NCERT Solutions for Class 12 Maths<\/a> Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise<\/h2>\n <\/p>\n
Question 1. \nShow that the line joining the origin to the point (2,1,1) is perpendicular to the line de-termined by the points (3, 5, – 1), (4, 3, -1). \nSolution: \nThe direction ratios of the line joining (0, 0, 0) and (2, 1, 1) are 2, 1,1 The direction ratios Of the line joining (3, 5, -1) and (4, 3, -1) are 1, -2, 0 \n\u2234 a1<\/sub>a2<\/sub> + b1<\/sub>b1<\/sub> + C1<\/sub>C2<\/sub> \n= 2(1) + 1(-2) + 1 (0) = 0 \nThe lines are perpendicular.<\/p>\nQuestion 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 m1<\/sub>n2<\/sub> – m2<\/sub> n1<\/sub>, n1<\/sub> l2<\/sub> – n2<\/sub> l1<\/sub>, l1<\/sub> m2<\/sub> – l2<\/sub> m1<\/sub> \nSolution:. \nLet \\(\\hat{a}\\) and \\(\\hat{b}\\) be unit vectors in the direction of the lines \n\u2234 \\(\\hat{a}=l_{1} \\hat{i}+m_{1} \\hat{j}+n_{1} \\hat{k}\\) \n\\(\\hat{b}=l_{2} \\hat{i}+m_{2} \\hat{j}+n_{2} \\hat{k}\\) \nSince a and b are mutually perpendicular, then \\(\\hat{a}\\) x \\(\\hat{b}\\) is a unit vector perpendicular to \\(\\hat{a}\\) and \\(\\hat{b}\\) \n \n\u2234 The direction cosines are m1<\/sub>n2<\/sub> – m2<\/sub> n1<\/sub>, n1<\/sub> l2<\/sub> – n2<\/sub> l1<\/sub>, l1<\/sub> m2<\/sub> – l2<\/sub> m1<\/sub><\/p>\nQuestion 3. \nFind the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. \nSolution: \na1<\/sub> = a, b1<\/sub> = b, c1<\/sub> = c \na2<\/sub> = b – c, b2<\/sub> = c – a, c2<\/sub> = a – b \nLet \u03b8 be the angle between the lines \n <\/p>\nQuestion 4. \nFind the equation of a line parallel to x- axis and passing the origin. \nSolution: \nThe direction ratios of any line parallel to x-axis is 1, 0, 0. Hence the equation of the line passing through the origin having direction ratios 1,0,0 is \n\\(\\frac{x-0}{1}=\\frac{y-0}{0}=\\frac{z-0}{0}\\) \ni.e., \\(\\frac{x}{1}=\\frac{y}{0}=\\frac{z}{0}\\)<\/p>\n
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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 line AB and CD. \nSolution: \nThe direction ratios of AB \n= 4 – 1, 5 – 2, 7 – 3 = 3, 3, 4 \nThe direction ratios of CD = 6, 6, 8 \nLet \u03b8 be the angle between AB and CD \n <\/p>\n
Question 6. \nIf the lines \\(\\frac{x-1}{-3}=\\frac{y-2}{2 k}=\\frac{z-3}{2}\\) and \\(\\frac{x-1}{3 k}=\\frac{y-1}{1}=\\frac{z-6}{-5}\\) are perpendicular, find the value of k. \nSolution: \nThe equation of the lines are \n\\(\\frac{x-1}{-3}=\\frac{y-2}{2 k}=\\frac{z-3}{2}\\) and \n\\(\\frac{x-1}{3 k}=\\frac{y-1}{1}=\\frac{z-6}{-5}\\) \nThe direction ratios of 1st<\/sup> line = – 3, 2k, 2 \nThe direction ratios of IInd<\/sup> line = 3k, 1, – 5 \nSince the two lines are perpendicular, \na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 0 \n– 3(3k) + 2k(1) + 2(- 5) = 0 \n– 9k + 2k – 10 = 0 \n– 7k = 10 \nk = \\(\\frac { -10 }{ 7 }\\)<\/p>\nQuestion 7. \nFind the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \\(\\vec{r} \\cdot(\\hat{i}+2 \\hat{j}-5 \\hat{k})+9\\) = 0. \nSolution: \nLet \\(\\hat{a}\\) be the position vector of the point (1, 2, 3) \na = \\(\\hat{i}+2 \\hat{j}+3 \\hat{k}\\) \nThe direction ratios of the normal to the plane are 1, 2, – 5. \nLet b = \\(\\hat{i}+2 \\hat{j}-5 \\hat{k}\\) \nLet \\(\\hat{r}\\) be the position vector of any point on the line. \nThe vector equation is \\(\\vec{r}=\\vec{a}+\\lambda \\vec{b}\\) \ni.e., \\(\\vec{r}=(\\hat{i}+2 \\hat{j}+3 \\hat{k})+\\lambda(\\hat{i}+2 \\hat{j}-5 \\hat{k})\\)<\/p>\n
Question 8. \nFind the equation of the plane passing through (a, b, c) and parallel to the plane \\(\\vec{r} \\cdot(\\hat{i}+\\hat{j}+\\hat{k})\\) = 2 \nSolution: \nThe given plane is \\(\\vec{r} \\cdot(\\hat{i}+\\hat{j}+\\hat{k})\\) = 2 \ni.e., x + y + z – 2 \nThe equation of any plane parallel to the given plane is x +y + z = k … (1) \nSince the required plane passes through the point (a, b, c), we get a + b + c = k \nThus (1) gives the required plane as x + y + z = a + b + c<\/p>\n
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Question 9. \nFind the shortest distance between the lines \\(\\vec{r}=6 \\hat{i}+2 \\hat{j}+2 \\hat{k}+\\lambda(\\hat{i}-2 \\hat{j}+2 \\hat{k})\\) and \\(\\vec{r}=-4 \\hat{i}-\\hat{k}+\\mu(3 \\hat{i}-2 \\hat{j}-2 \\hat{k})\\) \nSolution: \n <\/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. \nSolution: \nThe equation of the line through the points (5,1, 6) and (3,4, 1) is \n\\(\\frac{x-5}{3-5}=\\frac{y-1}{4-1}=\\frac{z-6}{1-6}=\\lambda\\) \ni.e., \\(\\frac{x-5}{-2}=\\frac{y-1}{3}=\\frac{z-6}{-5}=\\lambda\\) \ni.e., x = – 2\u03bb. + 5, y = 3\u03bb + 1, z = – 5\u03bb + 6 \nSince this line crosses the YZ – plane, x coordinate is zero \ni.e., – 2\u03bb + 5 = 0 \u2234 \u03bb = \\(\\frac { 5 }{ 2 }\\) \n\\(y=3\\left(\\frac{5}{2}\\right)+1=\\frac{17}{2}\\) \n\\(z=-5\\left(\\frac{5}{2}\\right)+6=\\frac{-13}{2}\\) \n\u2234 The required point is \\(\\left(0, \\frac{17}{2}, \\frac{-13}{2}\\right)\\)<\/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. \nSolution: \nThe equation of the line through (5, 1, 6) and (3, 4, 1) is \n\\(\\frac{x-5}{-2}=\\frac{y-1}{3}=\\frac{z-6}{-5}=\\lambda\\) \ni.e., x = – 2\u03bb + 5, y = 3\u03bb + 1, z = – 5\u03bb + 6 \nSince this line crosses the ZX – plane, \ny coordinate is zero. \n <\/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. \nSolution: \nThe equation of the line through (3, – 4, – 5) and (2, – 3, 1) is \n \nThe coordinate of any point P on the line is (- \u03bb + 3, \u03bb – 4, 6\u03bb – 5) \nP is a point on the plane 2x + y + z = 7 \ni.e., 2(- \u03bb + 3) + (\u03bb – 4) + (6\u03bb – 5) = 7 \n– 2\u03bb + 6 + \u03bb – 4 + 6\u03bb – 5 = 7 \n5\u03bb – 3 = 7 \u2234 \u03bb = 2 \n\u2234 P is (- 2+ 3, 2 – 4, 6(2) – 5) = (1, – 2, 7) \n\u2234 The coordinates of the point which crosses the plane is (1, -2, 7)<\/p>\n
