{"id":21446,"date":"2020-01-27T10:19:39","date_gmt":"2020-01-27T04:49:39","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21446"},"modified":"2021-07-02T15:27:00","modified_gmt":"2021-07-02T09:57:00","slug":"2nd-puc-maths-question-bank-chapter-11-miscellaneous-ex","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-11-miscellaneous-ex\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Miscellaneous Exercise"},"content":{"rendered":"

Students can Download Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Miscellaneous Exercise<\/h2>\n

Question 1.
\nShow that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).
\nAnswer:
\nDr\u2019s of the line joining the origin to the point (2,1,1 )is (2 – 0, 1 – 0, 1 – 0)
\n\u2234 (2, 1,1)
\nDr\u2019s of the line joining (3, 5,-1) and (+4, 3,-1)
\nis (4 – 3 ,3 – 5,1 – 1)\u00a0 (1, -2, 0)
\n1 x 2 – 2 x 1 + 1 x 0 = 0
\nhence a1<\/sub>b1<\/sub> + a2<\/sub>b2<\/sub> + a3<\/sub>b3<\/sub> = 0
\n\u2234 lines are perpendicular.<\/p>\n

Question 2.
\nIf l1<\/sub>, m1<\/sub> , n1<\/sub> and l2<\/sub>, m2<\/sub>, n2<\/sub> are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are
\nm1<\/sub>n2<\/sub> – m2<\/sub>n1<\/sub>, n1<\/sub>l2<\/sub> – n2<\/sub>lp<\/sub> l1<\/sub>m2<\/sub> – l2<\/sub>m1
\n<\/sub>Answer:
\nThe given lines are respectively parallel to
\n\\(\\bar{a}_{1}=1, \\bar{i}+m_{1} \\bar{j}+n_{1} \\bar{k} \\text { and } \\bar{a}_{2}=1_{2} \\bar{i}+m_{2} \\bar{j}+n_{2} \\bar{k} \\bar{a}_{1} \\times \\bar{a}_{2}\\) is a vector which is normal to both \\(\\bar { a } _{ 1 }\\quad and\\quad \\bar { a } _{ 2 }\\) and is of magnitude unity.
\n\"2nd<\/p>\n

Question 3.
\nFind the angle between the lines whose direction ratios are
\n(a, b, c) and (b – c, c – a, a – b.)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind the equation of a line parallel to x-axis and passing through the origin.
\nAnswer:
\nDc\u2019s of x-axis (1, 0, 0)
\nDc\u2019s of any line parallel to x-axis is also (1,0,0).
\nThe required equation passes through (0,0,0) and having Dc\u2019s (1, 0,0) is
\n\\(\\frac{x-0}{1}=\\frac{y-0}{0}=\\frac{z-0}{0}\\)<\/p>\n

Question 5.
\nIf the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (- 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
\nAnswer:
\nDr\u2019s of AB = (3, 3, 4), Direction CD = (6,6, 8)
\n\\(\\frac{3}{6}=\\frac{3}{6}=\\frac{4}{8} \\) Dr\u2019s are proportional. Hence they are parallel
\n\u2234 Angle between them = 0\u00b0<\/p>\n

Question 6.
\nIf the lines
\n\\(\\frac { x-1 }{ -3 } =\\frac { y-2 }{ 2k } =\\frac { z-3 }{ 2 } \\quad and\\quad \\frac { x-1 }{ 3k } =\\frac { y-1 }{ 1 } =\\frac { z-6 }{ -5 } \\)
\nare perpendicular, find the value of k.
\nAnswer:
\nHere Dr\u2019s of the lines are (-3, 2k, 2) and (3k, 1,-5) .
\nsince the lines are perpendicular
\n-3 x 3k + 2k x 1 x 2 x -5 = 0
\n\\(-9 k+2 k-10=0 \\Rightarrow-7 k=10 \\Rightarrow k=\\frac{-10}{7}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \\(\\overrightarrow{\\mathbf{r}} \\cdot(\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}-5 \\hat{\\mathbf{k}})+9=\\mathbf{0}\\)
\nAnswer:
\nDr\u2019s of the normal to the plane is (1,2,-5)
\nSince the required line is ‘\u22a5’\u00a0to the plane, these Dr\u2019s are equal.
\n\u2234 Dr\u2019s of the line (1,2, -5) and passing through (2,3)
\n\u2234 \\(\\text { Line is } \\frac{x-1}{1}=\\frac{y-2}{2}=\\frac{z-3}{-5}\\)
\nhence vector equation of the line is
\n\\(\\overline { { r } } =\\overline { { a } } +\\lambda \\overline { { b } } \\quad \\overrightarrow { { r } } =(\\overrightarrow { { i } } +2\\overrightarrow { { j } } +3\\overrightarrow { { k } } )+\\lambda (\\overrightarrow { { i } } +2\\overrightarrow { { j } } -5\\overrightarrow { { k } } )\\)<\/p>\n

