{"id":17490,"date":"2022-05-24T21:00:28","date_gmt":"2022-05-24T15:30:28","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17490"},"modified":"2022-05-16T15:31:03","modified_gmt":"2022-05-16T10:01:03","slug":"mcq-questions-for-class-11-maths-chapter-12","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-12\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Introduction to Three Dimensional Geometry Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Objective Questions.<\/p>\n

Introduction to Three Dimensional Geometry Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Introduction to Three Dimensional Geometry Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Introduction to Three Dimensional Geometry Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Introduction to Three Dimensional Geometry Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nThe projections of a directed line segment on the coordinate axes are 12, 4, 3. The DCS of the line are
\n(a) 12\/13, -4\/13, 3\/13
\n(b) -12\/13, -4\/13, 3\/13
\n(c) 12\/13, 4\/13, 3\/13
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 12\/13, 4\/13, 3\/13
\nLet AB be the given line and the DCs of AB be l, m, n. Then
\nProjection on x-axis = AB . l = 12 (Given)
\nProjection on y-axis = AB . m = 4 (Given)
\nProjection on z-axis = AB . n = 3 (Given)
\n\u21d2 (AB\u00b2) (l\u00b2 + m\u00b2 + n\u00b2) = 144 + 16 + 9
\n\u21d2 (AB\u00b2) = 169 {since l\u00b2 + m\u00b2 + n\u00b2 = 1}
\n\u21d2 AB = 13
\nHence, DCs of AB are 12\/13, 4\/13, 3\/13<\/p>\n<\/details>\n

\n

Question 2.
\nThe angle between the planes r . n1<\/sub> = d1<\/sub> and r . n1<\/sub> = d2<\/sub> is
\n(a) cos \u03b8 ={|n1<\/sub>| \u00d7 |n2<\/sub>|}\/ (n1<\/sub>. n2<\/sub>)
\n(b) cos \u03b8 = (n1<\/sub> . n2<\/sub>)\/{|n1<\/sub>| \u00d7 |n2<\/sub>|}\u00b2
\n(c) cos \u03b8 = (n1<\/sub> . n2<\/sub>)\/{|n1<\/sub>| \u00d7 |n2<\/sub>|}
\n(d) cos \u03b8 = (n1<\/sub> . n2<\/sub>)\u00b2 \/{|n1<\/sub>| \u00d7 |n2<\/sub>|}<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) cos \u03b8 = (n1<\/sub> . n2<\/sub>)\/{|n1<\/sub>| \u00d7 |n2<\/sub>|}
\nThe angle between the planes r . n1<\/sub> = d1<\/sub> and r . n2<\/sub> = d2<\/sub> is defined as
\ncos \u03b8 = (n1<\/sub> . n2<\/sub>)\/{|n1<\/sub>| \u00d7 |n2<\/sub>|}<\/p>\n<\/details>\n

\n

Question 3.
\nFor every point P(x, y, z) on the xy-plane
\n(a) x = 0
\n(b) y = 0
\n(c) z = 0
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) z = 0
\nThe perpendicular distance of P(x, y, z) from xy-plane is zero.<\/p>\n<\/details>\n

\n

Question 4.
\nThe locus of a point P(x, y, z) which moves in such a way that x = a and y = b, is a
\n(a) Plane parallel to xy-plane
\n(b) Line parallel to x-axis
\n(c) Line parallel to y-axis
\n(d) Line parallel to z-axis<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Line parallel to x-axis
\nSince x = 0 and y = 0 together represent x-axis, therefore x = a and y = b represent a line parallel to x-axis.<\/p>\n<\/details>\n