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Question 13. \nFind the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y +3z = 5 and 3x + 3y + z = 0. \nSolution: \nThe plane passes through (- 1, 3, 2) \n\u2234 x1<\/sub> = – 1, y1<\/sub> = 3, z1<\/sub> = 2 \nThe direction ratios of the normal to the planes are 1, 2, 3 and 3, 3, 1. \nHence the equation of the plane is \n\\(\\left|\\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\\\ a_{1} & b_{1} & c_{1} \\\\ a_{2} & b_{2} & c_{2} \\end{array}\\right|\\) = 0 \n\\(\\left|\\begin{array}{ccc} x+1 & y-3 & z-2 \\\\ 1 & 2 & 3 \\\\ 3 & 3 & 1 \\end{array}\\right|\\) = 0 \n– 7(x + 1) + 8(y – 3) – 3(z – 2) = 0 \ni.e., – 7x – 7 + 8y – 24 – 3z + 6 = 0 \n– 7x + 8y – 3z – 25 = 0 \n7x – 8y + 3z + 250 = 0<\/p>\nAnother method: \nThe equation of a plane passing through the point(- 1, 3, 2) is \na(x + 1) + b(y – 3) + c(z – 2) = 0 … (1) \nwhere a, b and c are the direction ratios of the normal to the plane. \nSince (1) is perpendicular to the planes \nx + 2y + 3z = 5 and 3x + 3y + z = 0, we get \na + 2b + 3c = 0 … (2) and \n3a + 3b + c = 0 … (3) \n\u2234 By the rule of cross multiplication, we get \n\\(\\frac{a}{\\left|\\begin{array}{ll} 2 & 3 \\\\ 3 & 1 \\end{array}\\right|}=\\frac{0}{\\left|\\begin{array}{ll} 3 & 1 \\\\ 1 & 3 \\end{array}\\right|}=\\frac{c}{\\left|\\begin{array}{ll} 1 & 2 \\\\ 3 & 3 \\end{array}\\right|}\\) \ni.e., \\(\\frac{a}{-7}=\\frac{b}{8}=\\frac{c}{-3}\\) \nSince a, b, c are the direction ratios of the normal to the plane, we can take a = – 7, b = 8, c = – 3. \n\u2234 (1) \u2192 7(x+ 1) + 8(y – 3) – 3(z – 2) = 0 \ni.e., – 7x + 8y – 3z – 25 = 0 \nor 7x – 8y + 3z + 25 = 0<\/p>\n
Question 14. \nIf the points (1, 1, p) and (-3, 0, 1) are equidistant from the plane \\(\\vec{r} \\cdot(3 \\hat{i}+4 \\hat{j}-12 \\hat{k})+13\\) = 0, then find the value of p. \nSolution: \nThe equation of the plane is, \n <\/p>\n
Question 15. \nFind the equation of the plane passing through the line of intersection of the planes \\(\\vec{r} \\cdot(\\hat{i}+\\hat{j}+\\hat{k})\\) = 1 and \\(\\vec{r} \\cdot(2 \\hat{i}+3 \\hat{j}-\\hat{k})+4\\) = 0 and parallel to x- axis. \nSolution: \nEquation of the given planes are \nx + y + z – 1 = 0 and 2x + 3y – z + 4 = 0 \nThe equation of the plane passing through the line of intersection of the planes is \n(x + y + z – 1) + (2x + 3y – z + 4) = 0 \n(1 + 2\u03bb)x + (1 + 3\u03bb)y + (1 – \u03bb )z + (4 – 1) = 0 … (1) \nThe direction ratios of the normal to the plane are 1 + 2\u03bb , 1 + 3\u03bb , 1 – \u03bb \nThe direction ratios of x-axis are 1, 0, 0 \nSince the required plane is parallel to the x-axis, the normal to the plane is perpendicular to x-axis. \n <\/p>\n
