{"id":6804,"date":"2024-01-16T12:39:34","date_gmt":"2024-01-16T07:09:34","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=6804"},"modified":"2024-01-17T09:31:04","modified_gmt":"2024-01-17T04:01:04","slug":"ts-inter-1st-year-maths-1a-solutions-chapter-4-ex-4b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-solutions-chapter-4-ex-4b\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)"},"content":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions<\/a> Chapter 4 Addition of Vectors Ex 4(b) to find a better approach to solving the problems.<\/p>\n

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

I.
\nQuestion 1.
\nFind the vector equation of the line passing through the point 2i\u0305 + 3j\u0305 + k\u0305 and parallel to the vector 4i\u0305 – 2j\u0305 + 3k\u0305.(March 2015-A.P) (May, March ’01) (V.S.A)
\nAnswer:
\nLet a = 2i\u0305 + 3j\u0305 + k\u0305 and b = 4i\u0305 – 2 j\u0305 + 3k\u0305
\nThe vector equation of the line passing through the point a\u0305 and parallel to the vector b\u0305 is
\nr\u0305 = a\u0305 + tb\u0305 where t is a scalar.
\nr\u0305 = (2i\u0305 + 3j\u0305 + k\u0305) + t (4i\u0305 – 2j\u0305 + 3k\u0305)
\n\u21d2 r\u0305 = (2 + 4t) i\u0305 + (3 – 2t) j\u0305 + (1 + 3t) k\u0305<\/p>\n

Question 2.
\nOABC is a parallelogram. If \\(\\overline{\\mathrm{OA}}=\\overline{\\mathrm{a}}\\) and \\(\\overline{\\mathrm{OC}}=\\overline{\\mathrm{c}}\\). Find the vector equation of the side BC. (March 2015-T.S) (V.S.A)
\nAnswer:
\nOABC is a parallelogram.
\n\"TS
\n\u2234 The vector equation of side BC is r\u0305 = (1 – t)c\u0305 + t(a\u0305 + c\u0305)
\n= (1 – t + t)c\u0305 + ta\u0305
\n= c\u0305 + t a\u0305 where t \u2208 R.<\/p>\n

Question 3.
\nIf a\u0305, b\u0305, c\u0305 are the position vectors of the vertices A, B and C respectively of a \u0394ABC, then find the vector equation of the median through the vertex A. (March 2013) (V.S.A)
\nAnswer:
\n\"TS
\nLet \\(\\overline{\\mathrm{OA}}=\\overline{\\mathrm{a}}, \\overline{\\mathrm{OB}}=\\overline{\\mathrm{b}}\\), and \\(\\overline{\\mathrm{OC}}=\\overline{\\mathrm{c}}\\)
\nVector equation of the median AD is (1 – t)
\na\u0305 + tb\u0305 = r\u0305
\nr\u0305 = (1 – t)a\u0305 + t\\(\\left(\\frac{\\overline{\\mathrm{b}}+\\overline{\\mathrm{c}}}{2}\\right)\\)<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the vector equation of the line joining the points 2i\u0305 + j\u0305 + 3k\u0305 and – 4i\u0305 + 3j\u0305 – k\u0305.(V.S.A)
\nAnswer:
\nLet a = 2i\u0305 + j\u0305 + 3k\u0305 and b = -4i\u0305 + 3 j\u0305 – k\u0305
\nThe vector equation of the line passing through the points a,b is
\nr\u0305 = (1 – t)a\u0305 + tb\u0305, t \u2208 R
\n= a\u0305 + t (b\u0305 – a\u0305)
\n= (2i\u0305 + j\u0305 + 3k\u0305) + t (-4i\u0305 + 3j\u0305 – k\u0305 – 2i\u0305 – j\u0305 – 3k\u0305)
\n= (2i\u0305 + j\u0305 + 3k\u0305) +t(-6i\u0305 + 2j\u0305 – 4k\u0305)<\/p>\n

