc iR7u2QDoNjo7Y 8D6KYY4pY9F1sAEJlN9fUttH56TAV9t7r ZGGSoOIju3DnSWZVJo7Z6Ygn61K 5YevJx cbGnA6jIMzFh5lWvhVOen4hyppi ZgIw. What can you conclude about the vectors sIBVAs uhXaDkV42 wLYE1Lt?

AnswerHGyKVMtuPlnQsUX1LZRItsDnbzU2lIUr9rdsKUGxsrxx4j PDAy0mjKlWfS

Then,

Either or  , or 31tVMqi3usBX6SUgPoKptflfqmDW71xZ5dpB X UImY0LRWOHy7GMmjzFrdwiFBTfVVzn4NvTkdXcshn7Y6FhGu20BnY9jrTqObEsr0lOjFb8UIKQCDuaKGu Rgc2Ev9vGCfd E

Either  or , or . But, MDnf0Yk09RXoBStelT7Djpa4zWm5YDFxkwO2vgj5VCJ2Z4zrmkhBxA4AjW Sjdcc2o ov Mgt7i09RGCRxskp8w 6n9x9vhB 6QXBmU5BXI qc6jrXLZovUmw1bcgV0MDcERYI and  WI9wrOz3M2rnfIJdTt3tczBrP5EGcRJOLXm0gFsDzIG4J114XLylT2canDJJPMVrTVmLgGI6RaoNOoFePvqt6E ic3lYnVfWykywitQaxgG0SOe7Xh8Y iwqaS1sbEbc38sA1P8 cannot be perpendicular and parallel simultaneously.

Hence, or .smGFcKmDYej5O5n1HPfNVLz7VzfGWmQofCekWIiTwXfc97 Xw9CVMc7Ce3hmr3AmoNfUpdNhxGXPmybZ9htYHC s68DVNrA mDpG1YiCTYdaVXdlm97Q

*Question 7:

Let the vectors NlZihaCVcKDtrV0dHehJxj 1ZOtH3kzETLNqzeT UtElglnppQPEWumXzemCcEd h4EIS4XzfLtfsUQGIKcYiqomKA2AZgO66YNnF blqiJ8MoRjkrTnEF4TaXeGrOOY6K oW w given as kHvN3X7JthLCoHIhNFqE3h5zTZT9xp. Then show that

Answer:

We have,

oFgTPvG XhfXM35WamPlpUqatLiRr

vLyCTrH7OV EJinDmHzIMVf7BLLn4IEjEX6NAyiap8o 4d n0rrjU14Y7w HuhptRYjK3FOg PVagiZBWGt4Q9xNa4HmzFFmOv1vLIlz8lD3tlcDw8wSiHOnG9Ye5F9odExFp AcEIXBcwEtSQfalDZqmvfM17U40iCwbWwmvii7KjYG6WPBGqBjogfPNRv49ePvQSD gQtRDfTnyRBpDc32XBmq04R8tAi5okRQDBCsel7BTZKf7eIVaJzL 8CTVE2W6YOArk7vn4
TYRw6WCLJtO0tl89TNr6oL5EwpL qYCMjbfi7ITzl0BSs8xsks1PXZB2lqyTnnGFPJgTcAuEzAo0ch lzXhx2 s HFs9fbaCkAPP3D3kF3sA queFaBnOprKf ARlwp8UdjSI8c
pjOz9uFAArwYot69bxdXcOP0j69MZf68y6

On adding (2) and (3), we get:
8Sd9Ou8g60iqQhGjkofCn9VVSiLoYgf7yeGJSvJZGXOLYNxXDlWi33tkfMnhrL 5STx6LCqxItEVuAtXOFx4O twMd837o15rJX5KYWMUoRvUL IjRN2FcmHdrz9AmgU1lEz A

Now, from (1) and (4), we have:
qzUOIqa1VGCLbxp3HJ 1VquXr9PwiYvCIurXOrBndwvYnSK7Pd31zAokCqK3Hgz1d9CtK0XKS2WFICVfmAHgYw31OjoVKx5WhpblqcP7Rch 0hTkYE0MmMF7ewbKrir4VI5D7O8

Hence, the given result is proved.

*Question 8:

If either zQXMgCq5BBv4VJ yAMk3FcClKiorc7veqAxOt7G3wD1zXaalMy6 vxVRih8p61Wo8ZiKRKzm WRG ud9nHcSa5i D3cc0 gz AGyFjP29iytB9hav2jLwbPoWUnWpghjQvkQ4qI or yBEyEhAFrQwwLtYdRhX1VTgBxaZJeBlIo5B1egbb86KT2jntjPpQZRqmFhKl fmfxhRQSxcfGmXSN42Ye5u3Q3jQJwJ68V7enCFJzWekRrRG2d5ABjtPk1qJ4nGmRruVRTe omE, then HSl7B3x7uX1njIb55SI4fnP nEfJ7e8C7 qFBLno3HYn3kTWsE52vIfJPuVqGxTfmAemYGkcKBx6uzJE Q5ab1sM 20yeCQHc Y1qHqYx3walBLMaPQc1Xub Is the converse true? Justify your answer with an example.