Question 8.
\nFind the equation plane passing (a,b,c) and parallel to the plane
\n\\(\\overrightarrow{\\mathbf{r}}\\cdot(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})=2\\)
\nAnswer:
\nThe required plane is parallel to
\n\\(\\overrightarrow{\\mathbf{r}}\\cdot(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})\\) hence Dr\u2019s of the normal to the plane is (1, 1, 1) and through the points
\n(a, b, c)
\n\u2234 its equation is \\((\\overline{\\mathbf{r}}-\\overline{\\mathbf{a}}) \\cdot \\hat{\\mathbf{n}}=0\\)
\n(x – a) x 1 + (y – b) x 1 x (z – c).1 = 0
\n\u2234 x + y + z = a + b + c.<\/p>\n

Question 9.
\nFind the shortest distance between lines
\n\\(\\begin{matrix} & \\overrightarrow { { r } } =6\\hat { { i } } +2\\hat { { j } } +2\\hat { { k } } +\\lambda (\\hat { { i } } -2\\hat { { j } } +2\\hat { { k } } ){ and } \\\\ & \\overrightarrow { { r } } =-4\\hat { { i } } -\\hat { { k } } +\\mu (3\\hat { { i } } -2\\hat { { j } } -2\\hat { { k } } ) \\end{matrix}\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nFind the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) , crosses the YZ- plane.
\nAnswer:
\nEquation of line through (5, 1, 6) and (3,4,1) is
\n\"2nd
\n\u2234 Any points on the line
\n\"2nd<\/p>\n

Question 11.
\nFind the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX – plane.
\nAnswer:
\n\"2nd<\/p>\n

Question 12.
\nFind the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3,1) crosses the plane 2x + y + z = 7.
\nAnswer:
\nEquation of plane through (3, -4, -5) and
\n\\((2,-3,1) \\text { is } \\frac{x-3}{-1}=\\frac{y+4}{1}=\\frac{z+5}{6}\\)
\nany point on the line (3 – \u03bb, -4 + \u03bb, -5 + 6\u03bb) but this point lies on the plane 2x + y + z = 7 6 – 2\u03bb,-4 + \u03bb -5 + 6\u03bb = 7
\n+5\u03bb = +10 \u21d2\u00a0\u03bb = +2
\n\u2234 Required point is (1, -2,7).<\/p>\n

Question 13.
\nFind the equation of the plane passing through the point (- 1, 3, 2) and perpen\u00addicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
\nAnswer:
\nLet the equation of plane through (-1,3, 2) be
\nA (x + 1) + B (y – 3) + C (z-2) = 0 \u2192 (1)
\nBut this plane is ‘\u22a5’ to both given planes
\nA + 2B + 3C = 5 and \u2192 (2)
\n3A + 3B + C = 0 \u2192 (3)
\nsolving (1), (2) and (3)
\n\"2nd
\n\u21d2 (x +1)(-7) – (y – 3)(-8) + (z – 2)(-3) = 0
\n7x – 8y + 3z + 25 = 0
\nOR
\nEquation of plane through (x1<\/sub>, y1<\/sub>, z1<\/sub>) and is perpendicular a2<\/sub>x + a2<\/sub>y + a3<\/sub>z = c1
\n<\/sub>b1<\/sub>x + b2<\/sub>y + b3<\/sub>z = c2<\/sub> is of the
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nIf the points (1,1, p) and (- 3,0,1) be equidistant from the plane \\(\\overrightarrow{\\mathrm{r}} \\cdot(3 \\hat{\\mathrm{i}}+4 \\hat{\\mathrm{j}}-12 \\hat{\\mathrm{k}})+13=0 \\) then find the value of p.
\nAnswer:
\nThe equation of plane is 3x + 4y – 12z + 13 =0 distances from (1, 1, p) and (-3, 0, 1) are
\n\"2nd<\/p>\n

