10.1 Overview 10.1.1 A quantity that has magnitude as well as direction is called a vector. G10.1.2 The unit vector in the direction of aa10.1.3 Position vector of a point P (x, y, z) is given as JJJGˆ ˆ ˆ and itsOP = x+ jzkiy+ JJJG 2 22magnitude as |OP | = x+ yz , where O is the origin.+ 10.1.4 The scalar components of a vector are its direction ratios, and represent its projections along the respective axes. 10.1.5 The magnitude r, direction ratios (a, b, c) and direction cosines (l, m, n) of any vector are related as: a bcl= , m= , n= . r rr and is represented by GGais given by ||a . G GGGGG10.1.6 The sum of the vectors representing the three sides of a triangle taken in order is 0 10.1.7 The triangle law of vector addition states that “If two vectors are represented by two sides of a triangle taken in order, then their sum or resultant is given by the third side taken in opposite order”. 10.1.8 Scalar multiplication If ais a given vector and λ a scalar, then λ ais a vector whose magnitude is |λ a is same as that of aif λ is positive and, opposite to that of a | = |λ| GG| a|. The direction of λ a λ is negative. if 10.1.9 Vector joining two points If P1 (x1, y1,z1) and P2 (x2, y2,z2) are any two points, then JJJJG PP = (x − xiˆ + ( y − y ˆj + (z − z ˆ) )) k1221 21 21 JJJJG 2 22|PP |= (x − x ) + ( y − y ) + (z − z )12 21 2121 10.1.10 Section formula The position vector of a point R dividing the line segment joining the points P and Q whose position vectors are a and b GGG mbG na (i) in the ratio m : n internally, is given by mn GGmb – na (ii) in the ratio m : n externally, is given by –mn GG a .GbGa G along bG10.1.11 Projection of a G⎛ along b is and the Projection vector ofG||b GG a . b b.⎜⎟ ||b⎝⎠ ⎞ is 10.1.12 Scalar or dot product GThe scalar or dot product of two given vectors a them is defined as Gand band b having angle θ between them is given as having an angle θ between GGG = | a G| | b | cos θ. ba 10.1.13 Vector or cross product GGGGGThe cross product of two vectors a = | a | sin θ nˆ, Ga × b | | b G and bGGand a are two vectors and λ is nˆwhere nˆ is a unit vector perpendicular to the plane containing a Gform a right handed system. , b, G If a any scalar, then Gˆ ˆaiˆ aj ˆ ak ˆˆbi bjbk 10.1.14and b+ + + += = 1 2 3 1 2 3 GGˆ bj ˆ bk ˆ+ b (abi ) () (3 )a + + + + += a a1 1 2 2 3 λ a GG ˆˆjkˆ(λai) (λ2 ) (λ3 )+ += a a1 G . b= ab+ ab+ ab2 33a 1 12 iˆˆjkˆ GGabc111 = (bc – bc) iˆ + (ac – cc)ˆj + (ab – ab) ˆ122121121b21k× ba = abc222 G and bis given by G GG Angle between two vectors a a. b +ab 3ab11 +ab 22 3 cos θ = =2 2 2222++ b1 ++ b2 b3a1 a2 a3 GG|||| ab 10.2 Solved Examples Short Answer (S.A.) GExample 1 Find the unit vector in the direction of the sum of the vectors ˆGˆ2 iˆˆj−+kˆˆji++kand b2 3a= = – . GGSolution Let c (2 Gdenote the sum of a ˆk and b. We have kˆ = iˆ +kˆG5ˆˆji−+ iˆˆj−++2 )( 3 )c += NowG c =|| G c 1 15  c ==G (i +5k )= i + kThus, the required unit vector is .26 2626c JJJG Example 2 Find a vector of magnitude 11 in the direction