{"id":21429,"date":"2020-01-25T14:17:40","date_gmt":"2020-01-25T08:47:40","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21429"},"modified":"2021-07-02T15:27:00","modified_gmt":"2021-07-02T09:57:00","slug":"2nd-puc-maths-question-bank-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-11-ex-11-3\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.3"},"content":{"rendered":"

Students can Download Maths Chapter 11 Three Dimensional Geometry Ex 11.3 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 Ex 11.3<\/h2>\n

2nd PUC Maths Three Dimensional Geometry NCERT Text Book Questions and Answers Ex 11.3<\/h3>\n

Question 1.
\nIn each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin,
\n(a) z = 2
\nAnswer:
\nZ = 2
\nax + by + cz = P
\nox + oy + 1z = 2
\nDirections = 0,0,1
\n\"2nd<\/p>\n

(b) x + y + z = 1
\nAnswer:
\n\"2nd<\/p>\n

(c) 2x + 3y – z = 5
\nAnswer:
\n\"2nd<\/p>\n

(d) 5y + 8 = 0
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \\(3 i+5 j-6 \\hat{k}\\)
\nAnswer:
\n\"2nd<\/p>\n

\"2nd<\/p>\n

Question 3.
\nFind the Cartesian equation of the following planes:
\n(a) \\(\\overrightarrow{\\mathrm{r}} \\cdot(\\hat{\\mathrm{i}}+\\hat{\\mathrm{j}}-\\hat{\\mathrm{k}})=2\\)
\nAnswer:
\n\"2nd<\/p>\n

(b) \\(\\overrightarrow{\\mathbf{r}} \\cdot(2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}-4 \\hat{\\mathbf{k}})=\\mathbf{1}\\)
\nAnswer:
\n2x + 3y +4z = 1<\/p>\n

(c) \\(\\overrightarrow{\\mathbf{r}} \\cdot[(\\mathbf{s}-2 \\mathbf{t}) \\hat{\\mathbf{i}}+(\\mathbf{3}-\\mathbf{t}) \\hat{\\mathbf{j}}+(2 \\mathbf{s}+\\mathbf{t}) \\hat{\\mathbf{k}}]=\\mathbf{1} \\mathbf{5}\\)
\nAnswer:
\nx(s – 2t) + y((3-t) +z(2s +t) = 15<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIn the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.<\/p>\n

(a) 2x + 3y + 4z – 12=O
\nAnswer
\nDirection ratio of any line normal to
\n2x + 3y + 4z – 12 = O is (2,3,4)
\n\u2234 Equation of line through (O, O, O) and is ‘\u22a5’ to given lines is
\n\\(\\frac{x-0}{2}=\\frac{y-0}{3}=\\frac{z-4}{4}\\)
\nany point on the lines is (2\u03bb, 3\u03bb, 4\u03bb) but this lines on the given plane
\n\"2nd<\/p>\n

(b) (1, 1, 0), (1, 2, 1), (- 2, 2, – 1)
\nAnswer
\nThe given equation passes through (1,1,0) equation of plane is (0, 3 , 4)
\nEquation of a lines through the orgin and is ‘\u22a5’ to given lines is
\n\\(\\frac{x-0}{0}=\\frac{y-0}{3}=\\frac{z-0}{4}=\\lambda\\) and any points on the line is (0, 3\u03bb, 4\u03bb) but this point lines on the given plane.
\nAnswer:
\n\"2nd<\/p>\n

(c) x + y + z = 1
\nAnswer
\nDirection of any line normal to the given plane is (1,1,1)
\nEquation of a lines through (0,0,0) and is ‘\u22a5’ to the given lines is
\n\\(\\frac{x-0}{1}=\\frac{y-0}{1}=\\frac{z-0}{1}=\\lambda\\)
\nany point on this line (\u03bb,\u03bb,\u03bb) but this point lies on the plane x +y + 2 = 1
\n\\(\\lambda+\\lambda+\\lambda=1 \\Rightarrow \\lambda=\\frac{1}{3}\\)
\n\\( \\text { foot of perpendicular is }\\left(\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3}\\right)\\)<\/p>\n

(d) 5y + 8 = 0
\nAnswer:
\nDirections of normal to the plane is (0,5,0) equation of a line through (0,0,0) and ‘\u22a5’ to the plane is \\(\\frac{x-0}{0}=\\frac{y-0}{5}=\\frac{z-0}{0}=\\lambda\\)
\nany point on this line is (0,5\u03bb,0) but this point lines on the plane
\n\"2nd<\/p>\n