Question 5.
\nFind the vector equation of the plane passing through the points i\u0305 – 2j\u0305 + 5k\u0305, – 5j\u0305 – k\u0305 and -3 i\u0305 + 5j\u0305. (V.S.A)
\nAnswer:
\nLet a\u0305 = i\u0305 – 2 j\u0305 + 5k\u0305, b\u0305 = -5 j\u0305 – k\u0305, c = -3i\u0305 + 5j\u0305. (May 2014)
\nThe vector equation of the plane passing through the points a\u0305, b\u0305, c\u0305 is r = (1 – s – t)a\u0305 + sb\u0305 + tc\u0305 where s, t \u2208 R
\n= a\u0305 + s(b\u0305 – a\u0305) + t(c\u0305 – a\u0305)
\n= (i\u0305 – 2j\u0305 + 5k\u0305) + s(-5j\u0305 – k\u0305 – i\u0305 + 2j\u0305 – 5k\u0305) + t(-3i\u0305 + 5j\u0305 – i\u0305 + 2j\u0305 – 5k\u0305)
\n= i\u0305 – 2j\u0305 + 5k\u0305 + s(-i\u0305 – 3j\u0305 – 6k\u0305) + t(-4i\u0305 + 7j\u0305 – 5k\u0305)<\/p>\n

Question 6.
\nFind the vector equation of the plane passing through the points (0,0, 0), (0, 5, 0) and (2, 0, 1). (V.S.A)
\nAnswer:
\nThe vector equation of the plane through a, b,c is
\nr\u0305 = (1 – s – t)a\u0305 + sb\u0305 + tc\u0305 where s, t \u2208 R
\n\u21d2 r\u0305 = (1 – s – t) 0 + s(5j\u0305) + t(2i\u0305 + k\u0305)
\n= (5s) j\u0305 + t(2i\u0305 + k\u0305);s, t \u2208 R<\/p>\n

II.
\nQuestion 1.
\nIf a, b, c are noncoplanar find the point of intersection of the line passing through the points 2a\u0305 + 3b\u0305 – c\u0305, 3a\u0305 + 4b\u0305 – 2c\u0305 with the line joining points a\u0305 – 2b\u0305 + 3c\u0305, a\u0305 – 6b\u0305 + 6c\u0305. (S.A)
\nAnswer:
\nThe vector equation of the straight line passing through the points 2a\u0305 + 3b\u0305 – c\u0305 and 3a\u0305 + 4b\u0305 – 2c\u0305 is
\nr\u0305 = (1 – t) (2a\u0305 + 3b\u0305 – c\u0305) + t(3a\u0305 + 4b\u0305 – 2c\u0305) where t \u2208 1
\n\u21d2 r\u0305 = (2 + t)a\u0305 + (3 + t) b\u0305 + (-1 – t)c\u0305
\n= (2a\u0305 + 3b\u0305 – c\u0305) + t (a\u0305 + b\u0305 – c\u0305) ……………(1)
\nThe vector equation of the straight line passing through the points a\u0305 – 2b\u0305 + 3c\u0305 and a\u0305 – 6b\u0305 + 6c\u0305 is
\nr\u0305 = (a\u0305 – 2b\u0305 + 3c\u0305) (1 – s) + s (a\u0305 – 6b\u0305 + 6c\u0305) where s \u2208 R
\n\u21d2 r\u0305 = a\u0305 + (-2 – 4s) b\u0305 + (3 + 3s)c\u0305
\n= (a\u0305 – 2b\u0305 + 3c\u0305) + s (-4b\u0305 + 3c\u0305) …………..(2)
\nEquating coefficients a\u0305, b\u0305, c\u0305 in (1) and (2) we have
\n2 + t = 1 ………..(3)
\n3 + t = – 2 – 4s ……….(4)
\nand – 1 – t = 3 + 3s ………..(5)
\nSolving equations (3), (4) and (5) we get t = – 1, and s = – 1
\nHence from (1) and (2) the point of intersection of lines (1) and (2) is a\u0305 + 2b\u0305
\nAlso line (1) is parallel to a + b – c ancl (2) is parallel to -4b\u0305 + 3c\u0305
\nIf a\u0305 + b\u0305 – c\u0305 and 3c\u0305 – 4b\u0305 are parallel then two lines are same since they have common point otherwise they have only one point of intersection a\u0305 + 2b\u0305<\/p>\n