Answer:

Take any parallel non-zero vectors so thatHSl7B3x7uX1njIb55SI4fnP nEfJ7e8C7 qFBLno3HYn3kTWsE52vIfJPuVqGxTfmAemYGkcKBx6uzJE Q5ab1sM 20yeCQHc Y1qHqYx3walBLMaPQc1Xub.
oF5qYRIhj3daXscKQ6p6jWxoMXr0OQmhnb7T5WNl2OiAZzu33azy5Ip JBDv4mYBCqyQEKNEvhPLv dYIRn3jd8l 4gXlvkJNqj3mBk2PxFl IQnOwL0KqmepAARbJUAdL1HLS PYW23 5GQ5XBdnrprUSSTXQ0ppaowLHnkTWFEkbXscvmgZeZAO 9PvafpimECELCMOzNPDKBWyqbvtdvEZzjQvASk

It can now be observed that:
26WduYJ mIGuA1SbsuTktB8i1NCRqe202kNoTBE30DkW0AJbQ7gAfTZ0SKbC4bxUR0OqACsAGh6p

Hence, the converse of the given statement need not be true.

*Question 9:

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

C (1, 5, 5).

Answer

The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and

C (1, 5, 5).

The adjacent sides and G2sREvMgiy duWtQQS7aA9JB8WjklhFGgubtNEMdwGSYdwG1XH7Wdn5C0Hovl4WRE2kl 8TM9cavEif700veldJWAxjEfjuH 8wOkTn dXj4NaxHqv5B7XH uAgYzvtSFYrLiXQ of OABC are given as:

5poVk6o4e8ZpY3Pj62cYy0FeKcn9G2sREvMgiy duWtQQS7aA9JB8WjklhFGgubtNEMdwGSYdwG1XH7Wdn5C0Hovl4WRE2kl 8TM9cavEif700veldJWAxjEfjuH 8wOkTn dXj4NaxHqv5B7XH uAgYzvtSFYrLiXQyrYlhMEOk63wgMA6h4ISId8fvIm

ES8llfV6GyfTmwNJDZeeQbXHHRMH2w
nXJrcB1xePbXmzCdMJd15A6KN3z6ayTvJPbX5UUQoxubx2gZFtVOb1x7dA8tT0O EARBcGJ qLYDTw0zXVx1V0tI0zqCkxeumK5 uUrsvAzawAA0OXK1rR7j7sb5TkCZZD797nM

7EIKn oLnLi ZbbysVSf56W41uGzhiQF9SVZMTAQ3McDJ88Wf90CS27dyl3LkCRkbjwDv4MQThQGQollxcVyQq4KmynfhPnDjKKuXql4rL4Xd43HRMmQDJ5TUesZfCsU wWGukU

Hence, the area of OABC

*Question 10:

Find the area of the parallelogram whose adjacent sides are determined by the vector

43i 7oZorANPQOzIUhuuMLJ2Ukx9uIAqqyNJTQSje7ZrDP HPrdacepdqfRGr7AcQcgNXRH4X7eVEg2J6D2BCGvb7T6Kbyqg7dKVifRMhpIetKidP3InjQupPOO EwzX13 FYMU

Answer

The area of the parallelogram whose adjacent sides are is Adjacent sides are given as:

Hence, the area of the given parallelogram is GNs5LneCHm45TqkB1cr7Q5h1mWFqYw0Qq1UZ vcYnO0vB75iw3Zw3v1s bk8zSQHHdWmhkdj0UWJyYjSkaeamAOflYD.

*Question 11:

Let the vectors and be such that and then, is a unit vectors, if the angle between and is

(A) (B) (C) (D)

Answer

It is given that and

We know that , where is a unit vector perpendicular to both. and and θ is the angle between and

Now is a unit vector if

A6RS9ZgwoAccBh8wrAiRgUOPQwJDPHDJG8NwoDjUP 99znYVz9tOdQF0mub6jmu4iH2pCy7XF

Hence, is a unit vector if the angle between and is . The correct answer is B.YxyNVIpFE3HeS8Q8pxCV5Gkllmij1YbXoarrIq44nTZ8dOWHjnb6 DxFueoWyCrsxBYMc 3QkKcd2Vr sUb43IH2KBUS7y6lwbL AJNYX1xjxiGiGbL5ff9g zZTRCG9qddpmbK9wbTNTPmPF IGplbKFgHYmOckMhyoSvQaFV Kum3yCD WCfZIVnN0mTSYaFza0Jw1074FWuSAYlfoWJLgI6tMH6t iNms2Cvqk7YDiNgiQHkI

*Question 12:

