{"id":19864,"date":"2022-08-10T10:48:57","date_gmt":"2022-08-10T05:18:57","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=19864"},"modified":"2023-03-30T14:27:06","modified_gmt":"2023-03-30T08:57:06","slug":"inter-1st-year-maths-1a-addition-of-vectors-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-1st-year-maths-1a-addition-of-vectors-important-questions\/","title":{"rendered":"Inter 1st Year Maths 1A Addition of Vectors Important Questions"},"content":{"rendered":"

Students get through Maths 1A Important Questions<\/a> Inter 1st Year Maths 1A Addition of Vectors Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 1st Year Maths 1A Addition of Vectors Important Questions<\/h2>\n

Question 1.
\nFind unit vector in the direction of vector a = 2i + 3j + k. (Mar. 14)
\nSolution:
\nThe unit vector in the direction of a vector a is given by \\(\\hat{a}\\) = \\(\\frac{1}{|a|} a\\)
\nNow |a| = \\(\\sqrt{2^{2}+3^{2}+1^{2}}\\) = \\(\\sqrt{14}\\).
\n\u2234 \\(\\hat{a}\\) = \\(\\frac{1}{\\sqrt{14}}\\)(2i + 3j + k)
\n= \\(\\frac{2}{\\sqrt{14}} i\\) + \\(\\frac{3}{\\sqrt{14}} \\mathrm{j}\\) + \\(\\frac{1}{\\sqrt{14}} k\\)<\/p>\n

Question 2.
\nFind a vector in the direction of vector a = i – 2j that has magnitude 7 units.
\nSolution:
\nThe unit vector in the direction of the given vector a is
\n\\(\\hat{a}\\) = \\(\\frac{1}{|\\mathrm{a}|} \\mathrm{a}\\) = \\(\\frac{1}{\\sqrt{5}}(i-2 j)\\) = \\(\\frac{1}{\\sqrt{5}} \\mathrm{i}\\) – \\(\\frac{2}{\\sqrt{5}} \\mathrm{j}\\)
\nTherefore, the vector having magnitude equal to 7 and in the direction of a is
\n7a = 7\\(\\left(\\frac{1}{\\sqrt{5}} \\mathrm{i}-\\frac{2}{\\sqrt{5}} \\mathrm{j}\\right)\\) = \\(\\frac{7}{\\sqrt{5}} \\mathrm{i}\\) – \\(\\frac{14}{\\sqrt{5}} j\\)<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the unit vector in the direction of the sum of the vectors.
\na = 2i + 2j – 5k and b = 2i + j + 3k.
\nSolution:
\nThe sum of the given vectors is
\na + b (= c, say) = 4i + 3j – 2k and
\n|c| = \\(\\sqrt{4^{2}+3^{2}+(-2)^{2}}\\) = \\(\\sqrt{29}\\).
\n\u2234 \\(\\hat{c}\\) = \\(\\frac{a+b}{|a+b|}\\) = \\(\\frac{4 i+3 j-2 k}{\\sqrt{29}}\\)<\/p>\n

Question 4.
\nWrite direction ratios of the vector a = i + j – 2k and hence calculate its direction cosines.
\nSolution:
\nNote that direction ratios a, b, c of a vector r = xi + yj + zk are just the respective components x. y and z of the vector. So, for the given vector, we have a = 1, b = 1, c = – 2, Further, if l, m and n the direction cosines of the given vector, then
\nl = \\(\\frac{a}{|r|}\\) = \\(\\frac{1}{\\sqrt{6}}\\), m = \\(\\frac{b}{|r|}\\) = \\(\\frac{1}{\\sqrt{6}}\\),
\nn = \\(]\\frac{c}{|r|}\\) = \\(\\frac{2}{\\sqrt{6}}\\) as |r| = \\(\\sqrt{6}\\).
\nThus, the direction cosines are
\n(\\(\\frac{1}{\\sqrt{6}}\\), \\(\\frac{1}{\\sqrt{6}}\\), \\(-\\frac{2}{\\sqrt{6}}\\)).<\/p>\n