Question 2.
\nABCD is a trapezium in which AB and CD are parallel. Prove by vector methods that the mid points of the sides AB, CD and the intersection of the diagonals are collinear. (E.Q)
\nAnswer:
\nLet A be the origin and \\(\\overline{\\mathrm{AB}}\\) = b\u0305
\n\u2234 \\(\\overline{\\mathrm{DC}}\\) = sb\u0305 (\u2235 \\(\\overline{\\mathrm{DC}} \\| \\overline{\\mathrm{AB}}\\))
\n\\(\\overline{\\mathrm{DC}}\\) = c\u0305 – d\u0305 = sb
\n\u21d2 d\u0305 = c\u0305 – sb\u0305
\n\u21d2 c\u0305 – d\u0305 = sb\u0305
\n\"TS
\nEquation of diagonal AC is
\nr\u0305 = (1 – t)o + tc\u0305
\n= tc\u0305 for t \u2208 R …………..(2)
\nEquation of diagonal BD is
\nr\u0305 = (1 – s)b\u0305 + sd\u0305 for s \u2208 R …………(3)<\/p>\n

Let R be the point of intersection of diagonals AC and BD.
\nFrom (2) and (3) t c = (1 – s) b\u0305 + s d\u0305
\n\u21d2 t c\u0305 = (1 – s) b\u0305 + s (c – sb\u0305) from (1)
\n\u21d2 tc\u0305 = (1 – s)\u03bb (c – d\u0305) + sd\u0305
\n= (1 – s) \u03bbc\u0305 – [\u03bb(1 – s) – s]d\u0305
\nEquating coefficients of c\u0305 and d\u0305 on both sides
\nt = (1 – s) \u03bb and \u03bb (1 – s) – s = 0
\n\u21d2 s = \u03bb (1 – s)
\n\u21d2 s(1 + \u03bb) = \u03bb
\n\u21d2 s = \\(\\frac{\\lambda}{1+\\lambda}\\)
\n\u2234 t = (1 – s)\u03bb = (1 – \\(\\frac{\\lambda}{1+\\lambda}\\))\u03bb
\n= \\(\\frac{\\lambda}{1+\\lambda}\\)
\nPosition vector of the point of intersection ‘R’ is
\n\"TS
\nFrom (4) and (5)
\n\\(\\overline{\\mathrm{RM}}\\) = \u03bb \\(\\overline{\\mathrm{NR}}\\)
\n\u21d2 M, R, N are collinear.
\nSo the mid points of parallel sides of a trapezium and the point of intersection of the diagonals are collinear.<\/p>\n

