{"id":2893,"date":"2026-01-19T15:57:07","date_gmt":"2026-01-19T10:27:07","guid":{"rendered":"https:\/\/www.manabadi.co.in\/boards\/?p=2893"},"modified":"2026-01-19T15:57:10","modified_gmt":"2026-01-19T10:27:10","slug":"ts-inter-1st-year-maths-1a-solutions-chp-4-addition-of-vectors-ex-4a","status":"publish","type":"post","link":"https:\/\/www.manabadi.co.in\/boards\/ts-inter-1st-year-maths-1a-solutions-chp-4-addition-of-vectors-ex-4a\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)"},"content":{"rendered":"\n

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)<\/h2>\n

TS Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a)<\/h2>\n

I.
\nQuestion 1.
\nABCD is a parallelogram. If L and M are the middle points of BC CD respectively then find
\ni) AL and AM in terms of AB and AD
\nii) \u03bb, if AM = \u03bb AC\u2212AL(V.S.A)<\/h3>\nAnswer:\n\nABCD is a parallelogram and hence AB=DC and BC=AD\nWe have BC = 12BC\n(\u2235 L is the mid point of BC)\n= 1\/2 AD (\u2235 BC = AD)\n\n

Question 2.
\nIn AABC, P, Q and R are the mid points of the sides AB, BC and CA respectively. If D is any point
\ni) Then express DA+DB+DC in terms of DP,DQ and DR.
\nii) If PA+QB+RC=α, then find a.(V.S.A)<\/h3>\nAnswer:\nLet D be the origin.\n\nand DA\u00af=a , DB=b and DC=c\nP.V. of P is mid point of AB=DP=a+b\/2\nP.V. of Q is mid point of BC=DQ\u00af=b+c\/2\nP.V. of R is mid point of AC=DR\u00af=c+a\/2\n\n

Question 3.
\nLet a\u0305 = i\u0305 + 2 j\u0305 + 3k\u0305 and b\u0305 = 3 i\u0305 + j\u0305. Find the unit vector in the direction of a+b. (V.S.A)<\/h3>\nAnswer:\nUnit vector in the direction of a+b is\n\n

Question 4.
\nIf the vectors -3i\u0305 + 4j\u0305 + \u03bbk\u0305 and \u03bci\u0305 + 8j\u0305 + 6k\u0305 are collinear vectors then find \u03bb and \u03bc. (May 2014, \u201912, Mar. \u201914)<\/h3>\nAnswer:\nIf the vectors a1 i\u0305 + b1 j\u0305 + c1k\u0305 and a2 i\u0305 + b2 j\u0305 + c2k\u0305 are collinear then\n\n

Question 5.
\nABCDE is a pentagon. If the sum of the vectors AB,AE,BC,DC,ED and AC is \u03bb AC , then find the value of \u03bb. (S.A)<\/h3>\nAnswer:\n\nGiven ABCDE is a pentagon and\n\n

Question 6.
\nIf the position vectors of the points A, B and C are -2i\u0305 + j\u0305 \u2013 k\u0305, -4i\u0305 + 2j\u0305 + 2k\u0305 and 6i\u0305 \u2013 3j\u0305 \u2013 13k\u0305 respectively and AB = \u03bbAC then find the value of \u03bb. (March 2011) (S.A)<\/h3>\nAnswer:\nLet O be the origin and given\n\n

Question 7.
\nIf OA\u00af=i\u00af+j\u00af+k; AB=3i\u00af\u22122j\u00af+k, BC=i\u00af+2j\u00af\u22122k and CD=2i\u00af+j\u00af+3k then find the vector OD\u00af. (March 2013) (V.S.A)<\/h3>\nAnswer:\nSince OA\u00af+AB+BC+CD=OD\u00af\n\u21d2 OD\u00af = (i\u0305 + j\u0305 + k\u0305) + (3i\u0305 \u2013 2j\u0305 + k\u0305) + (i\u0305 + 2 j\u0305 \u2013 2k\u0305) + (2 i\u0305 + j\u0305 + 3k\u0305)\n= 7i\u0305 + 2j\u0305 + 3k\u0305\n

Question 8.
\nIf a\u0305 = 2i\u0305 + 5j\u0305 + k\u0305 and b\u0305 = 4i\u0305 + mj\u0305 + nk\u0305 are collinear vectors then find m and n. (May 2011) (V.S.A)<\/h3>\nAnswer:\nSince a\u0305 = 2i\u0305 + 5j\u0305 + k\u0305 and\nb\u0305 = 4i\u0305 + mj\u0305 + nk\u0305 are collinear\n\u21d2 2\/4=5\/m=1\/n\n\u21d2 1\/2=5\/m and 1\/2=1\/n \u21d2 m = 10 and n = 2\n

Question 9.
\nLet a\u0305 = 2i\u0305 + 4j\u0305 \u2013 5k\u0305, b\u0305 = i\u0305 + j\u0305 + k\u0305 and c\u0305 = i\u0305 + 2k\u0305. Find the unit vector in the opposite direction of a\u0305 + b\u0305 + c\u0305. (March 2015-A.P)(May 2012; Mar. \u201904, \u201912; Board Model Paper) (V.S.A)<\/h3>\nAnswer:\n\n

