{"id":35096,"date":"2022-08-25T10:22:02","date_gmt":"2022-08-25T04:52:02","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35096"},"modified":"2022-08-25T10:22:02","modified_gmt":"2022-08-25T04:52:02","slug":"inter-1st-year-maths-1b-the-plane-solutions-ex-7a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-the-plane-solutions-ex-7a\/","title":{"rendered":"Inter 1st Year Maths 1B The Plane Solutions Ex 7(a)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B The Plane Solutions Exercise 7(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B The Plane Solutions Exercise 7(a)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the equation of the plane if the foot of the perpendicular from origin to the plane is (1, 3, -5).
\nSolution:
\nOP is the normal to the plane D. Rs of op are 1, 3, -5
\nThe plane passes through P( 1, 3, -5) equation of the plane is
\n1(x – 1) + 3(y – 3) – 5(z + 5) = 0
\nx – 1 + 3y – 9 – 5z – 25 = 0
\nx + 3y – 5z – 35 = 0
\n\"Inter<\/p>\n

Question 2.
\nReduce the equation x + 2y – 3z – 6 = 0 of the plane to the normal form.
\nSolution:
\nEquation of the plane is x + 2y – 3z – 6 = 0
\ni.e., x + 2y – 3z = 6
\nDividing in the
\n\\(\\sqrt{1^{2}+2^{2}+(-3)^{2}}=\\sqrt{1+4+9}=\\sqrt{14}\\)
\nThe equation of the plane in the normal form is
\n\"Inter<\/p>\n

Question 3.
\nFind the equation of the plane. Whose intercepts on X, Y, Z – axis are 1, 2, 4 respectively.
\nSolution:
\nEquation of the plane in the intercept form is
\n\\(\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}\\) = 1
\nGiven a = 1, b = 2, c = 4.
\nEquation of the required plane in the intercept form is
\n\\(\\frac{x}{1}+\\frac{y}{2}+\\frac{z}{4}\\) = 1
\nMultiplying with 4, we get
\n4x + 2y + z = 4<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the co-ordinate axes.
\nSolution:
\nEquation of the plane is 4x + 3y – 2z + 2 = 0
\n– 4x – 3y + 2z = 2
\n\"Inter
\nx – intercept = -1\/2, y – intercept = -2\/3, z intercept =1.<\/p>\n

Question 5.
\nFind the d.c.’s of the normal to the plane x + 2y + 2z – 4 = 0.
\nSolution:
\nEquation of the plane is x + 2y + z- 4 = 0
\nd.r.’s of the normal are (1, 2, 2)
\nDividing with \\(\\sqrt{1+4+4}\\) = 3,
\nd.c.’s of the normal to the plane are (\\(\\frac{1}{3}, \\frac{2}{3}, \\frac{2}{3}\\))<\/p>\n

Question 6.
\nFind the equation of the plane passing through the point (-2, 1, 3) and having (3, -5, 4) as d.r.\u2019s of its normal.
\nSolution:
\nd.r.’s of the normal are (3, -5, 4) and the plane passes through (-2, 1, 3).
\nEquation of the plane is 3(x + 2) – 5(y – 1) + 4(z – 3) = 0
\n3x + 6 – 5y + 5 + 4z – 12 = 0
\n3x -5y+ 4z – 1 = 0.<\/p>\n

Question 7.
\nWrite the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form.
\nSolution:
\nEquation of the plane is
\n4x – 4y + 2z + 5 = 0
\n– 4x + 4y – 2z = 5
\n\\(-\\frac{4x}{5}, \\frac{4y}{5}, \\frac{2z}{5}\\)
\nIntercept form is
\n\"Inter
\nx – intercept = \\(\\frac{5}{4}\\), y – intercept = \\(\\frac{5}{4}\\), z – intercept = –\\(\\frac{5}{2}\\)<\/p>\n