\"TS<\/p>\n

Question 3.
\nIn a quadrilateral ABCD, if the midpoints of one pair of opposite sides and the point of intersection of the diagonals are collinear, using vector methods, prove that the quadrilateral ABCD is a trapezium. (S.A)
\nAnswer:
\n\"TS
\n\\(\\overline{\\mathrm{AB}}=\\overline{\\mathrm{b}}, \\overline{\\mathrm{AC}}=\\overline{\\mathrm{c}}\\) and \\(\\overline{\\mathrm{AD}}=\\overline{\\mathrm{d}}\\)
\nLet M, N be the mid points of one pair of opposite sides AB and CD of a quadrilateral ABCD.
\n\\(\\overline{\\mathrm{AM}}=\\frac{\\overline{\\mathrm{b}}}{2}\\)
\n\\(\\overline{\\mathrm{AN}}=\\frac{\\overline{\\mathrm{c}}+\\overline{\\mathrm{d}}}{2}\\)
\nLet P be the point of intersection of mid points of sides AB, CD and pair of diagonals AC, BD respectively.
\nLet \\(\\overline{\\mathrm{AP}}=\\overline{\\mathrm{r}}\\). Then equation of \\(\\overline{\\mathrm{AC}}\\) is
\nr\u0305 = t c\u0305 where t is a scalar ………….(1)
\nEquation of BD is
\nr\u0305 = (1 – s)b\u0305 + sd\u0305 for some scalars ………..(2)
\nand equation of line MN is
\nr\u0305 = (1 – \u03b1)\\(\\frac{\\bar{b}}{2}\\) + \u03b1\\(\\left(\\frac{\\bar{c}+\\bar{d}}{2}\\right)\\)
\nwhere \u03b1 is a scalar
\n\u21d2 2r\u0305 = (1 – \u03b1)b\u0305 + a(c\u0305 + d\u0305)
\n\u21d2 r\u0305 + r\u0305 = (1 – \u03b1)b\u0305 + a(c\u0305 + d\u0305)
\nFrom (1) and (2)
\ntc\u0305 + (1 – s)b\u0305 + sd\u0305 = (1 – \u03b1)b\u0305 + \u03b1(c\u0305 + d\u0305)
\nEquating coefficients of b, c, d we get
\n1 – s = 1 – oc \u21d2 s = a and .
\nt = a \u21d2 s = t = a
\nFrom (1) and (2),
\nt c\u0305 = (1 – s) b\u0305 + s d\u0305
\n\u21d2 sc\u0305 = (1 – s) b\u0305 + s d\u0305 (‘- t = s)
\n\u21d2 (1 -s) b\u0305 =s(c\u0305 – d\u0305)
\n\u21d2 b is parallel to c\u0305 – d\u0305
\n\u21d2 AB is parallel to CD
\n\u2234 ABCD is a trapezium.<\/p>\n