Area of a rectangle having vertices A, B, C, and D with position vectors

and  respectively is

(A) (B) 1 (C) 2 (D) 4

Answer:

The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:
cBb23rbXhZtQTFIQVeEROlxR XrKgzw6psB4PHE0L W7GT CEBP4w3O0vdr9bDUfz6K8x3O2fhKEYXh2AggUyo4yK4GN5ho4qSfTJzdaEH9sxbXwEuGrJkM3 N HCGRIDdokCU

The adjacent sides and G2sREvMgiy duWtQQS7aA9JB8WjklhFGgubtNEMdwGSYdwG1XH7Wdn5C0Hovl4WRE2kl 8TM9cavEif700veldJWAxjEfjuH 8wOkTn dXj4NaxHqv5B7XH uAgYzvtSFYrLiXQ of the given rectangle are given as:wq9P3oOkPlowQZHyots3Z7XEeCJqXKPgpcixulbQF xs AFj1sT3Nqti6RsPQysChTlLk ka8WFfkrG8BdsUWxqrXfb2 N 8s3RNwXOZDul9sijK4voPFJDCXXbMYX32FfncIA

Now, it is known that the area of a parallelogram whose adjacent sides areId 6UI0O0R is .

Hence, the area of the given rectangle is The correct answer is C.

Miscellaneous Solutions

Question 1:

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

Answer:

If is a unit vector in the XY-plane, then cRguolm2U1Rh043BTOe4EkbgZI5E0Y6qUSLgZMNIhOyo

Here, θ is the angle made by the unit vector with the positive direction of the x-axis. Therefore, for θ = 30°:B xckxAC18Jyp69h2c6 iAJfdb4duyq6 RA8Zhi5QbDMPU2uew4 mVudMKRDnwQejPf VauMxuLea8CU86cxWaLlEMyttERlAivm UPYyGGOfNbd2x 8Id4WA2zslGyZ2bCWjg

Hence, the required unit vector is .

Question 2:

Find the scalar components and magnitude of the vector joining the points

.pF1LZj3bjnaO 4cr3TgMKVvma19tuTKGx4VFWAm4HNrS0cJFAm2G5VWxtMpfE9vlhI8N4Di9xmGz4Fqg7vs owsXN79cTdxF4hwVySQaBekKBFsyCMHoW2

Answer:

The vector joining the points can be obtained by,

v6zStBOfcIYKtVXRb4cg0j7J0n4ET24t1fcrkMNfKbb1q

Hence, the scalar components and the magnitude of the vector joining the given points are respectively 

and UokgTSDws1D9t yM4.

Question 3:

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

Answer:

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

kk0DSWfjBg2m2PcxmjDvMzW9KeCpkp64tQ8pZ9kOsY ernRhjYRVDKhfmQY19z8uTKhfROJw0TJuSA6f6ZjTSnGIeNplS e9OaZQsmimCSRhkVFqg6bihkpu frpDm u5bWMeQE

Now, we have:
8aMhBOWKO XiMyI6qUK7i95BEmsn2yTbnTPgLMP1k5r5wYNoKGVVUH ymvpMv080ToeY IL tLLMkNn3vOdCfV pgg5m1HcPLCu6 hf13hDtk6cfN77JuqbymKvmRByt5QKT 8Y

By the triangle law of vector addition, we have:
A93dS Pmd6cdbTcvarG4n8VpcGS2IFHhYGgrXhONy sDrnWIy7m2mmxiS9ESVSvNRul53rlFdNLm3GXLma5i1r0fxJnjRCCDKNoXmFGbHlk56p2ExiBy90o WY8vYwciyKoeu48

Hence, the girl’s displacement from her initial point of departure is

oEDwdBj3GMO YU9vSOkUvF7zuwRMYu0D6yS 0ZcTp8h8UNrE1 3uLW8YGtD uovy0n6YzpD95WJUfMbyCHASr6nOcNmAxzNU4rx7WJZUEHTMDm2 VIdwHYBOG4G88ftiAbp20cU

Question 4:

If then is it true that Justify your answer.

AnswerC2qPUWOX8a Cg 8LSzMsiMYzI2rHvSu 0AaQ2 A5FijrVvtFgGdyy7ZmnoPtg3nNeZCzkWb1zEH9i6idTjvPdJ0RqctSBT90Dyi8cE1fYJGUH6IKZf7cq82QLm24h63uodSeIkk
Xz6 zmrFgUXP fTEdboNJxz6W84eu 2CtQZtwR0ebD1Ocm65FRe23hb9jmME5ilNB9mw O6SznvifbfXf1sgiXTtj0WNp1kaDndTW1Py9tuo7r Nx b3Y X485NTbkLvgnRG02c

Now, by the triangle law of vector addition, we have 1WsReK9RV5djRlVt1qd3vxPhfgULnLexHXuz7SC62cMxTWoCl5xcmJ5fow1erbUXaWQAgRRoCSHPQdItczjdZsfDOT5ZHfuaalmMUEFzE2qPB85vmSjGOSOUgc lgiOD8bZNCpo.