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Question 16. \nIf O is 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. \nSolution: \n\\(\\overrightarrow{\\mathrm{OP}}=\\hat{i}+2 \\hat{j}-3 \\hat{k}\\) \nDirection ratios of OP are 1, 2, -3 \nOP is perpendicular to the plane \n\u2234 Direction ratios of the normal to the plane \nare a = 1, b = 2, c = -3 \nP( 1, 2, – 3) is a point on the plane. \nEquation of the plane is \na(x – x1<\/sub>) + b(y – y1<\/sub>) + c(z – z1<\/sub>) = 0 \ni.e, 1(x – 1) + 2(y – 2) – 3(z + 3) = 0 \nx – 1 + 2y – 4 – 3z – 9 = 0 \nx + 2y – 3z – 14 = 0 \nThe equation of the plane is x + 2y – 3z – 14 = 0<\/p>\nQuestion 17. \nFind the equation of the plane which contains the line of intersection of the planes \\(\\vec{r} \\cdot(\\hat{i}+2 \\hat{j}+3 \\hat{k})-4\\) = 0, \\(\\vec{r} \\cdot(2 \\hat{i}+\\hat{j}-\\hat{k})+5\\) = 0 and which is perpendicular to the plane \\(\\vec{r} \\cdot(5 \\hat{i}+3 \\hat{j}-6 \\hat{k})+8\\) = 0. \nSolution: \nThe equation of the given planes are x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0. \nThe equation of the plane containing the line of in\u00cdersection of the above plane is \n(x + 2y + 3z – 4) + \u03bb(2x + y – z + 5) = 0 \n(1 + 2\u03bb)x + (2 + \u03bb)y (3 – \u03bb)z + (5\u03bb – 4) = 0 … (1) \nSince (1) is perpendicular to the plane \n5x + 3y – 6z + 8 = 0, we get \n5(1 + 2\u03bb) + 3(2 + \u03bb) – 6(3 – \u03bb) = 0 \n5 + 10\u03bb + 6 + 3\u03bb – 18 + 6\u03bb = 0 \n19\u03bb – 7 = 0 \n\u2234 \u03bb = \\(\\frac { 7 }{ 19 }\\) \n <\/p>\n
Question 18. \nFind the distance of the point (- 1, – 5, – 10) from the point of intersection of the line \\(\\vec{r}=2 \\hat{i}-\\hat{j}+2 \\hat{k}+\\lambda(3 \\hat{i}+4 \\hat{j}+2 \\hat{k})\\) and the plane \\(\\vec{r} \\cdot(\\hat{i}-\\hat{j}+\\hat{k})\\) = 5 \nSolution: \nThe cartesian equation of the line \\(\\vec{r}=2 \\hat{i}-\\hat{j}+2 \\hat{k}+\\lambda(3 \\hat{i}+4 \\hat{j}+2 \\hat{k})\\) is \n\\(\\frac{x-2}{3}=\\frac{y+1}{4}=\\frac{z-2}{2}\\) = \u03bb … (1) \nThe cartesian equation of the plane \n\\(\\vec{r} \\cdot(\\hat{i}-\\hat{j}+\\hat{k})\\) is x – y + z = 5 … (2) \nThe coordinate of any point on this line is P(2 + 3\u03bb, – 1 + 4\u03bb, 2 + 2\u03bb). \nLet the line intersects the plane at P, then P is a point on the plane. \n\u2234 (2) \u2192 (2 + 3\u03bb) – (-1 + 4\u03bb) + (2 + 2\u03bb) = 5 \n2 + 3\u03bb + 1 – 4\u03bb + 2 + 2\u03bb = 5 \n\u2234 \u03bb = 0 \n\u2234 P is (2, – 1,2) and the given point is Q(- 1, – 5, – 10) \n\u2234 PQ = \\(\\sqrt{(-1-2)^{2}+(-5–1)^{2}+(-10-2)^{2}}\\) \n= \\(\\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}\\) \n= \\(\\sqrt{9+16+144}=\\sqrt{169}\\) \n= 13<\/p>\n
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Question 19. \nFind the vector equation of the line passing through (1,2,3) and parallel to the planes \\(\\vec{r} \\cdot(\\hat{i}-\\hat{j}+2 \\hat{k})=5 \\text { and } \\vec{r} \\cdot(3 \\hat{i}+\\hat{j}+\\hat{k})=6\\) \nSolution: \nThe planes are \\(\\hat{r}\\). \\(\\hat{n}\\)2<\/sub> = d1<\/sub> \nand \\(\\hat{r}\\). \\(\\hat{n}\\)2<\/sub> = d2<\/sub> \nwhere \\(\\hat{n}\\)1<\/sub> = \\(\\hat{i}-\\hat{j}+2 \\hat{k}\\) and \\(\\vec{n}_{2}=3 \\hat{i}+\\hat{j}+\\hat{k}\\) \nSince the required line is parallel to the planes, it is