Exercise 11.3 Class 12 NCERT Solutions Question 2.<\/strong> \nFind the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \\(3\\hat { i } +5\\hat { j } -6\\hat { k }\\) \nSolution: \n <\/p>\n <\/p>\n
Question 3. \nFind the Cartesian equation of the following planes. \n(a) \\(\\overrightarrow { r } \\cdot (\\hat { i } +\\hat { j } -\\hat { k) }\\) = 2 \n(b) \\(\\overrightarrow { r } \\cdot (\\hat { 2i } +3\\hat { j } -4\\hat { k) }\\) = 1 \n(c) \\(\\overrightarrow { r } \\cdot [(s-2t)\\hat { i } +(3-t)\\hat { j } +(2s+t)\\hat { k) }\\) = 15 \nSolution: \nLet \\(\\vec{r}=x \\hat{i}+y \\hat{j}+z \\hat{k}\\) be the position vector of any point on the plane. \n\\(\\overrightarrow { r } \\cdot (\\hat { i } +\\hat { j } -\\hat { k) }\\) = 2 \n\\((x \\hat{i}+y \\hat{j}+z \\hat{k}) \\cdot(\\hat{i}+\\hat{j}-\\hat{k})\\) \nx + y – z = 2 is the cartesian equation.<\/p>\n
b. Let \\(\\vec{r}=x \\hat{i}+y \\hat{j}+z \\hat{k}\\) \n\u2234 \\(\\vec{r} \\cdot(2 \\hat{i}+3 \\hat{j}-4 \\hat{k})\\) = 1, \n\\((x \\hat{i}+\\hat{y j}+z \\hat{k}) \\cdot(2 \\hat{i}+3 \\hat{j}-4 \\hat{k})\\) = 1 \n2x + 3y – 4z = 1 is the cartesian equation.<\/p>\n
c. Let \\(\\vec{r}=x \\hat{i}+y \\hat{j}+z \\hat{k}\\) \n\\(\\vec{r} \\cdot((s-2 t) \\hat{i}+(3-t) \\hat{j}+(2 s+t) \\hat{k})\\) = 15 \n\\((x \\hat{i}+y \\hat{j}+z \\hat{k}) \\cdot((s-2 t) \\hat{i}+(3-t) \\hat{j}\\) + \\((2 s+t) \\hat{k})\\) = 15 \n(s – 2t)x + (3 – t)y + (2s + t)z = 15, is the cartesian equation.<\/p>\n
Question 4. \nIn the following cases find the coordinates of the foot of perpendicular drawn from the origin \n(a) 2x + 3y + 4z – 12 = 0 \n(b) 3y + 4z – 6 = 0 \n(c) x + y + z = 1 \n(d) 5y + 8 = 0 \nSolution: \nLet A be the foot of the perpendicular drawn from the orgin O. \nEquation of the plane is 2x + 3y + 4z – 12 = 0 \n\u2234 The d.r\u2019s of the normal to the plane are 2, 3, 4 \nEquation of the line OA is \n\\(\\frac{x-x_{1}}{a}=\\frac{y-y_{1}}{b}=\\frac{z-z_{1}}{c}\\) \n\\(\\frac{x-0}{2}=\\frac{y-0}{3}=\\frac{z-0}{4}\\) \ni.e. \\(\\frac{x}{2}=\\frac{y}{3}=\\frac{z}{4}=\\lambda\\) \n\u2234 Any point on the line OA is (2\u03bb, 3\u03bb, 4\u03bb) Let it be A. \nA satisfies the equation 2x + 3y + 4z – 12 = 0 \n4\u03bb + 9\u03bb + 16\u03bb – 12 = 0 \n29\u03bb – 12 = 0 \n <\/p>\n
Another Method: \nThe equation of the plane is 2x + 3y + 42 = 12 … (1) \nDividing (1) by \\(\\), we get \n\\(\\sqrt{2^{2}+3^{2}+4^{2}}=\\sqrt{29}\\) is of the form lx + my + nz = d \nThe direction cosines of the normal to the plane are l = \\(\\frac{2}{\\sqrt{29}}\\), m = \\(\\frac{3}{\\sqrt{29}}\\), n = \\(\\frac{4}{\\sqrt{29}}\\) \nand the distance from the origin to the plane d = \\(\\frac{12}{\\sqrt{29}}\\) \nFoot of the perpendicular = (Id, md, nd) \n= \\(\\left(\\frac{2}{\\sqrt{29}}\\left(\\frac{12}{\\sqrt{29}}\\right), \\frac{3}{\\sqrt{29}}\\left(\\frac{12}{\\sqrt{29}}\\right), \\frac{4}{\\sqrt{29}}\\left(\\frac{12}{\\sqrt{29}}\\right)\\right)\\) \n= \\(\\left(\\frac{24}{29}, \\frac{36}{29}, \\frac{48}{29}\\right)\\)<\/p>\n