Question 5.
\nFind the vector and cartesian equations of the planes<\/p>\n

(a) that passes through the point (1, 0, – 2) and the normal to the plane is \\(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}\\)
\nAnswer:
\nEquation of plane through \\(\\overline{\\mathbf{a}}\\) and normal to \\(\\overline{\\mathbf{x}} \\) is of the form \\((\\overline{\\mathbf{r}}-\\overline{\\mathbf{a}}) \\cdot \\overline{\\mathbf{x}}=\\mathbf{0}\\)
\n\\([(x i+y j+z \\bar{k})-(i+0-2 j)] \\cdot[i+j-\\bar{k}]=0\\) which is the vector equation
\n(x – 1) x 1 + y x 1 + (z + 2) (- 1 ) = 0
\nx + y- z- 3 = 0 which is cartesian equation.<\/p>\n

\"KSEEB<\/p>\n

(b) that passes through the point (1,4, 6) and the normal vector to the plane is
\n\\(\\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nFind the equations of the planes that passes through three points.
\n(a) (1, 1, – 1), (6, 4, – 5), (- 4, – 2, 3)
\nAnswer:
\nEquation of plane is 3 point form is
\n\"2nd
\nplane is 0 mean the plane is not fixed by the given points. That is points are collinear.<\/p>\n

(b) (1, 1, 0), (1, 2, 1), (- 2, 2, – 1)
\nAnswer:
\nThe given equation passes through (1,1,0) equation of plane is
\n(x-1) + b (y – 1) + c (z-0) = 0
\nbut this plane passes through (1,2, 1)
\na (1 – 1) + b (2 – 1) + c (1- 0)
\n= 0 b + c = 0 \u2192 (1)
\nalso it passes through (-2, 2,-1)
\na (-2 -1) + b (2 – 1) + c (-1 – 0)
\n= 0 -3a + b – c = 0 \u2192 (2)
\nsolving (1), (2)
\n\"2nd
\n\u2234 equation of plane
\n(x -1) + (y – 1) -3 (z – 0) = 0
\n2x + 3y – 3z – 5 = 0
\n[Note: Part (a) and part (b) are of same type question but in two different method. However students can choose any one method according to their wish.]<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the intercepts cut off by the plane 2x + y – z = 5.
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\nFind the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane. Answer:
\nEquation of ZOX plane is y = 0
\nDirection ratio of normal to this plane is (0,1,0)
\nsince the required plane is parallel to y = 0
\nDirection ratio of its normal is also (0,1,0).
\nIntercepts on y-axis is 3,
\n\u2234 The point (0, 3,0) lies on the plane.
\n\u2234 equation of plane
\n(x – 0) + 1 (y – 3) + 0 (z – 0) = 0
\n\u2234 y = 3.<\/p>\n

Question 9.
\nFind the equation of the plane through the intersection of the planes 3x-y + 2z-4 = 0 and x + y + z- 2 = 0 and the point (2,2,1).
\nAnswer:
\nLet the plane be
\nP1 <\/sub>: 3x – y + 2z – 4 = 0,
\nP2<\/sub> : x + y – z – 2 = 0
\nplane through the intersection be P1<\/sub> + \u03bb P2<\/sub> =0
\n(3x – y + 2z – 4) +\u03bb (x + y + z – 2) = 0
\nbut it passes through (2,2,1)
\n6 – 2+ 2 – 4) + X (2 + 2 + 1 – 2) = 0 \u21d2 2 + 3\u03bb = 0
\n\\(\\lambda=\\frac{-2}{3}\\)
\n\u2234 Equation of plane is
\n\\((3 x-y+2 z-4)+\\frac{-2}{3}(x+y+z-2)=0\\)
\n9x – 3y + 6z – 12 – 2x – 2y -2z + 4 = 0
\n7x – 5y + 4z – 8 = 0<\/p>\n

Question 10.
\nFind the vector equation of the plane passing through the intersection of the planes
\n\\(\\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}+2 \\hat{\\mathrm{j}}-3 \\hat{\\mathrm{k}})=7, \\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}+5 \\hat{\\mathrm{j}}+3 \\hat{\\mathrm{k}})=9 \\) and through the point (2,1,3).
\nAnswer:
\nLet the planes be
\nP1 <\/sub>: 2x + 2y -3z – 7 = 0,
\nP2 <\/sub>: 2x +5y + 3z = 9
\n\u2234 Required equation be P1 <\/sub>+ \u03bb\u00a0P2<\/sub> = 0
\n(2x + 2y – 3z – 7) + \u03bb (2x + 5y + 3z – 9) = 0,
\npasses through (2, 1, 3)
\n(4 + 2 – 9 – 7) + \u03bb (4 + 5 + 9 – 9) = 0
\n\\(-10+\\lambda 9=0 \\quad \\Rightarrow \\lambda=\\frac{10}{9}\\)
\nRequired plane is
\n(2x + 2y – 3z – 7) + \u03bb (2x + 5y + 3z – 9) = 0
\n18x + 18y -27z – 63 + 20x + 50y + 50z – 90 = 0
\n38x + 68y + 3z – 153 = 0<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
\nAnswer
\nLet P1 <\/sub>: x + y + z – 1 = 0 and
\nP2 <\/sub>: 2x + 3y + 4z -5 = 0
\nEquation of plane is P1<\/sub> + \u03bb P2<\/sub> = 0
\n(x + y + z – 1) + \u03bb (2x + 3y + 4z – 5) = 0
\nx (1+2\u03bb) + y(1+ 3X) +z(1 + 4),)-1 -5\u03bb = 0
\nDr\u2019s of the normal to the plane is …
\n1+ 2\u03bb ,1+3\u03bb ,1+4\u03bb
\nsince the plane is ‘\u22a5’ to x – y + z = 0
\n(1 + 2\u03bb) (1) + (1+3\u03bb) (-1) + (1+4\u03bb) 1= 0
\n\"2nd<\/p>\n