It is clearly known that represent the sides of OABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

sd8C8jQDIvdbSdRfv5NL7HlQwhji3 rIw

Hence, it is not true that .

Question 5:

Find the value of x for which is a unit vector.

Answer

is a unit vector

UIiObEqTYJHSD00hAIDQCRTveJQAaYe35aAt

Hence, the required value of x is

Question 6:

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors5OSKoupi6HWW7kGK7kLjVB07VkCLiwXyhbyFisJsyq qDoRF88QWUhS V34YQXExJG4t6C5S IigwEoB9N74vCp avkVlZLc8Ucf q2rFqb gmhc3E0VazOEnfjI34jfpjAY2o

Answer:

We have,

olhlIjdXqnjhViyeLSm1 1VzIU6GnstOOYZEX xv nEYYQzxBVbfdfIsS8PnlWST a3pxHjben8Cdtk584Ea5MWzETOSKeQPXlIuOkiW38e5Cj2sz0dsHl aRlSwvC92NN3pqX0

Let be the resultant of

FsAgIs56o5jbsHjY sYCVdQ5791vyKYJTymtHtUHG9ZA0pfCccmtIK13c6qNhJB 4cVqgizf uCXK2KRuSstC1

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors Id 6UI0O0Ris

NcEB8dlo7vJia8ne7elbmt5g5QsGD7c5Z7H6bn8KfM0zJu0Uld6IZ4d kuURguw

Question 7:

If d69v tJQpDPs7RdEU2uZRix86o hN6bOjns41IO42YnWUgo86sHCmG1j4gNG8yct4SLUwY81g6wOfDvl78uCPEt9Lg 2bHENSjvrb8IwG1SYO77jZLPtAttJ7bR5JI7V u8j2V8, find a unit vector parallel to the

Vector EE9Ns8XBb8CMo4vy5WSazUZBNpD38muMB8wjuKI qUT1NmsvCbyypSWYwBV I4uoZFDg3SM6k9n0ShIf1gxiDfB8nFe T5rSa5tK79YgMwmmMV7L82yStcs1Ogo 3LsBs9eyl5E.

Answer:

We have,

d69v tJQpDPs7RdEU2uZRix86o hN6bOjns41IO42YnWUgo86sHCmG1j4gNG8yct4SLUwY81g6wOfDvl78uCPEt9Lg 2bHENSjvrb8IwG1SYO77jZLPtAttJ7bR5JI7V u8j2V8

o87Rdk5PuWWNRgmafsnbNdjvOG97 kOOtTlbNBMT90QZpT4qNif9oMzAyOLyy0XGGWLY851QKFE mKQ9oFKZHCvcjYBOUlf4GCCplpjRsjsSawZDnR6kNEkVWbn5WQny8hkVQiI

Hence, the unit vector along EE9Ns8XBb8CMo4vy5WSazUZBNpD38muMB8wjuKI qUT1NmsvCbyypSWYwBV I4uoZFDg3SM6k9n0ShIf1gxiDfB8nFe T5rSa5tK79YgMwmmMV7L82yStcs1Ogo 3LsBs9eyl5E is
zcXbjHwoLCOuFu2ulib N84tBXj2Sq0YlwTUj36riHluHyT IblB3IUUUZjb8houb nvk3g6sDtx9DKYWb0UIetCjzpRn3mWOds36BIWQrtJXxt MqtDZ2e26lh42O0e8w4RlWs

Question 8:

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

Answer:

The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).
2Zb13cLw6L7 AaC1Bn47RPQ7NWD NRZZaqeJ Q7dHFTFuaF8511DRa53Q0 xJGzt

Thus, the given points A, B, and C are collinear.

Now, let point B divide AC in the ratio Xtu2D9E6oW4PvT9iidn9Eh5Tr2OEckhJxOAnAX64hME0CvEo IyBkofYapnJLU7GLIbwIDgGjRopTdVuL1IrgdpRkHciVXthdus8Rmk5uKysJ4NZJ9 JKjaeC eRSFaEXU4F6U. Then, we have:
p5YGqjUOYFmT9F9

On equating the corresponding components, we get:
05r rYKM6 kIcKGzvsCGCfXy7kEPdUBCxoT9WbWY0t3Agc QSv YmY6gCVdTajcZwnlacvsDe yjEuh ggOhzKQoT6 n2wlfwnQDuZpHpT8MODHADBObWUCaeNoQyoIekBQVOE

Hence, point B divides AC in the ratio KHcxQTp4 yR5cNLOBp4pcymttkMdZVs8BzLkXJephK RMiM2PP88ZMDXy iqumFGFZePOBaLJOFfdnV9ILKNExLSFMfv0tdt4Th7XNgbpfz0OsQuWdhA2Q0vTD wc4GvuatOoSE

Question 9:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are externally in the ratio 1:2. Also, show that P is the mid-point of the line segment RQ.