Question 8.
\nFind the angle between the planes x + 2y + 2z-5 = 0 and 3x + 3y + 2z – 8 = 0.
\nSolution:
\nEquation of the planes are
\nx + 2y + 2z – 5 = 0
\n3x + 3y + 2z – 8 = 0
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the equation of the plane passing through the point (1,1,1) and parallel to the plane x +.2y + 3z – 7 = 0.
\nSolution:
\nEquation of the given plane is x + 2y + 3z – 7 = 0.
\nEquation of the parallel plane is x + 2y + 3z = k.
\nThis plane passes through the point P (1, 1, 1)
\n1 + 2 + 3 = k \u21d2 k = -6
\nEquation of the required plane is x + 2y + 3z = 6<\/p>\n

Question 2.
\nFind the equation of the plane passing through (2, 3, 4) and perpendicular to x – axis.
\nSolution:
\nThe plane is perpendicular to x – axis
\n\u2234 x – axis is the normal to the plane
\nd.c.’s of x -axis are 1, 0, 0
\nEquation of the required plane is x = k
\nThis plane passes through the point P(2, 3, 4)
\n\u2234 2 = k
\nEquation of the required plane is x = 2.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that 2x + 3y + 7 = 0 represents a plane perpendicular to XY – plane.
\nSolution:
\nEquation of the given plane is 2x + 3y + 7 = 0
\nEquation of xy – plane is z = 0
\na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 2.0 + 3.0 + 0.1 = 0
\n+0+0=0
\nThe plane 2x + 3y + 7 = 0 is perpendicular to XY – plane.<\/p>\n

Question 4.
\nFind the constant k so that the planes x – 2y + kz = 0 and 2x + 5y – z = 0 are at right angles. Find the equation of the plane through (1, -1, -1) and perpen-dicular to these planes.
\nSolution:
\nEquations of the given planes are x – 2y + kz = 0 and 2x + 5y – z = 0
\nThese the planes are perpendicular
\n1.2 – 2.5 + k (-1) = 0
\n2 – 10 = k \u21d2 k = -8
\nEquation of the planes
\nx – 2y – 8z = 0 ………. (1)
\n2x + 5y – z = 0 ……….. (2)
\nThe required plane passes through (1, -1, -1)
\n\u2234 Equation of the plane can be taken as
\na(x – 1) + b(y + 1) + c(z + 1) = 0 ………… (3)
\nThis plane is perpendicular to the planes (1) and (2)
\na – 2b – 8c = 0
\n2a + 5b – c = 0
\n\"Inter
\nSubstituting in (3), equation of the required planes
\n42 (x – 1) – 15(y + 1) + 9(z + 1) = 0
\n42x – 42 – 15y – 15 + 9z + 9 = 0
\n42x – 15y + 9z – 48 = 0.<\/p>\n

Question 5.
\nFind the equation of the plane through (-1, 6, 2) and perpendicular to the join of (1, 2, 3) and (-2, 3, 4).
\nSolution:
\nThe plane is perpendicular to the line joining A(1, 2, 3) and B(-2, 3, 4).
\nd.r.’s of AB are 1 + 2, 2 – 3, 3 – 4
\ni.e., 3, -1, -1
\n\"Inter
\nAB is normal to the plane and the plane passes through the point P(-1, 6, 2)
\nEquation of the required plane is 3(x + 1) – 1(y – 6) -1(z – 2) = 0
\n3x + 3 – y + 6- z + 2 = 0
\n3x – y – z + 11 = 0<\/p>\n

Question 6.
\nFind the equation of the plane bisecting the line segment joining (2, 0, 6) and (-6, 2, 4) and perpendicular to it.
\nSolution:
\nA (2, 0, 1), B(-6, 2, 4) are the given points ‘o’ is the mid point of AB
\n\"Inter
\nCo-ordinates of O are
\n\\(\\left(\\frac{2-6}{2}, \\frac{0+2}{2}, \\frac{6+4}{2}\\right)\\) = (-2, 1, 5)
\nThe plane is perpendicular to AB
\nd.r.’s of the normal to the plane are
\n2 + 6, 0 – 2, 6 – 4
\n8, -2, 2
\nEquation of the required plane is
\n+8 (x + 2) – 2(y – 1) + 2 (z – 5) = 0
\n8x + 16 – 2y + 2 + 2z – 10 = 0
\n8x – 2y + 2z + 8 = 0<\/p>\n