III.
\nQuestion 1.
\nFind the vector equation of the plane which passes through the points 2i\u0305 + 4j\u0305 + 2k\u0305, 2i\u0305 + 3j\u0305 + 5k\u0305 and parallel to the vector 3 i\u0305 – 2 j\u0305 + k\u0305. Also find the point where this plane meets the line joining the points 2 i\u0305 + j\u0305 + 3k\u0305 and 4 i\u0305 – 2 j\u0305 + 3k\u0305. (March 2012) (E.Q)
\nAnswer:
\nVector equation of the plane which passes through the points a\u0305 = 2i\u0305 + 4j\u0305 + 2k\u0305, b\u0305 = 2i\u0305 + 3j\u0305 + 5k\u0305 and parallel to vector c\u0305 = 3i\u0305 – 2j\u0305 + k\u0305 is
\nr\u0305 = (1 – t)a\u0305 + tb\u0305 + sc\u0305 where t, s e R
\n\u21d2 r\u0305 = (1 – t) (2i\u0305 + 4j\u0305 + 2k\u0305) + t(2i\u0305 + 3j\u0305 + 5k\u0305) + s(3i\u0305 – 2j\u0305 + k\u0305)
\n\u21d2 r\u0305 = (2 – 2t + 2t + 3s) i\u0305 + (4 – 4t + 3t – 2s) j\u0305 + (2 – 2t + 5t + s) k\u0305
\n\u21d2 r\u0305 = (2 – 2t + 2t + 3s) i\u0305 + (4 – 4t + 3t – 2s) j\u0305 + (2 – 2t + 5t + s) k\u0305
\n\u21d2 r\u0305 = (2 + 3s) i\u0305 + (4 – t – 2s) j\u0305 + (2 + 3t + s)k\u0305 …………(1)
\nVector equation of the line passing through the points c\u0305 = 2i\u0305 + j\u0305 + 3k\u0305 and d\u0305 = 4 i\u0305 – 2 j\u0305 + 3k\u0305 is r\u0305 = (1 – a)d\u0305 + ac\u0305 where a e R
\n\u21d2 r\u0305 = (1 – \u03b1)(2i\u0305 + j\u0305 + 3k\u0305) + \u03b1(4i\u0305 – 2j\u0305 + 3k\u0305)
\n\u21d2 r\u0305 = (2 – 2\u03b1 + 4\u03b1) i\u0305 + (1 – \u03b1 – 2\u03b1) j\u0305 + (3 – 3\u03b1 + 3\u03b1)k\u0305
\n\u21d2 r\u0305 = (2 + 2\u03b1) i\u0305 + (1 – 3\u03b1) j\u0305 + 3k\u0305 (2)
\nLet 7 be the point of intersection of (1) and (2)
\n(2 + 3s)i\u0305 + (4 – t – 2s) j\u0305 + (2 + 3t + s) k\u0305
\n= (2 + 2\u03b1) i\u0305 + (1 – 3\u03b1) j\u0305 + 3k\u0305
\nv Since i\u0305, j\u0305, k\u0305 are non coplanar,
\n2 + 3s = 2 + 2\u03b1 \u21d2 2\u03b1 – 3s = 0 ………………(3)
\n4 – 1 – 2s = 1 – 3\u03b1 \u21d2 3\u03b1 – 2s -1 = – 3 …………(4)
\n2 + 3t + s = 3 \u21d2 s + 3t = 1 ………………(5)
\nFrom (5), t = \\(\\frac{1-\\mathrm{s}}{3}\\)
\n\u2234 From (4) 3\u03b1 – 2s – \\(\\left(\\frac{1-s}{3}\\right)\\) = -3
\n\u21d2 9\u03b1 – 6s – 1 + s = -9
\n9\u03b1 – 5s + 8 = 0 (6)
\nSolving (6) & (3) equations
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the vector equation of the plane passing through the points 4 i\u0305 – 3 j\u0305 – k\u0305 , 3i\u0305 + 7j\u0305 – 10k\u0305 and 2i\u0305 + 5j\u0305 – 7k\u0305 and show that the point i\u0305 + 2 j\u0305 – 3k\u0305 lies in the plane. (March 2013) (S.A.Q.)
\nAnswer:
\nVector equation of the plane passing through
\nA(4i\u0305 – 3j\u0305 – k\u0305 ), B (3i\u0305 + 7j\u0305 – 10k\u0305 ) and C(2i\u0305 + 5j\u0305 – 7k\u0305 ) is
\nr\u0305 = (1 – s – t) (4i\u0305 – 3 j\u0305 – k\u0305 ) + s(3i\u0305 + 7j\u0305 – 10k\u0305 ) + t(2i\u0305 + 5j\u0305 – 7k\u0305 )
\nLet D (i\u0305 + 2j\u0305 – 3k\u0305 ) lies on the plane, then
\n(i\u0305 + 2j\u0305 – 3k\u0305 ) = (1 – s – t)(4i\u0305 – 3j\u0305 – k\u0305 ) + s (3i\u0305 + 7j\u0305 – 10k\u0305 ) + t (2i\u0305 + 5j\u0305 – 7k\u0305 )
\nSince i\u0305 , j\u0305 ,k\u0305 are non coplanar, equating coefficients of i\u0305 , j\u0305 , k\u0305 both sides.
\n4(1 – s – t) + 3s + 2t = 1
\n\u21d2 4 – 4s – 4t + 3s + 2t = 1
\n\u21d2 s + 2t = 3 …………(1)
\n– 3 (1 – s – t) + 7s + 5t = 2
\n\u21d2 -3 + 3s + 3t + 7s + 5t = 2
\n\u21d2 10s + 8t = 5
\nAlso – (1 – s – t) – 10s – 7t = – 3
\n\u21d2 – 1 + s + t – 10s – 7t = – 3
\n\u21d2 9s + 6t = 2
\nFrom (1), 3s + 6t = 9
\nSolving (1) & (3) equations 6s = – 7
\n\u21d2 s = – 7\/6
\n\"TS
\ns = \\(\\frac{-7}{6}\\) t = \\(\\frac{25}{12}\\). satisfy (1), (2), (3).
\nand D lies on the plane passing through A, B, C.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(b) to find a better approach to solving the problems. TS Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(b) I. Question 1. Find the vector equation of the line passing through the point 2i\u0305 + 3j\u0305 + k\u0305 … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6804"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6804"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6804\/revisions"}],"predecessor-version":[{"id":6855,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6804\/revisions\/6855"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6804"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6804"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6804"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}