Question 5.
\nConsider two points P and Q with position vectors OP = 3a – 2b and OQ = a + b. Find the position vector of a point R which divides the line joining P and Q in the ratio 2 :1,
\n(i) internally and
\n(ii) externally.
\nSolution:
\ni) The position vector of the point R dividing the join of P and Q internally in the ratio 2 : 1 is
\nOR = \\(\\frac{2(a+b)+(3 a-2 b)}{2+1}\\) = \\(\\frac{5 a}{3}\\)<\/p>\n

ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2 : 1 is
\nOR = \\(\\frac{2(a+b)-(3 a-2 b)}{2-1}\\) = 4b – a.<\/p>\n

Question 6.
\nShow that the points A (2i – j + k), B(i – 3j – 5k), C(3i – 4j – 4k) are the vertices of a right angled triangle.
\nSolution:
\nWe have AB = (1 – 2)i + (-3 + 1)j + (-5 – 1)
\nk = -i – 2j – 6k
\nBC = (3 – 1)i + (-4 + 3)j + (-4 + 5)k
\n= 2i – j + k. and CA = (2 – 3)i + (-1 + 4)j + (1 +4)k = -i + 3j + 5k.
\nwe have |AB|2<\/sup> = |BC|2<\/sup> + |CA|2<\/sup>.<\/p>\n

\"Inter<\/p>\n

Question 7.
\nLet A. B, C, D be four points with position vectors \\(\\overline{\\mathbf{a}}\\) + 2\\(\\overline{\\mathbf{b}}\\), 2\\(\\overline{\\mathbf{a}}\\) – \\(\\overline{\\mathbf{b}}\\), \\(\\overline{\\mathbf{a}}\\) and 3\\(\\overline{\\mathbf{a}}\\) + \\(\\overline{\\mathbf{b}}\\) respectively. Express the vectors \\(\\overline{\\mathbf{A C}}\\), \\(\\overline{\\mathbf{D A}}\\), \\(\\overline{\\mathbf{B A}}\\) and \\(\\overline{\\mathbf{B C}}\\) interms of \\(\\overline{\\mathbf{a}}\\) and \\(\\overline{\\mathbf{b}}\\).
\nSolution:
\nThe position vectors of A, B, C, D with respect to the origin O are
\n\"Inter<\/p>\n

Question 8.
\nLet ABCDEF be a regular hexagon with centre O. Show that \\(\\overline{\\mathbf{A B}}\\) + \\(\\overline{\\mathbf{A C}}\\) + \\(\\overline{\\mathbf{A D}}\\) + \\(\\overline{\\mathbf{A E}}\\) + \\(\\overline{\\mathbf{A F}}\\) = 3\\(\\overline{\\mathbf{A D}}\\) = 6\\(\\overline{\\mathbf{A O}}\\) (Mar. ’16, ’15)
\nSolution:
\nLet ABCDEF be a regular hexagon with centre O.
\nThen \\(\\overline{\\mathbf{A B}}\\) = \\(\\overline{\\mathbf{B C}}\\) = \\(\\overline{\\mathbf{C D}}\\) = \\(\\overline{\\mathbf{D E}}\\) = \\(\\overline{\\mathbf{E F}}\\) = \\(\\overline{\\mathbf{F A}}\\)
\n\"Inter<\/p>\n

Question 9.
\nIn \u0394ABC, If \\(\\overline{\\mathbf{a}}\\), \\(\\overline{\\mathbf{b}}\\), \\(\\overline{\\mathbf{c}}\\) are position vectors of the vertices A, B and C respectively, then prove that the position vector of the centroid G is \\(\\frac{1}{3}(\\bar{a}+\\bar{b}+\\bar{c})\\).
\nSolution:
\nLet G be the centroid of \u0394ABC and AD is the median through the vector A.
\nThen AG : GD = 2 : 1. Let ‘O’ be the origin \\(\\overline{\\mathrm{OA}}\\) = \\(\\bar{a}\\), \\(\\overline{\\mathrm{OB}}\\) = \\(\\bar{b}\\), \\(\\overline{\\mathrm{OC}}\\) = \\(\\bar{c}\\) \u2235 D is Mid point of BC, the P.V. of D is \\(\\overline{O D}\\) = \\(\\frac{1}{2}(\\bar{b}+\\bar{c})\\)
\n\"Inter
\nG divides the median AD in the ratio 2 : 1
\n\u2234 The P.V. of G is
\n\"Inter<\/p>\n