\n

Question 5.
\nThe equation of the plane containing the line 2x – 5y + z = 3, x + y + 4z = 5 and parallel to the plane x + 3y + 6z = 1 is
\n(a) x + 3y + 6z + 7 = 0
\n(b) x + 3y – 6z – 7 = 0
\n(c) x – 3y + 6z – 7 = 0
\n(d) x + 3y + 6z – 7 = 0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) x + 3y + 6z – 7 = 0
\nLet the equation of the plane is
\n(2x – 5y + z – 3) + \u03bb(x + y + 4z – 5) = 0
\n\u21d2 (2 + \u03bb)x + (\u03bb – 5)y + (4\u03bb + 1)z – (3 + 5\u03bb) = 0
\nSince the plane is parallel to x + 3y + 6z – 1 = 0
\n\u21d2 (2 + \u03bb)\/1 = (\u03bb – 5)\/3 = (1 + 4\u03bb)\/6
\n\u21d2 6 + 3\u03bb = \u03bb – 5
\n\u21d2 2\u03bb = -11
\n\u21d2 \u03bb = -11\/2
\nAgain,
\n6\u03bb – 30 = 3 + 12\u03bb
\n\u21d2 -6\u03bb = -33
\n\u21d2 \u03bb = -33\/6
\n\u21d2 \u03bb = -11\/2
\nSo, the required equation of plane is
\n(2x – 5y + z – 3) + (-11\/2) \u00d7 (x + y + 4z – 5) = 0
\n\u21d2 2(2x – 5y + z – 3) + (-11) \u00d7 (x + y + 4z – 5) = 0
\n\u21d2 4x – 10y + 2z – 6 – 11x – 11y – 44z + 55 = 0
\n\u21d2 -7x – 21y – 42z + 49 = 0
\n\u21d2 x + 3y + 6z – 7 = 0<\/p>\n<\/details>\n

\n

Question 6.
\nThe coordinate of foot of perpendicular drawn from the point A(1, 0, 3) to the join of the point B(4, 7, 1) and C(3, 5, 3) are
\n(a) (5\/3, 7\/3, 17\/3)
\n(b) (5, 7, 17)
\n(c) (5\/3, -7\/3, 17\/3)
\n(d) (5\/7, -7\/3, -17\/3)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (5\/3, 7\/3, 17\/3)
\nLet D be the foot of perpendicular and let it divide BC in the ration m : 1
\nThen the coordinates of D are {(3m + 4)\/(m + 1), (5m + 7)\/(m + 1), (3m + 1)\/(m + 1)}
\nNow, AD \u22a5 BC
\n\u21d2 AD . BC = 0
\n\u21d2 -(2m + 3) – 2(5m + 7) – 4 = 0
\n\u21d2 m = -7\/4
\nSo, the coordinate of D are (5\/3, 7\/3, 17\/3)<\/p>\n<\/details>\n

\n

Question 7.
\nThe coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ plane is
\n(a) (0, 17\/2, 13\/2)
\n(b) (0, -17\/2, -13\/2)
\n(c) (0, 17\/2, -13\/2)
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) (0, 17\/2, -13\/2)
\nThe line passing through the points (5, 1, 6) and (3, 4, 1) is given as
\n(x – 5)\/(3 – 5) = (y – 1)\/(4 – 1) = (z – 6)\/(1 – 6)
\n\u21d2 (x – 5)\/(-2) = (y – 1)\/3 = (z – 6)\/(-5) = k(say)
\n\u21d2 (x – 5)\/(-2) = k
\n\u21d2 x – 5 = -2k
\n\u21d2 x = 5 – 2k
\n(y – 1)\/3 = k
\n\u21d2 y – 1 = 3k
\n\u21d2 y = 3k + 1
\nand (z – 6)\/(-5) = k
\n\u21d2 z – 6 = -5k
\n\u21d2 z = 6 – 5k
\nNow, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)
\nThe equation of YZ-plane is x = 0
\nSince the line passes through YZ-plane
\nSo, 5 – 2k = 0
\n\u21d2 k = 5\/2
\nNow, 3k + 1 = 3 \u00d7 5\/2 + 1 = 15\/2 + 1 = 17\/2
\nand 6 – 5k = 6 – 5 \u00d7 5\/2 = 6 – 25\/2 = -13\/2
\nHence, the required point is (0, 17\/2, -13\/2)<\/p>\n<\/details>\n

\n

Question 8.
\nIf P is a point in space such that OP = 12 and OP inclined at angles 45 and 60 degrees with OX and OY respectively, then the position vector of P is
\n(a) 6i + 6j \u00b1 6\u221a2k
\n(b) 6i + 6\u221a2j \u00b1 6k
\n(c) 6\u221a2i + 6j \u00b1 6k
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 6\u221a2i + 6j \u00b1 6k
\nLet l, m, n be the DCs of OP.
\nThen it is given that l = cos 45 = 1\/\u221a2
\nm = cos 60 = 1\/2
\nNow, l\u00b2 + m\u00b2 + n\u00b2 = 1
\n\u21d2 1\/2 + 1\/4 + n\u00b2 = 1
\n\u21d2 n\u00b2 = 1\/4
\n\u21d2 n = \u00b11\/2
\nNow, r = |r|(li + mj + nk)
\n\u21d2 r = 12(i\/\u221a2 + j\/2 \u00b1 k\/\u221a2)
\n\u21d2 r = 6\u221a2i + 6j \u00b1 6k<\/p>\n<\/details>\n