Answer:

It is given that

v0a7NFcyctyRHUopHxGP MkF9zV7gD4dvNmgzZeZ20CRx8Zq84pAAKDky uQ2a255UidzoW5JMDSLwHhyVs6dS5JdpkXflYpAZTfTkWG vVYf FH7TBGuE49I10rtwAb8yhqLLA.

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2. Then, on using the section formula, we get:

Therefore, the position vector of point R is sYA2CdhvGDFufa j2U7COA4GI8thvCh1ZMwUm63hUpUTWhf0OgnumWYQiTk28nsarzirIkiDxQSGB0j9kEMS708Mkd9sBhFNKA1jP. Position vector of the mid-point of RQ=dYIGHwvtEGISOP4xMToAMfeLTAmcEYqzOPUDOMPkWT4KWUS7sDzubiZ0kmP41A 09T0IxU XXM7uXrzZmzgH611AomLOd7RRQi7MsJ21DnrqC0BlcQ0Zp7CEXiij1YJv74mXru4

Ig7ICWlStfv5LXELfMVQ

Hence, P is the mid-point of the line segment RQ.

Question 10:

The two adjacent sides of a parallelogram are gyPIvYnNshKcT8o9Lr2eU8A11DoN45pIPP1DmOusbwcNUhFkyHl9YlrzOscxC8NpHC66A0eboBp2 and IJsIPU Bj1gSaCnyDAxzBDr92lHTEecEcKl3IzG2F7J ayKumlf26tX2rob2mmGIFUMduledR0Sfx0oDPLxaCOdyO9k0jZ1m 4DacVrM93UGz m oHcneems mCVIQK89mugLNc. Find the unit vector parallel to its diagonal. Also, find its area.

Answer

Adjacent sides of a parallelogram are given as: 8Kv hfpha9DS5gj8Lc13q and Then, the diagonal of a parallelogram is given by tZxcbz4mMOV KjSR i bB1vNXGACvWb7Qv0xKFRFh08xqlPbEKjRNCUEvvkRkmX4YokTkM MdDQ7FiNpeeqS8cZO EuJujUYisll2koX4Nf5Sb Pk4XwUcLK8O Sko6qhVPKtvI.

Thus, the unit vector parallel to the diagonal is
JpixzU69 MLIQGjCFieUCtljGafuaVz2ZmPEJ5Ie2Q

HL8fFQB0YLT5jqKq2BUKkxK562dBCpZw3ouHzD g SZumPBnl193 HMZ9kcq126EXgRQ
w3zLE gioCf7RhIrDqjuJTAtLpFz8XHx h5QMHgLnH5sM0vpmaRtt2jRHt1BRZyzjMjtyOAVa4FLPHXGXT4HFYNNmwPrsngUyhHSvsY0p7Awh1tsztXhMv6AN4EyBd52I8znJsI

Hence, the area of the parallelogram is kjAvkmtfEvKFCNQNSVQUds4 46QOBjV1w7xfw8 yMJEVMqCfwMl410tDxPFZ9Sljnmu1GpYpXc xXXHdigxLO6L45lWgB2tC850lZVMl5eKrF bwCZe3K6V7rG square units.MV3C5 nclZCVGfw3Ayva7STks CuX4 Nz1y8YbwuQSvd3LqFBd F92VsbyZDGANzeBMO3C9AZQ9v4nqYn9KF xAoPvbzo1gFxrAMrV LX5MfcqL9kf7EhdLXEhRPIghRIvAfSTYojR43sWTn 1yttT7zqSerzrwltMErVM7Y3 8x0KY6gelcDQ1vOeAhnLywV LffGZU8djZqHENsEH 2zqeZIP8UX

*Question 11:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Answer

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.

QDIX6XrJrQTBu dJSFJTuOKYnL9TOiHwlPMDG4hzo06eqssc64ZRxbXZEk8cKIIMY

Hence, the direction cosines of the vector which are equally inclined to the axes aregzygV2BA5Br3sAkfU7P jxYfX9AEgKVhT5GfExT2zwXvLKzzzi5Vz709FnOAow3S5FKNW4EPsY2hQAobXZNFt6CiMV1 L7gYf oWvf1swPmVRFpqi9Vg2WN0m4VwKcuL2o7PuGg

*Question 12:

Let aWd4th8LcAO9Ky 0DwT cseimn6ewayvgnW8HhM UM359Nq 0DqjAX6MtVH0SOknX3AaRY50RAkoWEBTMsoKMRu9E6ALSd topcYm3YkO68X and 1iNRiIaK8eC4UPWtFRCgepvKUO4RcZtPrRAyoxPiT8UVqlT0YYltrLie7H EB9okNd1PG66xQaKGzNV9CatXSm14EXhAX tofTcSHiD6590Ii42j0dASyDCQp 3j rcl7wZRCbU. Find a vector which is perpendicular to both and LG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0, andywSqlZGY6lzwfZCJkKT9LNsFISRowKHp Zpd2OKL 4m52zGHzWH bbjMF9I.