Question 7.
\nFind the equation of the plane passing through (0, 0, -4) and perpendicular to the line joining the points (1, -2, 2) and (-3, 1, -2).
\nSolution:
\nA(1, -2, 2), B(-3, 1, -2) are the given points
\nd.r.’s of AB are 1 + 3, -2 -1, 2 + 2 i.e., 4, -3, 4
\nAB is normal to the plane and the plane passes through the point P(0, 0, -4).
\nEquation of the required plane is
\n4(x – 0) – 3(y – 0) + 4(z + 4) = 0
\n4x – 3y + 4z + 16 = 0<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind the equation of the plane through (4, 4, 0) and perpendicular to the planes 2x + y + 2 z + 3 = 0 and 3x + 3y + 2z – 8 = 0.
\nSolution:
\nEquation of the plane passing through P(4, 4, 0) is
\na(x – 4) + b(y – 4) + c(z – 0) = 0 ………. (1)
\nThis plane is perpendicular to
\n2x + y + 2z – 3 = 0
\n3x + 3y + 2z – 8 = 0
\n\u2234 2a + b + 2c = 0 ………… (2)
\n3a + 3b + 2c = 0 ………….(3)
\n\"Inter
\nSubstituting in (1), equation of the required plane is
\n-4 (x – 4) + 2(y – 4) + 3(z – 0) = 0
\n-4x + 16 + 2y – 8 + 3z = 0
\n-4x + 2y + 3z + 8 = 0<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nFind the equation of the plane through the points (2, 2, -1), (3, 4, 2), (7, 0, 6).
\nSolution:
\nA(2, 2, -1), B(3, 4, 2), C(7, 0, 6) are the given points.
\nEquation of the plane passing through A(2, 2, -1) is
\na(x – 2) + b(y – 2) + c(z + 1) = 0
\nThis plane passes through B(3, 4, 2) and C(7, 0, 6)
\na(3 – 2) + b(4 – 2) +c(2 + 1) = 0
\na + 2b + 3c = 0 ……….. (2)
\na(7 – 2) + b(0 – 2) + c(6 + 1) = 0
\n5a – 2b + 7c = 0 ……….. (3)
\nFrom (2) and (3)
\n\"Inter
\nSubstituting in (1) equation of the required plane is
\n5(x- 2) + 2(y- 2) – 3(z +1) = 0
\n5x – 10 + 2y – 4 – 3z – 3 = 0
\n5x + 2y – 3z -17 = 0 or 5x + 2y- 3z = 17<\/p>\n

Question 2.
\nShow that the points (0, -1, 0), (2, 1, -1), (1, 1, 1), (3, 3, 0) are coplanar.
\nSolution:
\nEquation of the plane through A(0, -1, 0) is
\nax + b(y + 1) + cz = 0 …….. (1)
\nThis plane passes through B(2, 1, -1) and C(1, 1, 1)
\n2a + 2b – c = 0 …………. (2)
\na + 2b + c = 0 ………….. (3)
\n(2) – (3) gives a – 2c = 0 \u21d2 a = 2c \u21d2 \\(\\frac{a}{2}=\\frac{c}{1}\\)
\n(2) + (3) gives 3a + 4b = 0 \u21d2 3a = – 4b
\n\u21d2 \\(\\frac{a}{4}=\\frac{b}{-3}\\)
\n\u2234 \\(\\frac{a}{4}=\\frac{b}{-3}=\\frac{c}{2}\\)
\nSubstitutes in (1) equation of the plane ABC is
\n4x – 3(y + 1) + 2(z – 0) = 0
\n4x – 3y + 2z – 3 = 0
\n4x – 3y + 2z – 3 = 4.3 – 3.3. + 0.3
\n= 12 – 9 – 3 = 0
\nThe given points A, B, C, D are coplanar.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the equation of the plane through (6, – 4, 3), (0, 4, -3) and cutting of inter-cepts whose sum is zero.
\nSolution:
\nSuppose a, b, c are the intercepts of the plane.
\nEquation of the plane is \\(\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}\\) = 1
\nGiven a + b + c = 0
\nc = – (a + b)
\nThe plane passes through
\nP(6, -4, 3), Q(0, 4, -3)
\n\"Inter
\nc = -a – b = -3 – b
\n4(-3 – b) – 3b = b(-3 – b)
\n-12 – 4b – 3b = – 3b – b\u00b2
\nb\u00b2 – 4b – 12 = 0
\n(b – 6) (b + 2) = 0 \u21d2 b = 6 – 2<\/p>\n