b. The equation of the plane is , 3y + 4z = 6 … (1) \nDividing (1) by \\(\\sqrt{3^{2}+4^{2}}\\) = 5, we get \n\\(\\frac{3}{5} y+\\frac{4}{5} z=\\frac{6}{5}\\), is in the form lx + my + nz = d \n\u2234 l = 0, m = \\(\\frac { 3 }{ 5 }\\), n = \\(\\frac { 4 }{ 5 }\\), d = \\(\\frac { 6 }{ 5 }\\) \nThe foot of the perpendicular from the origin = (Id, md, nd) \n= \\(\\left(0\\left(\\frac{6}{5}\\right), \\frac{3}{5}\\left(\\frac{6}{5}\\right), \\frac{4}{5}\\left(\\frac{6}{5}\\right)\\right)\\) \n= \\(\\left(0, \\frac{18}{25}, \\frac{24}{25}\\right)\\)<\/p>\n
c. The equation of the plane is \nx + y + z = 1 … (1) \n <\/p>\n
d. Equation of the plane is \n5y = – 8 or – 5y = 8 \nDividing (1) by \\(\\sqrt{(-5)^{2}}\\) = 5, we get \n\\(\\frac{-5}{5} y=\\frac{8}{5} \\quad \\text { or } \\quad-y=\\frac{8}{5}\\), is in the form \nlx + my + nz = d \nl = 0, m = – 1, n = 0, d = \\(\\frac { 8 }{ 5 }\\) \nThe foot of the perpendicular from the origin = (Id, md, nd) \n= \\(\\left(0\\left(\\frac{8}{5}\\right),-1\\left(\\frac{8}{5}\\right), 0\\left(\\frac{8}{5}\\right)\\right)\\) \n= \\(\\left(0, \\frac{-8}{5}, 0\\right)\\)<\/p>\n
<\/p>\n
Question 5. \nFind the vector and cartesian equation of the planes \n(a) that passes through the point (1,0, -2) and the normal to the plane is \\(\\hat { i } +\\hat { j } -\\hat { k }\\) \n(b) that passes through the point (1,4,6) and the normal vector to the plane is \\(\\hat { i } -2\\hat { j } +\\hat { k }\\) \nSolution: \n(a) The plane passes through the point (1, 0, – 2) \n\u2234 \\(\\vec{a}=\\hat{i}+0 \\hat{j}-2 \\hat{k}\\) \nThe normal vector to the plane \\(\\vec{n}=\\hat{i}+\\hat{j}-\\hat{k}\\) \nThe vector equation of the plane passing through \\(\\vec{a}\\) and perpendicular to \\(\\vec{n}\\) is (r – a).n = 0 \n \nThe cartesian equation is x + y – z – 3 = 0<\/p>\n
(b) The plane passes through (1, 4, 6) and the direction ratios of the normal to the plane are 1, -2, 1. \nThe cartesian equation of the plane passing through (x1<\/sub>, T1<\/sub>, Z1<\/sub>) and perpendicular to the line with direction ratios a, b, c is \na(x – x1<\/sub>) + b(y – y1<\/sub>) + c(z – z1<\/sub>) = 0 \nHence the cartesian equation of the plane is 1 (x – 1) – 2(y – 4) + 1(z – 6) = 0 \nx – 2y + z + 1 = 0 \nThe vector equation of the plane is \n\\(\\vec{r} \\cdot(\\hat{i}-2 \\hat{j}+\\hat{k})+1=0\\)<\/p>\nQuestion 6. \nFind the equations of the planes that passes through three points \n(a) (1, 1, – 1) (6, 4, – 5), (- 4, -2, 3) \n(b) (1, 1, 0), (1, 2, 1), (-2, 2, -1) \nSolution: \nLet the equation of plane passing through (1, 1, – 1) be \na(x – 1) + b(y – 1) + c(z + 1 ) = 0 … (1) \nSince (6, 4, – 5) is a point on (1), we get \na(6 – 1) + b(4 – 1) + c(- 5 + 1) = 0 \ni.e., 5a + 3b – 4c = 0 … (2) \nSince (- 4, – 2, 3) is a point on (1), we get \na(- 4 – 1) + b(- 2 – 1) + c(3 + 1) = 0 \n– 5a – 3b + 4c = 0 \n5a + 3b – 4c = 0 … (3) \nFrom (2) and (3), by the rule of cross multiplication, we get \n\\(\\frac{a}{-12+12}=\\frac{b}{-20+20}=\\frac{c}{15-15}\\) \n\\(\\frac{a}{0}=\\frac{b}{0}=\\frac{c}{0}\\) \na = 0, b = 0 and c = 0 \nHence, there is no unique plane passing through the given points.<\/p>\n