Answer

Let

S sVcsw InL6n0 XbKuWjPJ4MMvDw.

Since is perpendicular to both , we have:

QlFwUUhhirphtxew5YGVru60UZ6FbED0PBvHYCZ4f2 6hj5mVrv3yBsqbZgh0vwelRHkZAoIXAPOb M1g9nqauG2cQ5hq8wgZG4sPKeiXiyc49Jhgj8jpSbXSIUZFU p2eaZQOk

Also, it is given that:
WHwe1uACQjSu85evVzjpA7MsV9C7w528lULSVAYVnIVUGauzf0BfaiE Ve9wAxEIrG4appdDBHn5vQDojDsZHOTxBBbOeLGu8Tn6

On solving (i), (ii), and (iii), we get:

jrpbNsYlQA9H JsfiR08KUE0 5 plu8u1pLSACvIpZW6AvCBrFBk2RysY3l hk1qlU3Th4whD9KtFf1h9ZOREbNoI90Pa YVGLUXWW14m5mW4sbSmqGlX8J0jCBC6tsj9Ya 5eA

Hence, the required vector is.

*Question 13:

The scalar product of the vector with a unit vector along the sum of vectors

and is equal to one. Find the value of  

Answer

aB1rsuo7GjrS7buOXOqB6twXkdlkWFy5Pf pubsGMzAG3W9zJV zKBunrFxNT4Ze6pE3w5YImxGY7 mtIxd6OaertpwcEC2pNFTHhxk1Oq NqTFo9TJ XHO7oTfqaolQqwqt87UaBQfVU825IHPu94YNBo opO Z8zBdRhFcMM2OlYXOuguxLQind OXWS DFIOU4 X2Wb0xK5 niYJTeFCUWG4ltZV8Zw6 j

Scalar product of with this unit vector is 1.

tcHRmSKkAk p4Dj8WV8 xjd4nTSmE1aughg6MElD0aKdBlXpDO9SxH0JBO2eKsDjZAEdwgxSnSt9ZkDLgfWzqaU8qT wunNs3GOT5C0ne1bmqW6HNiCeaXF4lbNGeYfyllTLaz0

Hence, the value of λ is 1.

*Question 14:

If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to

Answer

Since  URE6VJikaoLtFVxKnYHS4JBVXlgyOh8rNx bNdG50SzxRRYfQMhmYOoaRQrYoYZCrV8RDK64f munO523w BHovj1L6dS SXpskbHLJL8IgCk7CUEkTy8PICd cOO4A1CLqDAVc are mutually perpendicular vectors, we have
tLGY6JjcIAOqUVL00 o5hmmgrt J8niF5SqsCNDyLsAKMhIHN1v6RgBL9egHl53xpTjHUAKD4K cUeL5cyhmSaya9mAYznmMLMePEUL4P9d0PX5tdwS9II6XooHAR PNiKtGO0g

It is given that:
BJQzss75mnwn5ZTJa6dWu KXnRkoELC Enz74riGItBFpWitCSwYVlHlxNU3I7un3l7iOSqpQZ7oOUDG51bkhAMNXY9A JYqF0HqwRw NbH YYWIodKZK3RNqV0JMrUrUK6MJtg

Let vector be inclined to  URE6VJikaoLtFVxKnYHS4JBVXlgyOh8rNx bNdG50SzxRRYfQMhmYOoaRQrYoYZCrV8RDK64f munO523w BHovj1L6dS SXpskbHLJL8IgCk7CUEkTy8PICd cOO4A1CLqDAVc at angles respectively. Then, we have:

cInulzOGjr9ybXkTBlUFtTTyq0M yN tDaV3zoPSMkCEEazVAm LRsNOSPsWOODxzMPzn1XRYN0pIrzVdV8eI9T13uBDmpG6gmF4fKTAGLrVDBSnW 4FLaeCtwANtvURSsEyS g

Now, as , .jlMk1kPZ7AlBc zUeFeNLiXOBveSU5nsJXiDu wpH9Mib4PvbLWvi2Jv GUJE5t8H7GOdmKg8KnfggZbk2NgJrpdq0J PRR0YxnGwbXd9wxLCNSbj7EBXXdZ 5BYtNYmj0FKLnwcurdmVai53z5G3NdLaQS61RbdpMEGB5jzLXM4iZWRVzDT4 cI4srsJhK0KZAOiGT3h H P8BvyBlLr8UndQEFCGjyiO4SWeMiBCEI46p1jT CHsw1PgvAM4e6mIf0 nNHVbFVT4