Case i) b = 6
\nc = -3 – b = -3 – 6 = -9
\nEquation of the plane is
\n\\(\\frac{x}{3}+\\frac{y}{6}+\\frac{z}{9}\\) = 1
\n6x + 3y – 2z = 18<\/p>\n

Case ii) b = -2
\nc = -3 – b = -3 + 2 = -1
\nEquation of the plane is
\n\\(\\frac{x}{3}+\\frac{y}{-2}+\\frac{z}{-1}\\) = 1<\/p>\n

Question 4.
\nA plane meets the co-ordinate axes in A, B, C. If the centroid of \u2206ABC is (a, b, c). Show that the equation of the plane is
\n\\(\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}\\) = 1
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3 are the intercepts of the plane ABC.
\nEquation of the plane is the intercept form is
\n\\(\\frac{x}{\\alpha}+\\frac{y}{\\beta}+\\frac{z}{\\gamma}=1\\) ………… (1)
\n\"Inter
\nCo-ordinates of A are (\u03b1, 0, 0), B are (0, \u03b2, 0) and C are (0, 0, \u03b3)
\nG is the centroid of \u2206ABC
\nCo-ordinates of Gare =
\n\"Inter
\n\u03b1 = 3a, \u03b2 = 3b, \u03b3 = 3c
\nSubstituting in (1), equation of the plane ABC is
\n\"Inter<\/p>\n

Question 5.
\nShow that the plane through (1, 1, 1) , (1, -1, 1) and (-7, -3, -5) is parallel . to Y-axis.
\nSolution:
\nEquation of the plane through A (1, 1, 1) can be taken as
\na(x – 1) +b(y – 1) + c(z – 1) = 0 ……….. (1)
\nThis plane passes through B(1, -1, 1) and C(-7, -3, -5)
\n0 – 2b + 0 = 0 \u21d2 b – 0
\nEquation of zy – plane is y = 0
\n0.x + 1.y + 0.z = 0
\na.0 + 0.1 + c.0 = 0
\nThe required plane is perpendicular to zx – plane hence it is parallel to Y – axis.<\/p>\n

\"Inter<\/p>\n

Question 6.
\nShow that the equations ax + by + r =0, by + cz + p = 0, cz + ax + q = 0 represent planes perpendicular to xy,yz, zx planes respectively.
\nSolution:
\nEquation of the given plane is
\nax + by + c = Q
\nd.rs of the normal are (a, b, c)
\nEquation of XY – plane is z = 0
\nd.rs of the normal are (0, 0, 1)
\na.0 + b.0 + 0.1 = 0
\n\u2234 ax + by + r = 0 is perpendicular to xy – plane.
\nSimilarly we can show that by + cz + p = 0 is perpendicular to yz – plane and cz + ax + q – 0 is perpendicular to zx – plane.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Plane Solutions Exercise 7(a) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B The Plane Solutions Exercise 7(a) I. Question 1. Find the equation of the plane if the foot of the perpendicular from origin to …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35096"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35096"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35096\/revisions"}],"predecessor-version":[{"id":35097,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35096\/revisions\/35097"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35096"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35096"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35096"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}