Question 15.
\nFind the equation of the plane passing through the line of intersection of the planes
\nand parallel to x-axis.
\nAnswer:
\nEquation of planes are x + y + z – 1= 0 and 2x + 3y – z + 4 = 0 any plane through their enter section is of the form.
\n(x + y + z – 1)+\u03bb (2x + 3y – z + 4) = 0
\n(1+ 2\u03bb) x + y(1+3\u03bb) + z(1- \u03bb) – 1 + 4\u03bb) = 0
\nbut this plane is parallel to x-axis whose direction are (1,0,0)
\n\"2nd<\/p>\n

Question 16.
\nIf O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.
\nAnswer:
\nDirections of OP = (1, 2, -3)
\n\u2234 Equation of plane through (1, 2, -3) and
\nDr\u2019s of normal is (1,2, -3) is A (x – x1<\/sub>) + B (y – y1<\/sub>) + C (z – z1<\/sub>) = 0
\n1 (x – 1) + 2(y – 2) -3(z + 3) = 0
\nx – 1 + 2 (y – 2) -3 (z + 3) = 0
\nx + 2y – 3z – 14 = 0<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nFind the equation of the plane which contains the line of intersection of the planes \\(\\overrightarrow{\\mathrm{r}} \\cdot(\\hat{\\mathrm{i}}+2 \\hat{\\mathrm{j}}+3 \\hat{\\mathrm{k}})-4=0, \\quad \\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}+\\hat{\\mathrm{j}}-\\hat{\\mathrm{k}})+5=0 \\) and which is perpendicular to the plane \\(\\overrightarrow{\\mathrm{r}} \\cdot(5 \\hat{\\mathrm{i}}+3 \\hat{\\mathrm{j}}-6 \\hat{\\mathrm{k}})+8=0\\)
\nAnswer:
\n(x + 2y + 3z – 4) + \u03bb (2x + y – z + 5) = 0
\nx(1+2\u03bb)+ \u03bb (2+\u03bb) +z (3 – \u03bb) – 4 + 5 \u03bb =0
\nthe given plane is ‘\u22a5’ to the plane
\n\"2nd<\/p>\n

Question 18.
\nFind the distance of the point (- 1, – 5, -10) from the point of intersection of the line
\n\\(\\overrightarrow { { r } } =2\\hat { { i } } -\\hat { { j } } +2\\hat { { k } } +\\lambda (3\\hat { { i } } +4\\hat { { j } } +2\\hat { { k } } )and\\quad plane\\quad \\overrightarrow { { r } } \\cdot (\\hat { { i } } -\\hat { { j } } +\\hat { { k } } )=5\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 19.
\nFind the vector equation of the line passing through (1,2,3) and parallel to the planes
\n\\(\\overrightarrow{\\mathbf{r}} \\cdot(\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}})=5 \\text { and } \\overrightarrow{\\mathbf{r}} \\cdot(3 \\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})=6\\)
\nAnswer:
\nEquation of the line through (1,2,3) is \\(\\frac{x-1}{a}=\\frac{y-2}{b}=\\frac{z-3}{c}\\) but this is plane is parallel to the plane
\nx – y + 2z = 5 and 3x + y + z = 6
\n1 x a – 1 x b + 2 x c = 0 \u21d2 a – b + 2c = 0
\nand 3 x a + 1 x b + 1 x c = 0
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nFind the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
\n\\(\\begin{aligned}&\\frac{x-8}{3}=\\frac{y+19}{-16}=\\frac{z-10}{7} \\text { and }\\\\&\\frac{x-15}{3}=\\frac{y-29}{8}=\\frac{z-5}{-5}\\end{aligned}\\)
\nAnswer:
\n\"2nd
\n\\(\\overline{\\mathbf{r}}=(\\overrightarrow{\\mathbf{i}}+2\\overrightarrow{\\mathrm{j}}-4 \\overline{\\mathbf{k}})+\\lambda(2 \\overrightarrow{\\mathrm{i}}+3 \\overrightarrow{\\mathrm{j}}+6 \\overrightarrow{\\mathrm{k}})\\)<\/p>\n