Question 10.
\nIn \u0394ABC, if ‘O’ is the circum centre and H is the orthocentre, then show that
\ni) \\(\\overline{O A}\\) + \\(\\overline{O B}\\) + \\(\\overline{O C}\\) = \\(\\overline{O H}\\)
\nii) \\(\\overline{H A}\\) + \\(\\overline{H B}\\) + \\(\\overline{H C}\\) = 2\\(\\overline{H O}\\)
\nSolution:
\nLet D be the midpoint of BC<\/p>\n

i) Take’O’as the origin, let \\(\\overline{O A}\\) = \\(\\bar{a}\\) \\(\\overline{O B}\\) = \\(\\bar{b}\\) and \\(\\overline{O C}\\) = \\(\\bar{c}\\)
\n\"Inter
\n(\u2235 AH =2 R cos A, OD = R cos A, R is the circum radius of \u0394 ABC and hence AH = 2 (OD))<\/p>\n

ii) \\(\\overline{\\mathrm{HA}}\\) + \\(\\overline{\\mathrm{HB}}\\) + \\(\\overline{\\mathrm{HC}}\\) = \\(\\overline{\\mathrm{HA}}\\) +2\\(\\overline{\\mathrm{HD}}\\)
\n= \\(\\overline{\\mathrm{HA}}\\) + 2(\\(\\overline{\\mathrm{HO}}\\) + \\(\\overline{\\mathrm{OD}}\\))
\n= \\(\\overline{\\mathrm{HA}}\\) + 2\\(\\overline{\\mathrm{HO}}\\) + 2\\(\\overline{\\mathrm{OD}}\\)
\n= \\(\\overline{\\mathrm{HA}}\\) + 2\\(\\overline{\\mathrm{HO}}\\) + \\(\\overline{\\mathrm{AH}}\\) = 2\\(\\overline{\\mathrm{HO}}\\)<\/p>\n

Question 11.
\nLet \\(\\bar{a}\\), \\(\\bar{b}\\), \\(\\bar{c}\\) and \\(\\bar{d}\\) be the position vectors of A, B,C and D respectively which are the vertices of a tetrahedron. Then Prove that the lines joining the vertices to the centroids of the opposite faces are concurrent (this point is called the centroid or the centre of the tetrahe dron)
\nSolution:
\nLet ‘O’ be the origin of reference
\nLet G1<\/sub>, G2<\/sub>, G3<\/sub> and G4<\/sub> be the centroids of \u0394BCD, \u0394CAD, \u0394ABD and \u0394ABC, respectively then OG1<\/sub> = \\(\\frac{\\overline{\\mathrm{b}}+\\overline{\\mathrm{c}}+\\overline{\\mathrm{d}}}{3}\\)
\n\"Inter
\nConsider the point P that divides AG1<\/sub> in the ratio 3 : 1
\n\"Inter
\nSimilarly we can show that the P.vs. of the points dividing BG2<\/sub>, CG3<\/sub> and DG4<\/sub> in the ratio 3 : 1 are equal to \\(\\frac{1}{4}\\) (\\(\\overline{\\mathrm{a}}\\) + \\(\\overline{\\mathrm{b}}\\) + \\(\\overline{\\mathrm{c}}\\) + \\(\\overline{\\mathrm{d}}\\)). Therefore P lies on each of AG1<\/sub>, BG2<\/sub>, CG3<\/sub> and DG4<\/sub>.<\/p>\n