Hence, the vector is equally inclined

*Question 15:

Prove that , if and only if are perpendicular, given

Answer:

*Question 16:

If θ is the angle between two vectors and LG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0, then uE7kfAWNybmzacB5X8UETx9VFVISy9HbItJvOFiAprkV2N8qZbL56A8kfjWOGJjdW0ydz0WMBNVLY only when

(A) (B) (C) InBvHP8Dt9w4kBNvPDPgkAg8y 3aIIZLibjOaoeTXbInOqBHE8HXcamiK1UhSj887TDFOiE2IeSk7A32X9h9FD6rnmI081Q cwGeC56YPEKD9Hd1nNQcDB K6YnWmz (D) UJYLLudEzzeO2VyiJsn8DaN0P2zTp9mLM6mXgc4fajkZOverG mpMe1Bo411R1RfVdksq e7Tvp 5VITL83nqM z b LJye IQfZj436skQZJRHW

Answer:

Let θ be the angle between two vectors andLG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0.

Then, without loss of generality, and LG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0are non-zero vectors so that

bDYVMMpvw6X1AU 44AYaZEf3zn4bgD9mW2KPSTN2q2HFqUM21h3ZzOQcCfoMtGOLhDI48shtNn9 z8btxL0ooAFPg 1Ry10SKUXXSd6kKTzJn0sOwhF8vXa0J9x1ZeDzpCk EEQ
juEtoki7g6tTsLtAkTjvOBAZ7Pe5dj8YWZxwZOTBYRJUx4p4kuMDVJYzwGMaY80j8ydTo2I61eJ dUJ2cAPgKH0UPfS6kGky30sNDCm6Bpvi1t0ggSUd3uwPeUUp0Ncl9EXyplA
qEwkPlrBI9uhcP1ZU2APSDpPrLGdhMiXy1490Qg2RmjgWvdPfK1mjHcmKEBnJYvQSm1rz4W fIsbGl2wX6oKqIHu4P48J 31 p2BM2t6QLqic 65Z3E37tPTPjTC2UqCfQ PwFo

Hence when The correct answer is B.

*Question 17:

Let and LG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0 be two unit vectors and θ is the angle between them. Then n BLxRCD zs Ekv7OpUL8sEbJVpeF2I0yd3mOTS1VBA7WMw6N8AAeZoiYPDE8Ix9Xfq17QtWVEdq3y jFQex7pcm5r2rV9U8S6EfWNPBNcPEudPoQoQZ9TuIswD is a unit vector if

(A) (B)

(C) (D)

Answer:

Let R2SR6fJEZ7YnxPZ4xz3tzJI1hXkiF8 it3aWPqpN1tFJ5t3SRT4OU pkBUF6CTNJaSrKCmkqIrWrGCbne4qvWcY3EqTnm f JbaAONZb and  LG4Rx6yElHTuEasYxr0ifr153opaTqbSXkyDYylDUUme lJuyqiDefsDxHzdE2HT7cQTcZ8o8bKYpWQ vcp7qH736aFku18 gQZU5eYxHbmVwADp5fyZfQCcty s9bxIsXn Bv0 be two unit vectors and θ be the angle between them.

Then,

VKj6J2ISdZKqfL66oCyBcWaHIi zGTXkdi2RsKxZn4i6W74skKDE3H y9Fen8W70JhkC7SUyCh59 pRMsygGr0ka5pZp n G aciQCTqQ2lCjyf6Sa02ntwVJufj7BEuG709Y

Now, n BLxRCD zs Ekv7OpUL8sEbJVpeF2I0yd3mOTS1VBA7WMw6N8AAeZoiYPDE8Ix9Xfq17QtWVEdq3y jFQex7pcm5r2rV9U8S6EfWNPBNcPEudPoQoQZ9TuIswD is a unit vector if

7SUr th Ip5WZEp6hG9cKrg88Th D83GXNKE5iS1TphXG9Mk0lYSUiMiLQ5rrWYcGoEjwbfPSfmR wiV9q6ySH9gNrQoWKuG7FKE5NQKZrU17HHlDkVrEddE0n0ly7 Vvp5oGdo

Hence, is a unit vector if  .  The correct answer is D.n BLxRCD zs Ekv7OpUL8sEbJVpeF2I0yd3mOTS1VBA7WMw6N8AAeZoiYPDE8Ix9Xfq17QtWVEdq3y jFQex7pcm5r2rV9U8S6EfWNPBNcPEudPoQoQZ9TuIswD

*Question 18:

The value of is

(A) 0 

(B) –1 

(C) 1 

(D) 3

Answer:
W9BsDoltjd8Yzj0XUFe5ROktZgA9hdlQnhxLF5LfnwOsURNNoVQ7sI7JpC1xj HNOoOgnekEmFMp56u6lItppKk5I3tw0qIckirGw3sAMILpAf1NyFyJ7J8 j09yXb2LoxemN o

The correct answer is C.