Question 12.
\nFind the angle,between the planes whose vector equations are
\n\\(\\overrightarrow{\\mathrm{r}} \\cdot(2 \\hat{\\mathrm{i}}+2 \\hat{\\mathrm{j}}-3 \\hat{\\mathrm{k}})=5 \\text { and } \\overrightarrow{\\mathrm{r}} \\cdot(3 \\hat{\\mathrm{i}}-3 \\hat{\\mathrm{j}}+5 \\hat{\\mathrm{j}}+5 \\hat{\\mathrm{k}})=3\\)
\nAnswer:
\nDr\u2019s of the normal to the planes are (2,2,-3) and (3,-3, 5)
\nLet \u03b8 be angle between them
\n\"2nd<\/p>\n

Question 13.
\nIn the following cases, determine whether the given planes are parallel or perpen\u00addicular, and in case they are neither, find the angles between them.<\/p>\n

(a) 7x + 5y + 6z + 30 = 0 and
\n3x – y – 10z + 4 = 0
\nAnswer
\nDr\u2019s of the normal to the planes are (7,5,6) and (3,-1,-10) since Dr\u2019s are neither proportional nor the product = 0
\nThey are neither parallel nor ‘\u22a5’. Hence they interset.
\nLet \u03b8 be the angle between them
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
\nAnswer
\nDr\u2019s are (2,1,3), (1,-2, 0)
\nsince product of direction
\n\u2234 2 x 1 + 1 x -2 + 3 x 0 = 0
\nplanes are perpendicular. (\u03b8= 90\u00b0)<\/p>\n

(c) 2x – 2y + 4z + 5 = 0 and
\n3x – 3y + 6z – 1 = 0
\nAnswer
\nDr\u2019s are (2,-2,4)), (3, -3, 6)
\n\\(\\frac{2}{3}=\\frac{-2}{-3}=\\frac{4}{6} \\Rightarrow \\frac{2}{3}=\\frac{2}{3}=\\frac{2}{3}\\)
\nSince Dr\u2019s are proportional, planes are parallel. (\u03b8 = 0\u00b0)<\/p>\n

(d) 2x – y + 3z – 1 = 0 and
\n2x – y + 3z + 3 = 0
\nAnswer
\nDr\u2019s are (2,-1,3), (2,-1,3)
\nSince Dr\u2019s are proportional, planes are parallel. (\u03b8 = 0\u00b0)
\n\\(\\frac{2}{2}=\\frac{-1}{-1}=\\frac{3}{3} \\Rightarrow 1=1=1\\)
\nSince Dr\u2019s are proportional, planes are parallel. (\u03b8 = 0\u00b0)<\/p>\n

(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
\nAnswer
\nDr\u2019s are (4,8,1), (0, 1, 1)
\nDr\u2019s are neither proportional nor the product = 0
\nhence they intersect.
\nLet \u03b8 be the angle between them.
\n\"2nd
\n\"2nd<\/p>\n

Question 14.
\nIn the following cases, find the distance of each of the given points from the
\ncorres\u00adponding given plane.
\nPoint\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Plane<\/p>\n

(a) (0, 0, 0), 3x – 4y + 12 z = 3
\nAnswer:
\n\"2nd<\/p>\n

(b) (3,-2,1), 2x – y + 2z + 3 = 0
\nAnswer:
\n\"2nd<\/p>\n

(c) (2, 3,-5), x + 2y – 2z = 9
\nAnswer:
\n\"2nd<\/p>\n

(d) (- 6, 0, 0) 2x – 3y + 6z – 2 = 0
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 11 Three Dimensional Geometry Ex 11.3 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 Ex 11.3 2nd PUC Maths …<\/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\/21429"}],"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=21429"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21429\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21429"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21429"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21429"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}