Question 12.
\nLet OABC be a parallelogram and D the mid point of OA. Prove that the segment CD trisects the diagonal OB and is trisected by the diagonal OB.
\nSolution:
\nLet \\(\\overline{\\mathrm{OA}}\\) = \\(\\bar{a}\\), \\(\\overline{\\mathrm{OC}}\\) = \\(\\bar{b}\\), So that
\n\\(\\overline{\\mathrm{OB}}\\) = \\(\\bar{a}\\) + \\(\\bar{b}\\), \\(\\overline{\\mathrm{OD}}\\) = \\(\\frac{1}{2}\\)(\\(\\bar{a}\\))
\n\"Inter
\nLet M be the intersection of OB and CD.
\nLet OM : MB = k : 1 and CM : MD = l : 1
\n\"Inter
\n\u2234 l = 2 and k = \\(\\frac{1}{2}\\)
\n\u2234 CD trisects OB and OB trisects CD<\/p>\n

\"Inter<\/p>\n

Question 13.
\nLet \\(\\bar{a}\\), \\(\\bar{b}\\) be non – collinear vectors, if \\(\\bar{\\alpha}\\) = (x + 4y)\\(\\bar{a}\\) + (2x + y + 1)\\(\\bar{b}\\) and \\(\\bar{\\beta}\\) = (y – 2x + 2)\\(\\bar{a}\\) + (2x – 3y – 1)\\(\\bar{b}\\) are such that 3\\(\\bar{\\alpha}\\) = \\(\\bar{\\beta}\\), then find x andy.
\nSolution:
\nGiven 3\\(\\bar{\\alpha}\\) = 2\\(\\bar{\\beta}\\)
\n\u21d2 3(x + 4y)\\(\\bar{a}\\) + 3(2x + y + 1)\\(\\bar{b}\\) = 2(y – 2x + 2)\\(\\bar{a}\\) + 2(2x – 3y – 1)\\(\\bar{b}\\)
\nOn comparing the coeffs. of \\(\\bar{a}\\) and \\(\\bar{b}\\), we get
\n3x + 12y = 2y – 4x + 4 \u21d2 7x + 10y = 4 ——–\u2014 (1)
\nand 6x + 3y + 3 = 4x – 6y – 2 \u21d2
\n2x + 9y = -5 ——–\u2014(2)
\nSolve (1) and (2), we get
\nx = 2 and y = -1<\/p>\n

Question 14.
\nShow that the points whose position vectors are
\n-2\\(\\bar{a}\\) + 3\\(\\bar{b}\\) + 5\\(\\bar{c}\\), \\(\\bar{a}\\) + 2\\(\\bar{b}\\) + 3\\(\\bar{c}\\), 7\\(\\overline{\\mathrm{a}}\\) + \\(\\overline{\\mathrm{c}}\\) are non – collinear when \\(\\bar{a}\\), \\(\\bar{b}\\), \\(\\bar{c}\\) are non – coplanar vectors.
\nSolution:
\nLet O be the origin and let P, Q, R be the P.Vs. of the given points
\n\"Inter
\n\u21d2 The points P, Q, R are collinear.<\/p>\n

Question 15.
\nIf the points whose position vectors are 3\\(\\bar{i}\\) – 2\\(\\bar{j}\\) – \\(\\bar{k}\\), 2\\(\\bar{i}\\) + 3\\(\\bar{j}\\) – 4\\(\\bar{k}\\), –\\(\\bar{i}\\) + \\(\\bar{j}\\) + 2\\(\\bar{k}\\) and 4\\(\\bar{i}\\) + 5\\(\\bar{j}\\) + \u03bb\\(\\bar{k}\\) are coplanar, then show that \u03bb = \\(\\frac{-146}{7}\\)
\nSolution:
\nLet ‘O\u2019 be the origin and Let A, B, C and D be the given points. Then
\n\"Inter
\nEquating the corresponding coeffs.
\n-x – 4y = 1, 5x + 3y = 7, -3x + 3y = \u03bb + 1
\nSolving x + 4y + 1 = 0 and 5x + 3y – 7 = 0
\n\"Inter<\/p>\n

