{"id":32695,"date":"2021-08-31T15:25:30","date_gmt":"2021-08-31T09:55:30","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32695"},"modified":"2021-08-31T15:25:30","modified_gmt":"2021-08-31T09:55:30","slug":"ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise"},"content":{"rendered":"

These 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.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise<\/h2>\n

\"NCERT<\/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>\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 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\"NCERT
\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>\n

Question 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\"NCERT<\/p>\n

Question 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

\"NCERT<\/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 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\"NCERT<\/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>\n

Question 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

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT<\/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\"NCERT
\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

\"NCERT<\/p>\n

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>\n

Another 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\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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>\n

Question 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\"NCERT<\/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

\"NCERT<\/p>\n

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>\n

Question 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

\"NCERT<\/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}}\\).
\nSolution:
\nThe equation of a plane in the intercept from is \\(\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}\\) = 1.
\n\"NCERT<\/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\"NCERT
\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\"NCERT<\/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\"NCERT<\/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. 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 the …<\/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":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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