Question 10.
\nIs the triangle formed by the vectors 3i\u0305 + 5j\u0305 + 2k\u0305, 2i\u0305 \u2013 3j\u0305 \u2013 5k\u0305 and -5i\u0305 \u2013 2 j\u0305 + 3k\u0305 equilateral ? (V.S.A)<\/h3>\nAnswer:\nLet ABC be the triangle with AB = 3i\u0305 + 5 j\u0305 + 2k\u0305\nBC = 2i\u0305 \u2013 3 j\u0305 \u2013 5k\u0305\nCA = -5i\u0305 \u2013 2j\u0305 + 3k\u0305\n\n\u2234 The given vectors formed on equilateral triangle.\n

Question 11.
\nIf α, β and \u03b3 are the angles made by the vector 3i\u0305 \u2013 6j\u0305 + 2k\u0305 with the positive directions of the coordinate axes then find cos α, cos β, cos \u03b3. (S.A)<\/h3>\n

Answer:
\nUnit vectors along the coordinate axes are respectively i\u0305, j\u0305, k\u0305
\nLet p\u0305 = 3i\u0305 \u2013 6j\u0305 + 2k\u0305<\/p>\n\n

Question 12.
\nFind the angles made by the straight line passing through the points (1, -3, 2) and (3, -5, 1) with the coordinate axes. (S.A)<\/h3>\nAnswer:\nLet the vectors along the coordinate axes be i\u0305, j\u0305, k\u0305 respectively. Let O be the origin and the points A(1, -3, 2) and B(3, -5, 1).\ni. e. OA = i\u0305 \u2013 3 j\u0305 + 2k\u0305, OB\u00af = 3 i\u0305 \u2013 5 j\u0305 + k\u0305\nAB\u00af=OB\u00af\u2212OA = (3i\u0305 \u2013 5j\u0305 + k\u0305) \u2013 (i\u0305 \u2013 3j\u0305 + 2k\u0305) = 2i\u0305 \u2013 2j\u0305 \u2013 k\u0305\nLet α be the angle between AB\u00af and i\u0305 then\n\n

II.
\nQuestion 1.
\nIf a\u0305 + b\u0305 + c\u0305 = αd\u0305, b\u0305 + c\u0305 + d\u0305 = βa\u0305 and a\u0305, b\u0305, c\u0305 are non-coplanar vectors, then show that a\u0305 + b\u0305 + c\u0305 + d\u0305 = 0\u0305. (S.A)<\/h3>\n

Answer:
\nGiven a\u0305 + b\u0305 + c\u0305 = αd\u0305 \u2026\u2026\u2026\u2026\u2026. (1)
\nb\u0305 + c\u0305 + d\u0305 = βa\u0305 \u2026\u2026\u2026\u2026\u2026.. (2)
\nFrom (2), d\u0305 = pa\u0305 \u2013 b\u0305 \u2013 c\u0305
\nFrom (1), a\u0305 + b\u0305 + c\u0305 = a, (pa\u0305 \u2013 b\u0305 \u2013 c\u0305)
\n\u21d2 (1 \u2013 αβ)a\u0305 + (1 + a)b\u0305 + (1 + a)c\u0305 = 0
\n\u2234 a\u0305, b\u0305, c\u0305 are non coplanar vectors
\n1 \u2013 αβ = 0 \u21d2 αβ = 1 and
\n1 + α = 0 \u21d2 α = -l β = -1
\nHence from (1); a\u0305 + b\u0305 + c\u0305 = -d\u0305
\n\u21d2 a\u0305 + b\u0305 + c\u0305 + d\u0305 = 0<\/p>\n

Question 2.
\na\u0305, b\u0305, c\u0305 are non coplanar vectors. Prove that the following four points are coplanar.
\ni) -a\u0305 + 4b\u0305 \u2013 3c\u0305, 3a\u0305 + 2b\u0305 \u2013 5c\u0305 (May,\u201914,\u201912)
\n-3a\u0305 + 8b\u0305 \u2013 5c\u0305, \u2013 3a\u0305 + 2b\u0305 + c\u0305<\/h3>\n