opposite to that of PQ , where P and Q are the points (1, 3, 2) and (–1, 0, 8), respetively. Solution The vector with initial point P (1, 3, 2) and terminal point Q (–1, 0, 8) is given by JJJG PQ= (– 1 – 1) iˆ+ (0 – 3) ˆj + (8 – 2) kˆ = – 2 iˆ– 3 ˆj + 6 kˆ JJJGJJJG = – ˆ3ˆˆThus Q P PQ = 2i +− j 6 k JJJG 22 2⇒|QP| =2 3 (–6) =++ 4936 ++ = 49 =7 JJJGTherefore, unit vector in the direction of QP is given by JJJG m Q P 2 iˆ 3ˆj 6 kˆ QP JJJG | Q P| 7 JJJGHence, the required vector of magnitude 11 in direction of QP is 2iˆ 3ˆj 6 kˆ 22 33 66ˆˆ ˆ–11 m = 11 = i + k .QP7 j77 7 Example 3 Find the position vector of a point R which divides the line joining the twoJJJG JG JG JJJG JG JG points P and Q with position vectors OP 2 ab and OQ a –2 b , respectively, in the ratio 1:2, (i) internally and (ii) externally. Solution (i) The position vector of the point R dividing the join of P and Q internally in the ratio 1:2 is given by JGJG JG JG JGJJJJG 2(2 ab ) 1( a –2b )5a OR .12 3 (ii) The position vector of the point R′ dividing the join of P and Q in the ratio 1 : 2 externally is given by JG JG JJJJG +− 2(2 ab ) 1( a −2 b) JG JG OR′= =3a +4 b .21− Example 4 If the points (–1, –1, 2), (2, m, 5) and (3,11, 6) are collinear, find the value of m. Solution Let the given points be A (–1, –1, 2), B (2, m, 5) and C (3, 11, 6). Then JJJJGA B (2 1) iˆ(m 1) ˆj (5 – 2) kˆ = 3 iˆ ++ + m 1) ˆj ˆ=+ +++ (3k JJJJGand A C (3 1) i (11 1) ˆ (6 2)ˆ 4 i +12 ˆ 4ˆ=+ +++−ˆ jk =ˆ j +k . JJJGJJJJG Since A, B, C, are collinear, we have AB = λ A C , i.e., (3 iˆ(m 1) ˆj 3 ) ˆ iˆˆjkˆk λ(4+12 +4) ⇒ 3 = 4 λ and m + 1 = 12 λ Therefore m = 8. JGπExample 5 Find a vector r of magnitude 32units which makes an angle of and4 π with y and z - axes, respectively. 2π 1 π =Solution Here m = cos and n = cos = 0.42 2Therefore, l2 + m2 + n2 = 1 gives 1l2 + + 0 = 12 1 ⇒ l = ±2 JG Hence, the required vector r= 32 ˆ + ˆj nkˆ) is given by(li m+JG JG( 1ˆ 1ˆj 0 kˆ) = r= ±3iˆ +3ˆj.ir= 32 22 JGi− jk,ˆExample 6 If JG a= 2ˆˆ + ˆ b= iˆ+ ˆj– 2kˆ and JG c= iˆ+ 3ˆj– k, find λ such that GG G ais perpendicular to bc. Solution We have Gλ JGb+ c= λ ( ˆ+ ˆj– 2 kˆ ) + ( ˆ+ 3ˆj– kˆ)i i= ( λ + 1) iˆ+ (λ + 3 )ˆj– (2λ + 1) kˆJGJGG GGSince a G ⊥ (λ b+ c), a.(λ b+ c) = 0 ⇒(2 iˆ – ˆj+ kˆ) . [( λ + 1) iˆ + (λ + 3) ˆj– (2λ + 1) kˆ] = 0 ⇒2 (λ + 1) – (λ + 3) – (2λ + 1) = 0 ⇒λ = – 2. Example 7 Find all vectors of magnitude 10 3 that are perpendicular to the plane i 2ˆˆ ˆ jkˆof ˆjkand i3ˆ 4. JGSolution Let a G = iˆ2ˆjkˆ and b= iˆ 3ˆjk4 ˆ . Then iˆˆjk ˆ GGab 121 iˆ(8 3) ˆj(4 1) kˆ(3 2) = 5 iˆ – 5ˆj+ 5 kˆ–13 4 GG 222 2ab (5) (5) (5) 3(5) 53.⇒ JG Therefore, unit vector perpendicular to the plane of a G and bis given by Gab G iˆjk 5ˆ55ˆ GG 53ab JGHence, vectors of magnitude of 10 3 that are perpendicular to plane of a G and b 10 3 iˆjk ˆ55ˆ5 are 53 , i.e., 10(ˆˆ ˆ).ijkLong Answer (L.A.) Example 8 Using vectors, prove that cos (A – B) = cosA cosB + sinA sinB. Solution Let mOQ be unit vectors making angles A and B, respectively, with OP and mpositive direction of x-axis. Then ∠QOP = A – B [Fig. 10.1] JJJG JJJG We know OPm = icosA+ ˆ and mON + NQ iˆˆJJJJG JJJG OM + MP ˆ jsinA OQ= cosB + jsin B. mn mOP. OQ OP OQ cos A-BBy definition m m m1 OQ = cos (A – B) ... (1) ∵OP In terms of components, we have mmˆ ˆ ˆˆOP.OQ =(cosA ji jsin B) i sinA).