perpendicular to both \\(\\hat{n}\\)1<\/sub> and \\(\\hat{n}\\)2<\/sub>. \n\\(\\hat{n}\\)1<\/sub> x \\(\\hat{n}\\)2<\/sub> is perpendicular to both \\(\\hat{n}\\)1<\/sub> and \\(\\hat{n}\\)2<\/sub> \n\\(\\vec{n}_{1} \\times \\vec{n}_{2}=\\left|\\begin{array}{ccc} \n\\hat{i} & \\hat{j} & \\hat{k} \\\\ \n1 & -1 & 2 \\\\ \n3 & 1 & 1 \n\\end{array}\\right|=-3 \\hat{i}+5 \\hat{j}+4 \\hat{k}\\) \nThe required line passes through the point (1, 2, 3) and is parallel to \\(\\hat{n}\\)1<\/sub> x \\(\\hat{n}\\)2<\/sub>. \n\u2234 The equation of the line is \n\\(\\vec{r}=(\\hat{i}+2 \\hat{j}+3 \\hat{k})+\\lambda(-3 \\hat{i}+5 \\hat{j}+4 \\hat{k})\\)<\/p>\nQuestion 20. \nFind the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: \n\\(\\frac{x-8}{3}=\\frac{y+19}{-16}=\\frac{z-10}{7}\\) \n\\(\\frac{x-15}{3}=\\frac{y-29}{8}=\\frac{z-5}{-5}\\) \nSolution: \nLet \\(\\vec{b}_{1}\\) and \\(\\vec{b}_{2}\\) be the direction of the lines where \\(\\vec{b}_{1}=3 \\hat{i}-16 \\hat{j}+7 \\hat{k}\\) and \\(\\vec{b}_{2}=3 \\hat{i}+8 \\hat{j}-5 \\hat{k}\\). \nThe required line is perpendicular to both \\(\\vec{b}_{1}\\) and \\(\\vec{b}_{2}\\). Hence it is parallel to \\(\\vec{b}_{1}\\) x \\(\\vec{b}_{2}\\) \n\\(\\vec{b}_{1} \\times \\vec{b}_{2}=\\left|\\begin{array}{ccc} \n\\hat{i} & \\hat{j} & \\hat{k} \\\\ \n3 & -16 & 7 \\\\ \n3 & 8 & -5 \n\\end{array}\\right|=24 \\hat{i}+36 \\hat{j}+72 \\hat{k}\\) \nThe direction ratios of \\(\\vec{b}_{1}\\) x \\(\\vec{b}_{2}\\) = 24, 36, 72 \n= 2, 3, 6 \nThe required line passes through the point (1, 2, – 4 ) and has direction ratios 2, 3, 6. \nHence the equation is \n\\(\\vec{r}=\\hat{i}+2 \\hat{j}-4 \\hat{k}+\\lambda(2 \\hat{i}+3 \\hat{j}+6 \\hat{k})\\)<\/p>\n
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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}}\\). \nSolution: \nThe equation of a plane in the intercept from is \\(\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}\\) = 1. \n <\/p>\n
Question 22. \nDistance between the two planes: \n2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is \na. 2 units \nb. 4 units \nc. 8 units \nd. \\(\\frac{2}{\\sqrt{29}}\\) units \nSolution: \nd. \\(\\frac{2}{\\sqrt{29}}\\) units \n \nHence the planes are parallel. \nThe distance between the parallel planes is the difference between their distance from the origin. \nThe distance of the plane 2x + 3y + 4z = 4 \n <\/p>\n
Question 23. \nThe planes 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are \na. perpendicular \nb. parallel \nc. intersect y-axis \nd. passes through (0, 0, \\(\\frac { 5 }{ 4 }\\)) \nSolution: \nb. parallel \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Question 1. Show that the line joining the origin to the point (2,1,1) is perpendicular to …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise<\/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 Miscellaneous Exercise - 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