Question 16.
\nIn the two dimensional plane, Prove by using vector methods, the equation of the line whose intercepts on the
\naxes are ‘a\u2019 and \u2019b\u2019 is \\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1.
\nSolution:
\nLet A = (a, 0), B = (0, b)
\n\u2234 A = a\\(\\overline{\\mathbf{i}}\\), B = b\\(\\overline{\\mathbf{j}}\\)
\nThe equation of the line is
\n\\(\\overline{\\mathbf{r}}\\) = (1 – t)a\\(\\overline{\\mathbf{i}}\\) + t(b\\(\\overline{\\mathbf{j}}\\))
\nIf \\(\\overline{\\mathbf{r}}\\) = x\\(\\overline{\\mathbf{i}}\\) + y\\(\\overline{\\mathbf{j}}\\), then x = (1 – t)a and y = t b
\nBy eliminating \u2018t\u2019
\nx = \\(\\left(1-\\frac{y}{b}\\right) a\\)
\n\u21d2 \\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1<\/p>\n

Question 17.
\nUsing the vector equation of the straight line passing through two points, prove that the points whose position vectors are \\(\\overline{\\mathbf{a}}\\), \\(\\overline{\\mathbf{b}}\\) and (3\\(\\overline{\\mathbf{a}}\\) – 2\\(\\overline{\\mathbf{b}}\\)) are collinear.
\nSolution:
\nThe vector equation of the st. line passing through two pt.s \\(\\overline{\\mathbf{a}}\\) and \\(\\overline{\\mathbf{b}}\\) is
\n\\(\\overline{\\mathbf{r}}\\) = (1 – t) \\(\\overline{\\mathbf{a}}\\) + t\\(\\overline{\\mathbf{b}}\\) ; for some t \u2208 R
\nThis line also passes through the pt. which
\nP.v. is 3\\(\\overline{\\mathbf{a}}\\) – 2\\(\\overline{\\mathbf{b}}\\) then
\n3\\(\\overline{\\mathbf{a}}\\) – 2\\(\\overline{\\mathbf{b}}\\) = (1 – t)\\(\\overline{\\mathbf{a}}\\) + t\\(\\overline{\\mathbf{b}}\\)
\nEquating the corresponding coeffs,
\n1 – t = 3 and t = -2
\n\u2234 The three given points are collinear.<\/p>\n

\"Inter<\/p>\n

Question 18.
\nFind the equation of the line parallel to the vector 2\\(\\overline{\\mathbf{i}}\\) – \\(\\overline{\\mathbf{j}}\\) + 2\\(\\overline{\\mathbf{k}}\\) and which passes through the point A whose position vector is 3\\(\\overline{\\mathbf{i}}\\) + \\(\\overline{\\mathbf{j}}\\) – \\(\\overline{\\mathbf{k}}\\). If P is a point on this line such that AP = 15, find the position vector of R
\nSolution:
\nThe vector equation of the line passing through the point A whose P.v is
\n\"Inter
\n\"Inter<\/p>\n

Question 19.
\nShow that the line joining the pair of points 6\\(\\overline{\\mathbf{a}}\\) – 4\\(\\overline{\\mathbf{b}}\\) + 4\\(\\overline{\\mathbf{c}}\\) and the line joining the pair of points –\\(\\overline{\\mathbf{a}}\\) – 2\\(\\overline{\\mathbf{b}}\\) – 3\\(\\overline{\\mathbf{c}}\\), \\(\\overline{\\mathbf{a}}\\) + 2\\(\\overline{\\mathbf{b}}\\) – 5\\(\\overline{\\mathbf{c}}\\) intersect at the point -4\\(\\overline{\\mathbf{c}}\\) when \\(\\overline{\\mathbf{a}}\\), \\(\\overline{\\mathbf{b}}\\), \\(\\overline{\\mathbf{c}}\\) are non-coplanar vectors. (T.S) (Mar. ’16)
\nSolution:
\nThe vector equation o+ the-line joining the pair of points \\(\\bar{q}\\) = 6\\(\\bar{a}\\) – 4\\(\\bar{b}\\) + 4\\(\\bar{c}\\), and \\(\\bar{p}\\) = -4\\(\\bar{c}\\) is \\(\\bar{r}\\) = (1 – t)\\(\\bar{p}\\) + t\\(\\bar{q}\\) for t \u2208 R
\n\"Inter
\nLet \\(\\bar{r}\\) be the point of intersection of (1) and (2) then equating the coeffs of \\(\\bar{a}\\), \\(\\bar{b}\\) and \\(\\bar{c}\\), we get
\n6t = 2s – 1 \u21d2 6t – s = -1 ——- (3)
\n– 4t = 4s – 2 \u21d2 4t + 4s = 2 —— (4)
\n-4 + 8t = -2s – 3 \u21d2 8t + 2s = 1 —— (5)
\nFrom (3) and (4)
\n\"Inter
\nFrom (1) s = \\(\\frac{1}{2}\\)
\nSubstitute t = 0 and s = \\(\\frac{1}{2}\\) in (5)
\n8(0) + 2(\\(\\frac{1}{2}\\)) = 1 \u21d2 1 = 1
\nHence the lines (1) and (2) point of intersection put t = 0 in (1)
\n(i.e) \\(\\bar{r}\\) = -4\\(\\bar{c}\\)
\n\u2234 The two lines intersect at the point whose P.v. is -4\\(\\bar{c}\\)<\/p>\n