Another Method: \nThe plane passes through \n(1, 1, – 1),(6, 4, – 5), (- 4, – 2, 3) \nThe equation of the plane passing through \n\\(\\left(x_{1}, y_{1}, z_{1}\\right),\\left(x_{2}, y_{2}, z_{2}\\right) \\text { and }\\left(x_{3}, y_{3}, z_{3}\\right)\\) is \n \nR2<\/sub> and R3<\/sub> are proportional. \nHence we get 0 = 0 \nSo we cannot find the equation of the plane passing through these points in a unique way.<\/p>\n <\/p>\n
Question 7. \nFind the intercepts cut off by the plane 2x+y-z = 5. \nSolution: \nEquation of the plane is 2x + y- z = 5 x y z \nDividing by 5: \n\\(\\Rightarrow \\frac { x }{ \\frac { 5 }{ 2 } } +\\frac { y }{ 5 } -\\frac { z }{ -5 }\\) = 1 \n\u2234 The intercepts on the axes OX, OY, OZ are \\(\\frac { 5 }{ 2 }\\), 5, – 5 respectively.<\/p>\n
Question 8. \nFind the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane. \nSolution: \nAny plane parallel to ZOX plane is y=b where b is the intercept on y-axis. \n\u2234 b = 3. \nHence equation of the required plane is 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). \nSolution: \nGiven planes are: \n3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 \nAny plane through their intersection is \n3x – y + 2z – 4 + \u03bb(x + y + z – 2) = 0 \npoint (2, 2, 1) lies on it, \n\u2234 3 x 2 – 2 + 2 x 1 – 4 + \u03bb(2 + 2 + 1 – 2) = 0 \n\u21d2 \u03bb = \\(\\frac { -2 }{ 3 }\\) \nNow required equation is 7x – 5y + 4z – 8 = 0<\/p>\n
Question 10. \nFind the vector equation of the plane passing through the intersection of the planes \\(\\overrightarrow { r } \\cdot \\left( 2\\hat { i } +2\\hat { j } -3\\hat { k } \\right) =7,\\overrightarrow { r } \\cdot \\left( 2\\hat { i } +5\\hat { j } +3\\hat { k } \\right) =9\\) and through the point (2, 1, 3). \nSolution: \nHere \\(\\vec{n}_{1}=2 \\hat{i}+2 \\hat{j}-3 \\hat{k}, d_{1}=7\\) \n\\(\\vec{n}_{2}=2 \\hat{i}+5 \\hat{j}+3 \\hat{k}, d_{2}=9\\) \nThe required vector equation is \n \nis the required vector equation of the plane.<\/p>\n
Another method: \nThe equation of the planes are \n2x + 2y – 3z = 7 … (1) \n2x + 5y + 3z = 9 … (2) \nThe equation of the plane through the inter-section of (1) and (2) is (2x + 2y – 3z) + \u03bb(2x + 5y + 3z) = 7 + 9\u03bb … (3) \nSince the plane (3) passes through (2, 1, 3), we get \n(2(2) + 2(1) – 3(3)) + \u03bb (2(2) + 5(1) + 3(3)) = 7 + 9\u03bb \n– 3 + 18\u03bb = 7 + 9\u03bb \n9 \u03bb = 10 \u2234 \u03bb = \\(\\frac { 10 }{ 9 }\\) \n(3) gives the cartesian equation of the plane. \n(3) \u2192 (2x + 2y – 3z) + \\(\\frac { 10 }{ 9 }\\) (2x + 5y + 3z) \n= 7 + 9(\\(\\frac { 10 }{ 9 }\\)) \nMultiplying by 9, we get 18x + 18y – 27z + 20x + 50y + 30z = 63 + 90 38x + 68y + 3z = 153 \nThe vector equation is \n\\(\\bar{r} \\cdot(38 \\hat{i}+68 \\hat{j}+3 \\hat{k})\\) = 153<\/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 = o. \nSolution: \nThe equation of the required plane is (x + y + z – 1) + \u03bb(2x + 3y + 4z – 5) = 0 \n <\/p>\n
Question 12. \nFind the angle between the planes whose vector equations are \\(\\overrightarrow { r } \\cdot \\left( 2\\hat { i } +2\\hat { j } -3\\hat { k } \\right) =5,\\overrightarrow { r } \\cdot \\left( 3\\hat { i } -3\\hat { j } +5\\hat { k } \\right)\\) = 3 \nSolution: \nThe angle \u03b8 between the given planes is \n <\/p>\n
<\/p>\n