Answer:
\nLet 0 be the origin and A, B, C, D are the four points given by
\nOA = -a\u0305 + 4b\u0305 \u2013 3c\u0305, OB = 3a\u0305 + 2b\u0305 \u2013 5c\u0305
\nOC = -3a\u0305 + 8b\u0305 \u2013 5c\u0305, OD = -3a\u0305 + 2b\u0305 + c\u0305
\nAB\u00af=OB\u00af\u2212OA = (3a\u0305 + 2b\u0305 \u2013 5c\u0305) \u2013 (-a\u0305 + 4b\u0305 \u2013 3c\u0305)
\n= 4a\u0305 \u2013 2b\u0305 \u2013 2c\u0305
\nAB\u00af=OB\u00af\u2212OA = (-3a\u0305 + 2b\u0305 \u2013 5c\u0305) \u2013 (3a\u0305 + 2b\u0305 \u2013 5c\u0305)
\n= -6a\u0305 \u2013 4b\u0305 + 3c\u0305
\nAC\u00af=OC\u2212OA = (-3a\u0305 + 8b\u0305 \u2013 5c\u0305) \u2013 (-a\u0305 + 4b\u0305 \u2013 3c\u0305) = -2a\u0305 + 4b\u0305 \u2013 2c\u0305
\nAD=OD\u2212OA = (3a\u0305 + 2b\u0305 + c\u0305) \u2013 (-a\u0305 + 4b\u0305 \u2013 3c\u0305) = -2a\u0305 \u2013 2b\u0305 + 4c\u0305
\nLet a vector be expressed as a linear combination of other two.
\nSuppose AB\u00af = x(AC\u00af) + y (AD) where x, y are scalars.
\n\u2234 4a\u0305 \u2013 2b\u0305 \u2013 2c\u0305 = x (-2a\u0305 + 4b\u0305 \u2013 2c\u0305) + y(-2a\u0305 \u2013 2b\u0305 + 4c\u0305)
\n\nComparing coefficients of a\u0305, b\u0305, c\u0305 we get
\n(\u2235 a\u0305, b\u0305, c\u0305 are non coplanar vectors)
\n-2x \u2013 2y = 4 \u2026\u2026\u2026\u2026\u2026(1)
\n4x \u2013 2y = -2 \u2026\u2026\u2026\u2026\u2026(2)
\n-2x + 4y = -2 \u2026\u2026\u2026\u2026.(3)
\nSolving (1) and (2) we get 2x + 2y = \u2013 4 and 4x \u2013 2y = \u2013 2
\n6x = \u2013 6 \u21d2 x = -1
\nx + y = -2 \u21d2 y = -1
\nx = \u2013 1 and y = -1 satisfy equation (3).
\n\u21d2 A, B, C, D are coplanar and
\nAB\u00af,AC\u00af,AD are coplanar.
\nand AB\u00af,AC\u00af,AD are coplanar.
\n\u2234 The given points A, B, C, D are coplanar.
\n\nii) 6a\u0305 + 2b\u0305 \u2013 c\u0305, 2a\u0305 \u2013 b\u0305 + 3c\u0305, -a\u0305 + 2b\u0305 \u2013 4c\u0305, -12a\u0305 \u2013 b\u0305 \u2013 3c\u0305
\nAnswer:
\nLet O be the origin and A, B, C, D be the given points.
\nOA = 6a\u0305 + 2b\u0305 \u2013 c\u0305, OB\u00af = 2a\u0305 \u2013 b\u0305 + 3c\u0305
\nOC = -a\u0305 + 2b\u0305 \u2013 4c\u0305, OD = -12a\u0305 \u2013 b\u0305 \u2013 3c\u0305
\n\u2234 AB\u00af=OB\u00af\u2212OA
\n= (2a\u0305 \u2013 b\u0305 + 3c\u0305) \u2013 (6a\u0305 + 2b\u0305 \u2013 c\u0305)
\n= \u2013 4a\u0305 \u2013 3b\u0305 + 4c\u0305
\nAC\u00af=OC\u2212OA = (-a\u0305 + 2b\u0305 \u2013 4c\u0305) \u2013 (6a\u0305 + 2b\u0305 \u2013 c\u0305) = -7a\u0305 \u2013 3c\u0305
\nAD=OD\u2212OA = (-12a\u0305 \u2013 b\u0305 \u2013 3c\u0305) \u2013 (6a\u0305 + 2b\u0305 \u2013 c\u0305)= -18a\u0305 \u2013 3b\u0305 \u2013 2c\u0305
\n\u2234 Let a vector be expressed as a linear combination of other two.
\nSuppose AB\u00af = xAC\u00af + yAD
\n\u21d2 -4a\u0305 \u2013 3b\u0305 + 4c\u0305 = x(-7a\u0305 \u2013 3c\u0305) + y(-18a\u0305 \u2013 3b\u0305 \u2013 2c\u0305)
\nComparing coefficients of a\u0305, b\u0305, c\u0305 since a\u0305, b\u0305, c\u0305 are non coplanar,
\n-7x \u2013 18y = \u2013 4 \u2026\u2026\u2026\u2026(1)
\n-3y = -3 \u21d2 y = 1 \u2026\u2026\u2026.(2)
\n\u2234 -7x \u2013 18 = \u2013 4 \u21d2 \u2013 7x = 14 \u21d2 x = -2
\nComparing coefficient of c,
\n-3x \u2013 2y = 4 \u2026\u2026\u2026..(3)
\nx = \u2013 2 and y = 1 satisfy equation (3)
\nand hence A, B, C, D are coplanar.
\n

Alternate Method For Above Problem :
Use scalar triple product of vectors <\/span>A<\/span>B<\/span><\/span><\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>,<\/span>A<\/span>C<\/span><\/span><\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span>\u00af<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/nobr><\/span>