Question 20.
\nFind the point of intersection of the line \\(\\bar{r}\\) = 2\\(\\bar{a}\\) + \\(\\bar{b}\\) + t(\\(\\bar{b}\\) – \\(\\bar{c}\\)) and the plane \\(\\bar{r}\\) = \\(\\bar{a}\\) + x(\\(\\bar{b}\\) + \\(\\bar{c}\\)) + y(\\(\\bar{a}\\) + 2\\(\\bar{b}\\) – \\(\\bar{c}\\)) where \\(\\bar{a}\\), \\(\\bar{b}\\), \\(\\bar{c}\\) are non \u2014 coplanar vectors.
\nSolution:
\nLet \\(\\bar{r}\\) be the P.v. of the point P, the intersection of the line and the plane.
\nThen
\n2\\(\\bar{a}\\) + \\(\\bar{b}\\) + t(\\(\\bar{b}\\) – \\(\\bar{c}\\)) = \\(\\bar{a}\\) + x(\\(\\bar{b}\\) + \\(\\bar{c}\\)) + y(\\(\\bar{a}\\) + 2\\(\\bar{b}\\) – \\(\\bar{c}\\))
\n\u2235 \\(\\bar{a}\\), \\(\\bar{b}\\), \\(\\bar{c}\\) are non – coplanar vectors.
\nEquation th coeffs. of \\(\\bar{a}\\), \\(\\bar{b}\\), \\(\\bar{c}\\)
\n2 = 1 + y \u21d2 y = 1
\n1 + t = x + 2y
\n\u21d2 1 + t = x + 2(1)
\n\u21d2 t – x = 1 ——— (1)
\nand -t = x – y
\nx + t = y
\n\u21d2 x + t = 1 ——— (2)
\nFrom (1),(2) t = 1, x = 0, y = 1
\n\u2234 Point of intersection is
\n\\(\\bar{r}\\) = 2\\(\\bar{a}\\) + \\(\\bar{b}\\) + 1(\\(\\bar{b}\\) – \\(\\bar{c}\\))
\n\u21d2 \\(\\bar{r}\\) = 2\\(\\bar{a}\\) + \\(\\bar{b}\\) + 1(\\(\\bar{b}\\) – \\(\\bar{c}\\))
\n\u21d2 \\(\\bar{r}\\) = 2\\(\\bar{a}\\) + 2\\(\\bar{b}\\) – \\(\\bar{c}\\)
\n\u2234 P.v. of the point of intersection is 2\\(\\bar{a}\\) + 2\\(\\bar{b}\\) – \\(\\bar{c}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Addition of Vectors Important Questions which are most likely to be asked in the exam. Intermediate 1st Year Maths 1A Addition of Vectors Important Questions Question 1. Find unit vector in the direction of vector a = 2i + 3j + k. (Mar. … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":39448,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[13],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Addition-of-Vectors-Important-Questions.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19864"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=19864"}],"version-history":[{"count":12,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19864\/revisions"}],"predecessor-version":[{"id":20002,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19864\/revisions\/20002"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/39448"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=19864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=19864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=19864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}