( cosB = cosA cosB + sinA sinB ... (2) From (1) and (2), we get cos (A – B) = cosA cosB + sinA sinB. sinA sinB sinC Example 9 Prove that in a Δ ABC, , where a, b, crepresent theabc magnitudes of the sides opposite to vertices A, B, C, respectively. Solution Let the three sides of the triangle BC, CA and AB be represented by GGG,andc, respectively [Fig. 10.2].ab GGGG GG GWe have abc0. i.e., ab c Gwhich pre cross multiplying by a, and Gpost cross multiplying by b, gives GGGG ab c a×=× GGGGand abbcrespectively. Therefore, GGGGGG abbccaG GG G GG ab bc ca ⇒ G GG G GGabsin( –C) bcsin( –A) casin( –B) ⇒ ⇒ absin C = bcsinA = casinB Dividing by abc, we get sinC sinA sinB sinA sinB sinC i.e. cababc Objective Type Questions Choose the correct answer from the given four options in each of the Examples 10 to 21. Example 10 The magnitude of the vector ˆ ˆjkˆ is6i 23 (A) 5 (B)7 (C)12 (D) 1 Solution (B) is the correct answer. Example 11 The position vector of the point which divides the join of points withGG GGposition vectors aband 2abin the ratio 1 : 2 is G G GG G G32ab 5ab 4abG(A) (B) a (C) (D)3 33 Solution (D) is the correct answer. Applying section formula the position vector of the required point is G GGG GG2(ab) 1(2 ab )4ab 21 3 Example 12 The vector with initial point P (2, –3, 5) and terminal point Q(3, –4, 7) is (A) ˆˆ2ˆ (B) iˆj12 ˆij k 57ˆ k (C) ˆˆ2ˆ (D)ij k None of these Solution (A) is the correct answer. Example 13 The angle between the vectors ˆˆˆjkˆ isijand 25(A) (B) (C) (D)3 336 GG .abGGSolution (B) is the correct answer. Apply the formula cosθ = . .ab Example 14 The value of λ for which the two vectors 2ij kˆˆ ˆ and ˆ ˆˆ23i jk are perpendicular is (A) 2 (B)4 (C)6 (D) 8 Solution (D) is the correct answer. ˆˆExample 15 The area of the parallelogram whose adjacent sides are ikand 2ijkˆˆˆ is (A) 2 (B) 3 (C)3 (D) 4 Solution (B) is the correct answer. Area of the parallelogram whose adjacent sides 2are a G and b G is ˆabG . Example 16 If a G = 8, b 3 G and ab 12 , then value of GG .abG G is (A) 63 (B) 83 (C) 12 3 (D) None of these Solution (C) is the correct answer. Using the formula ab GG aG . b G |sinθ|, we get 6 πθ=± . Therefore, ab.G G = .cosab GG = 8 × 3 × 3 = 12 3 . Example 17 The 2 vectors ˆ + ˆ ˆ ˆ ˆ represents the two sides AB and jkand 3 ij k −+4 AC, respectively of a ΔABC. The length of the median through A is 34 48(B) (C) 18 (D) None of these(A) 22 JJJGSolution (A) is the correct answer. Median AD is given by JJJG 1 34 ˆˆ ˆAD = 3 ++5ij k = 2 2 GExample 18 The projection of vector a ijkalong bi 2ˆ2 is2ˆˆˆ Gˆ jk ˆ 21(A) (B) (C) 2 (D) 33 GGSolution (A) is the correct answer. Projection of a vector ab ison GG .(2ˆˆ ˆ).