Question 13. \nIn the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them. \n(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0 \n(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0 \n(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0 \n(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0 \n(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0. \nSolution: \na. Comparing with the general equation, we get \na1<\/sub> = 7 b1<\/sub> = 5 c1<\/sub> = 6 and \na2<\/sub> = 3 b2<\/sub> = – 1 c2<\/sub> = – 10 \na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 7(3) + 5(- 1) + 6(- 10) \u2260 0 \n\u2234 The lines are not perpendicular \n <\/p>\nb. a1<\/sub> = 2 b1<\/sub> = 1 c1<\/sub> = 3 and \na2<\/sub> = 1 b2<\/sub> = – 2 c2<\/sub> = 0 \na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 2 – 2 + 0 = 0 \n\u2234 The two lines are perpendicular.<\/p>\nc. a1<\/sub> = 2 b1<\/sub> = – 2 c1<\/sub> = 4 and \na2<\/sub> = 3 b2<\/sub> = – 3 c2<\/sub> = 6 \n\\(\\frac{a_{1}}{a_{2}}=\\frac{2}{3}, \\frac{b_{1}}{b_{2}}=\\frac{2}{3}, \\frac{c_{1}}{c_{2}}=\\frac{4}{6}=\\frac{2}{3}\\) \ni.e., \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\) \n\u2234 The two lines are parallel.<\/p>\nd. a1<\/sub> = 2 b1<\/sub> = – 1 c1<\/sub> = 3 and \na2<\/sub> = 2 b2<\/sub> = – 1 c2<\/sub> = 3 \n\\(\\frac{a_{1}}{a_{2}}=1, \\frac{b_{1}}{b_{2}}=1, \\frac{c_{1}}{c_{2}}=1\\) \n\u2234 \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\) \n\u2234 The two lines are parallel.<\/p>\ne. a1<\/sub> = 4 b1<\/sub> = 8 c1<\/sub> = 1 and \na2<\/sub> = 0 b2<\/sub> = 1 c2<\/sub> = 1 \na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 2(3) + – 2(3) + 4(6) \u2260 0 \n\u2234 The two lines are perpendicular. \n <\/p>\nQuestion 14. \nIn the following cases, find the distance of each of the given points from the corresponding given plane. \nPoint\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Plane \n(a) (0, 0,0) 3x – 4y + 12z = 3 \n(b) (3,-2,1) 2x – y + 2z + 3 = 0. \n(c) (2,3,-5) x + 2y – 2z = 9 \n(d) (-6,0,0) 2x – 3y + 6z – 2 = 0 \nSolution: \n(a) Given plane: 3x – 4y + 12z – 3 = 0 \n\u2234 The distance from the point (0, 0, 0) to \n <\/p>\n
b. The equation of the plane is \n2x – y + 2z + 3 = 0 \n\u2234 The distance from the point (3, – 2, 1) to the given plane \n= \\(\\left|\\frac{2(3)-1(-2)+2(1)+3}{\\sqrt{4+1+4}}\\right|=\\frac{13}{3}\\)<\/p>\n
c. The equation of the plane is x + 2y – 2z – 9 = 0 \na = 1, b = 2, c = – 2, d = 9 \n\u2234 The distance from the point (2, 3, – 5) to the plane \n= \\(\\left|\\frac{a x_{1}+b y_{1}+c z_{1}-d}{\\sqrt{a^{2}+b^{2}+c^{2}}}\\right|=\\left|\\frac{2+2(3)-2(-5)-9}{\\sqrt{1+4+4}}\\right|\\) \n= \\(\\frac{9}{3}\\) = 3<\/p>\n
d. The equation of the plane is 2x – 3y + 6z – 2 = 0 \n\u2234 The perpendicular distance from the point (- 6, 0, 0) to the plane \n= \\(\\left|\\frac{2(-6)-3(0)+6(0)-2}{\\sqrt{4+9+36}}\\right|\\) \n= \\(\\left|\\frac{14}{\\sqrt{49}}\\right|=\\frac{14}{7}\\) = 2<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-ex-11-3\/ NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3 Ex 11.3 Class 12 NCERT Solutions Question 1. In each of the following cases, determine the …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n