Question 21.
\nProve that if a plane has the intercepts a, b, c and is at a distance of p units from the origin then \\(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}=\\frac{1}{p^{2}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Choose the correct answer in exercise 22 and 23.<\/p>\n

Question 22.
\nDistance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
\n(A) 2 units
\n(B) 4 units
\n(C) 8 units
\n(D) \\(\\frac{2}{\\sqrt{29}} \\text { units }\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nThe planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
\n(A) Perpendicular
\n(B) Parallel
\n(C) passes through \\(\\left(0,0, \\frac{5}{4}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

2nd PUC Maths Three Dimensional Geometry Miscellaneous Exercise Extra Questions and Answers<\/h3>\n

Question 1.
\nWrite the distance of the following plane from the origin, 2x-y + 2z + 1 = 0 (CBSE\u00a0 2010)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nFind the points on the line \\(\\frac{x+2}{3}=\\frac{y+1}{2}=\\frac{z-3}{2}\\) at a distance of 5 units from the points (1,3,5) (CBSE\u00a0 2010)
\nAnswer:<\/p>\n

\"2nd<\/p>\n

Question 3.
\nFind the distance of the point P (6,5,9) from the plane determined by A (3, -1, 2) B (5, 2, 4) and C (-1, -1, 6) (CBSE 2010)
\nAnswer:
\nEquation of plane inn three point form is
\n\"2nd
\n(x – 3) 12 – (y + 1) (8 + 8) + (z – 2) (12) = 0
\n12x – 36 – 16y – 16 + 12z – 24 = 0
\n12x – 16y + 12z – 76 = 0
\n3x – 4y + 3z -19 = 0
\n\"2nd<\/p>\n

Question 4.
\nWrite the dc\u2019s of the line joining the points (1,0, 0) and (0,1,1) (CBSE 2011)
\nAnswer:
\nDirection ratios = (-1,1,1)
\n\"2nd<\/p>\n

Question 5.
\nWrite the vector equation of the line (CBSE 2011)
\n\\(\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}\\)
\nAnswer:
\nThe given lines passes through (5, -4,6)
\nand is parallel to the vector
\n\\((3 i+7 j+2 \\bar{k}) is\\)
\n\"2nd<\/p>\n

Question 6.
\nFind the distance of a point 3x – 4y + 122 = 3 from the origin (CBSE 2012)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the co-ordinate of the foot of \u22a5r<\/sup> and the perpendicular distance of the point P (3, 2,1) from the plane 2x -y + z + 1 = 0. Find also the image of the point in the plane (CBSE 2011)
\nAnswer:
\nDirection of a normal to the plane (2, -1, 1)
\nLet Q be the foot of \u22a5 from p to the plane
\nNormally line is parallel to plane PQ
\n\"2nd
\nEquation of PQ =
\n\\(\\frac{x-3}{2}=\\frac{y-2}{-1}=\\frac{z-1}{1}\\)
\nAny point on this line is (2\u03bb + 3, -\u03bb + 2, \u03bb +1)
\nIf this point is Q which lies on the plane
\n\"2nd<\/p>\n