*Question 19:

If θ is the angle between any two vectors and , then when θ is equal to

(A) 0 (B) (C) (D)

Answer:

cqIqCYs FbyOyxvRpUgLSKabowlhioyFxfhEKNRkBu4kzx6ia2Nm2Xq9dyIO9JWsMVZJ44kj7 r r4qGw58hqhJj ZOrVDPWayEjp3jmMQjIegWF4U sZPAGb8ASZNIRviOhBQE
S4dRVB 5 RTRWU9MmFwSKOcXK7Mi2mta7binhb JAXTaEOEZLOpLj0WUK jfnBJuyg9CiRHE7GVw4iw0Vs8nippGFOdqmllsZDkhq7GtV5HH2fXwqtGZwvh1oHVd7sUWwC7gyv8
tqie3a38Z7JaMKPH5qp5CiYDBtZ39DvaBt vYiRZifZbpdmSHofJE5bEpqCTnElT3vGq7UeCyWgDsIVizgHCkFmTwVkbxcNEz0 PVrlKc 1XZhcc5PYA 3Aj7 8XjJIBR SIEW8
L7IAU2 Oj7M2 Eu83YdCsQmTDfga95AS41nnyFSTWKfJ8KtWflaARUTVnHK45ARLPPYc5MwQ0mrns1KPM0hfcrMH6KJ58igWYcFpn6GwWVAGEhKT rjb1Vl1Ef1YFOvcANem7og
xJi7x5D2BGhBU IBpqwyYyylqthOT4FalxtckuOE7l9YWhAIejoyP8ABO9AlLIiuhpT8THqQ9

Conclusion

Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10, “Vector Algebra,” is a comprehensive study material designed to help students understand the fundamental concepts and principles of vector algebra. The solutions provide step-by-step explanations and solved examples that help students to develop a deeper understanding of the subject. The chapter covers a range of topics including scalar and vector products of two and three-dimensional vectors, triple scalar product, and vector triple product. With the help of these solutions, students can improve their problem-solving skills and gain the confidence to tackle complex vector algebra problems. Swastik Classes’ NCERT solutions are designed in accordance with the latest CBSE syllabus, making them useful for students preparing for board exams or competitive exams like JEE and NEET. Overall, Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10 is an excellent resource for students who want to excel in mathematics and build a strong foundation in vector algebra.

The finest thing the NCERT answers Chapter 10 vector algebra is that it conveys tough parts in vernacular and easy language so that students of all intellect levels can understand them.

Topics included on NCERT Class 12 Maths Chapter 10

Section no.Topics
10.1Introduction to Vectors
10.2Some Basic Concepts
10.3Types of Vectors
10.4Addition of Vectors
10.5Multiplication of a Vector by a Scalar
10.6Product of Two Vectors

Weightage of Math Class 12 Chapter 10 in CBSE Exam

ChaptersMarks
Vector Algebra5 Marks

Related Links

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NCERT Solution for Class XIIth Maths Chapter 13 ProbabilityNCERT Solution for Class XIIth Maths Chapter 2 Inverse Trigonometry
NCERT Solution for Class XIIth Maths Chapter 1 Relations and FunctionNCERT Solution for Class XIIth Maths Chapter 3 Matrices

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FAQS on NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

What is meant by vector algebra?

an algebra in which the components involved might be vectors and the assumptions and rules are based on vector behaviour.

How many exercises are in vector algebra?

5 exercises

What are 3 types of vectors?

  • Null Vector or Zero Vector

When the magnitude of a vector is 0 and the starting point and terminus of the vector are the same, the vector is said to be a Zero Vector. PQ, for example, is a line segment in which the coordinates of point P and point Q are the same. The symbol for a zero vector is 0. There is no fixed orientation for the zero vector.

  • Vector of a unit

When the magnitude of a vector is 1 unit in length, it is said to be a unit vector. If x is a vector of magnitude x, then the unit vector is denoted by x in the vector’s direction and has a magnitude equal to 1.

However, two-unit vectors cannot be equal since their directions may differ.

  • Travelling position

A position vector is defined as a point X in the plane. It just indicates the current location. Assume OX is a point in a plane with respect to its origin. If O is used as the reference origin and X is an arbitrary point in the plane, the vector is referred to as the point’s position vector.

What is an equal vector?

When the magnitude and direction of two or more vectors are the same, they are said to be equal.

What is the triangular law of addition?

The triangle rule of vector addition asserts that when two vectors are represented as two sides of a triangle with the same order of magnitude and direction, the magnitude and direction of the resulting vector is represented by the third side of the triangle.

What is the resultant of vector?

The vector sum of two or more vectors is the outcome. It’s the outcome of combining two or more vectors. When you put the displacement vectors A, B, and C together, you obtain vector R. Vector R may be found using an appropriately drawn, scaled vector addition diagram, as illustrated in the picture.

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