( ˆ 2ˆ 2) ˆ 2ab ijki jk G = = .144 3b GGExample 19 If aand bare unit vectors, then what is the angle between GG3abto be a unit vector? (A) 30° (B) 45° (C) 60° (D) Solution (A) is the correct answer. We have GG GGG 2 G22(3ab)3ab 23 a.b GG33⇒ . = cosθ =⇒θ = 30°.ab22 Example 20 The unit vector perpendicular to the vectors ijˆˆand ˆˆijright handed system is ˆˆij(A) kˆ(B) – kˆ(C) (D)2 ˆˆ ˆˆij ij Solution (A) is the correct answer. Required unit vector is ˆˆ ˆij ijˆ GGExample 21 If a3 and –1 k2 , then kalies in the interval (A) [0, 6] (B) [– 3, 6] (C) [ 3, 6] (D) 6 GG aand bfor 90° forming a ˆˆij 2 2kˆ ˆ= k.2 [1, 2] GSolution (A) is the correct answer. The smallest value of kawill exist at numerically GGka ka 03 0smallest value of k, i.e., at k= 0, which gives GThe numerically greatest value of kis 2 at which ka 6. 10.3 EXERCISE Short Answer (S.A.) 1. Find the unit vector in the direction of sum of vectors a ijkˆ and bG 2ˆˆ. G 2ˆˆjk GGˆˆˆ 2ˆˆˆ2. If aij kand 2, find the unit vector in the direction of2b ij k G GG(i)6 b (ii) 2ab JJJG 3. Find a unit vector in the direction of PQ , where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively. GG4. If aand b are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA. 5. Using vectors, find the value of ksuch that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear. G6. A vector r G is inclined at equal angles to the three axes. If the magnitude of ris G23 units, find r. 7. A vector r G has magnitude 14 and direction ratios 2, 3, – 6. Find the direction cosines and components of r G , given that r G makes an acute angle with x-axis. 2ˆˆ2ˆ ˆˆ3ˆ. 8. Find a vector of magnitude 6, which is perpendicular to both the vectors ijk and 4–ijk9. Find the angle between the vectors 2i jkˆˆˆ and 3iˆ 4ˆjkˆ. G GGGG G GG10. If abc0 , show that abbcca G . Interpret the result geometrically? 11. Find the sine of the angle between the vectors a G3ˆˆ2ˆ andij k b G 2ˆ 2ˆjk ˆ.i 4 12. If A, B, C, D are the points with position vectors ˆˆˆ , 2ij kˆˆ 3ˆ,JJJG23,3i kiˆ ˆ2 jkˆ , respectively, find the projection of AB i jkJJJG .ˆ ˆ along CD 13. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1). 14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area. Long Answer (L.A.) 22 2bc – a15. Prove that in any triangle ABC, cos A , where a, b, care the2bc magnitudes of the sides opposite to the vertices A, B, C, respectively. GGG16. If ab determine vertices a show,,c the of triangle, that GG1 GGGG 2 bccaab gives the vector area of the triangle. Hence deduce the GGcondition that the three points ,,cab G are collinear.Also find the unit vector normal to the plane of the triangle. G17. Show that area of the parallelogram whose diagonals are given by a G and bis GGab . Also find the area of the parallelogram whose diagonals are 2ijkˆˆˆ 2 and iˆ3ˆjkˆ. G G GG18. If a G = ˆˆbjk ˆ , find a vector c GG and ac3.ijkˆ and ˆ G such that acb .Objective Type Questions Choose the correct answer from the given four options in each of the Exercises from 19 to 33 (M.C.Q) 19. The vector in the direction of the vector iˆ 2ˆjk2 ˆ that has magnitude 9 is ˆ 2ˆ 2ˆi jk (A) i2ˆ 2ˆ (B)(C) 3(iˆ 2ˆ 2) ˆ (D) 9(iˆ 2ˆ 2) ˆˆ jk 3 jk jk G GG G20. The position vector of the point which divides the join of points 2ab 3and ab in the ratio 3 : 1 is G G GGG G3 2b 78 35a ab aa(A) (B) (C) (D) 2 444 21. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is (A) iˆ 12ˆ 4ˆ (B) iˆjk 4ˆjk 52ˆ (C) 52iˆˆjk ˆ (D) ˆˆˆ4 ijk GG22. The angle between two vectors aand bwith magnitudes 3 and 4, respectively, GGand . 3isab 2 5(A) (B) (C) (D)6 322 G23. Find the value of λ such that the vectors a G2iˆˆjk ˆ and biˆ 2ˆjk 3 ˆ are orthogonal 3 5(A) 0 (B)1 (C) (D)– 2 2 24. The value of λ for which the vectors ˆ ˆ ˆ ˆ ˆjkˆ are parallel is36i jk and 2 i4 2 352(A) (B) (C) (D)3 225 25. The vectors from origin to the points A and B are a G 23 jk i ,respectively, then the area of triangle OAB is iˆ ˆ 2ˆand b G 23ˆ ˆjk ˆ 1 229(A) 340 (B) 25 (C) 229 (D) 2 GGG 2 Gˆ226. For any vector a, the value of (aiˆ)2 (a ˆj)(ak ) is equal to (A) a G2 (B) 3a G2 (C) 4 a G2 (D) 2 a G2 G GGGGGb ab27. If a = 10, = 2 and ., then value of isab 12 (A) 5 (B)10 (C)14 (D) 16 28. The vectors ijˆˆ2,ˆ ˆ ˆjk ˆ ijki and2 ˆˆ kˆ are coplanar if (A) λ = –2 (B) λ = 0 (C) λ = 1 (D) λ = – 1 GGG GGGGG G GGG,, are unit vectors such that G , then the value of abbc ca 3 29. If abc abc 0 ... is (A) 1 (B) 3 (C) – (D) None of these2 GG30. Projection vector of aon b is GGGGGGGGG ab ab ˆab. .ab..GbG G G(A) 2 (B) (C) (D) 2 bb a ab GGG Gb 3,31. If GGare three vectors such that GG GGand a 2, c 5,,,ab c 0ab c GGGGGGthen value of . . . isabbc ca(A) 0 (B) 1 (C) – 19 (D) 38 GG32. If a 4 and 3 2 , then the range of a is (A) [0, 8] (B) [– 12, 8] (C) [0, 12] (D) [8, 12] 33. The number of vectors of unit length perpendicular to the vectors a G2iˆˆj2ˆ=++k Gˆˆand bjk is=+ (A) one (B) two (C) three (D) infinite Fill in the blanks in each of the Exercises from 34 to 40. GGG34. The vector a+ b bisects the angle between the non-collinear vectors aand Gbif ________ GGG GG35. If . 0, rb G 0, and rc 0 for some non-zero vector r G, then the value ofra .. G.( GG) is ________ab cG36. The vectors a 32ij 2kˆ and G m are the adjacent sides of ab –ik 2 parallelogram. The acute angle between its diagonals is ________. 37. The values of kfor which are _______. 38. The value of the expression G G2 2G G39. If = 144 andab a.b GG G G 1 Gka aand ka 2 ais parallel to a Gholds true G2 G GG +(.)ab2 is _______.ab× GG ba 4 , then is equal to _______. aii ajj a.kk 40. If a is any non-zero vector, then (. ) ˆˆ .ˆ ˆ ˆˆ State True or False in each of the following Exercises. G GG Ga b41. If , then necessarily it implies ab. GG Gequals _______. 42. Position vector of a point P is a vector whose initial point is origin. G GG G Gab ab 43. If , then the vectors a Gand bare orthogonal. G GG GG 2 G22 G44. The formula (ab) ab 2ab is valid for non-zero vectors a Gand b. GGG45. If a Gand bare adjacent sides of a rhombus, then a. b= 0.

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