Question 8.
\nFind the equation of the plane passing through the point P (1,1,1) and containing the line \\(\\overline{\\mathbf{r}}=(-3 \\mathbf{i}+\\mathbf{j}+5 \\overline{\\mathbf{k}})+\\lambda(\\mathbf{3} \\mathbf{i}-\\mathbf{j}-\\mathbf{5} \\overline{\\mathbf{k}})\\). Also show that the plane contains the line
\nAnswer:
\n\\(\\overline{\\mathbf{r}}=(-3 \\mathbf{i}+\\mathbf{j}+5 \\overline{\\mathbf{k}})+\\lambda(\\mathbf{3} \\mathbf{i}-\\mathbf{j}-\\mathbf{5} \\overline{\\mathbf{k}})\\)
\nEquation of plane through the point (1,1,1) is of the form
\nA (x – 1) + B (y – 1) + C (z – 1) = 0 .
\nThis plane containing the line
\n\"2nd
\n(x – 1) (4) – (y – 1) (20 – 12) +(z – 1) (4)
\n4x – 4 – 8y + 8 + 4z – 4 = 0
\n4x-8y + 4z = 0 \u21d2 x -2y + z = 0
\nsince the point (1,-2, -5) lies on the plane
\nx – 2y + z = 0 as 1 x 1 – 2x – 2 – 5 = 0 and
\n\\(\\left( i-2j-5\\bar { k } \\right) \\cdot \\left( i-2j\\quad +\\bar { k } \\right) =1+4-5=0\\) the plane x – 2y +z = 0 contains the lines
\n\\(\\overline{\\mathbf{r}}=(-i+2 j+5 \\overline{\\mathbf{k}})+\\mu(\\mathbf{i}-2 j-5 \\overline{\\mathbf{k}})\\)<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nFind the points on the line
\n\\(\\frac{x+2}{3}=\\frac{y+1}{2}=\\frac{z-3}{2}\\) at a distance of 5 units from the points (1,3,3) (CBSE 2011)
\nAnswer:
\nAny point on the line.
\n\"2nd<\/p>\n

Question 10.
\nFind the distance of the point P (6, 5, 9) from the plane determined by the points A (3, -1, 2), B (5, 2, 4) and C (-1, -1, 6) (CBSE 2011)
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nFind the shortest distance between the following lines whose vector equals
\n\\(\\begin{aligned}&\\overline{\\mathbf{r}}=(1-\\mathbf{t}) \\mathbf{i}+(\\mathbf{t}-\\mathbf{2}) \\mathbf{j}+(\\mathbf{3}-\\mathbf{2} \\mathbf{t}) \\overline{\\mathbf{k}} \\text { and }\\\\&\\overline{\\mathbf{r}}=(\\mathbf{s}+\\mathbf{1}) \\mathbf{i}+(2 \\mathbf{s}-\\mathbf{1}) \\mathbf{j}-(2 \\mathbf{s}+\\mathbf{1}) \\overline{\\mathbf{k}}\\end{aligned}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 12.
\nIf the lines
\n\\(\\frac { x-1 }{ -3 } =\\frac { y-2 }{ -2k } =\\frac { z-3 }{ 2 } ,\\frac { { x }-{ 1 } }{ { K } } =\\frac { { y }-{ 2 } }{ { 1 } } =\\frac { { z }-{ 3 } }{ { 5 } } { are }\\perp ^{ { r } },{ find\\quad }{ K }\\)
\nAnswer:
\nDirections of tines are (-3, -2K, 2) and (K, 1,5)
\nsince lines are \u22a5r<\/sup> a1<\/sub>b1<\/sub> + a2<\/sub>b2<\/sub> + a3<\/sub>b3<\/sub> = 0
\n-3 x K – 2K x 1 + 2 x 5 = 0
\n-3K – 2K + 10 = 0, -5K = -10, K =2<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nFind the length and the foot of the \u22a5r<\/sup> from the point P (7, 14, 5) to the plane 2x + 4y – z = 2. Also find the image of the point P in the plane. (CBSE 2012)
\nAnswer:
\nDirection of the normal to the plane is (2, 4,-1)
\n\"2nd
\n\u2234 Equation of PQ through (7,14,5) and parallel to a vector whose Directions are (2, 4, 1) is \\(\\frac{x-7}{2}=\\frac{y-14}{4}=\\frac{z-5}{-1}\\)
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Question 1. Show …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21446"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=21446"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21446\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21446"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21446"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21446"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}