\n

Question 9.
\nThe image of the point P(1,3,4) in the plane 2x – y + z = 0 is
\n(a) (-3, 5, 2)
\n(b) (3, 5, 2)
\n(c) (3, -5, 2)
\n(d) (3, 5, -2)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (-3, 5, 2)
\nLet image of the point P(1, 3, 4) is Q in the given plane.
\nThe equation of the line through P and normal to the given plane is
\n(x – 1)\/2 = (y – 3)\/-1 = (z – 4)\/1
\nSince the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
\nNow, the coordinate of the mid-point of PQ is
\n(r + 1, -r\/2 + 3, r\/2 + 4)
\nNow, this point lies in the given plane.
\n2(r + 1) – (-r\/2 + 3) + (r\/2 + 4) + 3 = 0
\n\u21d2 2r + 2 + r\/2 – 3 + r\/2 + 4 + 3 = 0
\n\u21d2 3r + 6 = 0
\n\u21d2 r = -2
\nHence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
\n= (-3, 5, 2)<\/p>\n<\/details>\n

\n

Question 10.
\nThere is one and only one sphere through
\n(a) 4 points not in the same plane
\n(b) 4 points not lie in the same straight line
\n(c) none of these
\n(d) 3 points not lie in the same line<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 4 points not in the same plane
\nSphere is referred to its center and it follows a quadratic equation with 2 roots. The mid-point of chords of a sphere and parallel to fixed direction lies in the normal diametrical plane.
\nNow, general equation of the plane depends on 4 constants. So, one sphere passes through 4 points and they need not be in the same plane.<\/p>\n<\/details>\n

\n

Question 11.
\nThe points on the y- axis which are at a distance of 3 units from the point ( 2, 3, -1) is
\n(a) either (0, -1, 0) or (0, -7, 0)
\n(b) either (0, 1, 0) or (0, 7, 0)
\n(c) either (0, 1, 0) or (0, -7, 0)
\n(d) either (0, -1, 0) or (0, 7, 0)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) either (0, -1, 0) or (0, 7, 0)
\nLet the point on y-axis is O(0, y, 0)
\nGiven point is A(2, 3, -1)
\nGiven OA = 3
\n\u21d2 OA\u00b2 = 9
\n\u21d2 (2 – 0)\u00b2 + (3 – y)\u00b2 + (-1 – 0)\u00b2 = 9
\n\u21d2 4 + (3 – y)\u00b2 + 1 = 9
\n\u21d2 5 + (3 – y)\u00b2 = 9
\n\u21d2 (3 – y)\u00b2 = 9 – 5
\n\u21d2 (3 – y)\u00b2 = 4
\n\u21d2 3 – y = \u221a4
\n\u21d2 3 – y = \u00b14
\n\u21d2 3 – y = 4 and 3 – y = -4
\n\u21d2 y = -1, 7
\nSo, the point is either (0, -1, 0) or (0, 7, 0)<\/p>\n<\/details>\n

\n

Question 12.
\nThe coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ plane is
\n(a) (0, 17\/2, 13\/2)
\n(b) (0, -17\/2, -13\/2)
\n(c) (0, 17\/2, -13\/2)
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) (0, 17\/2, -13\/2)
\nThe line passing through the points (5,1,6) and (3,4,1) is given as
\n(x – 5)\/(3 – 5) = (y – 1)\/(4 – 1) = (z – 6)\/(1 – 6)
\n\u21d2 (x – 5)\/(-2) = (y – 1)\/3 = (z – 6)\/(-5) = k(say)
\n\u21d2 (x – 5)\/(-2) = k
\n\u21d2 x – 5 = -2k
\n\u21d2 x = 5 – 2k
\n(y – 1)\/3 = k
\n\u21d2 y – 1 = 3k
\n\u21d2 y = 3k + 1
\nand (z-6)\/(-5) = k
\n\u21d2 z – 6 = -5k
\n\u21d2 z = 6 – 5k
\nNow, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)
\nThe equation of YZ-plane is x = 0
\nSince the line passes through YZ-plane
\nSo, 5 – 2k = 0
\n\u21d2 k = 5\/2
\nNow, 3k + 1 = 3 \u00d7 5\/2 + 1 = 15\/2 + 1 = 17\/2
\nand 6 – 5k = 6 – 5 \u00d7 5\/2 = 6 – 25\/2 = -13\/2
\nHence, the required point is (0, 17\/2, -13\/2)<\/p>\n<\/details>\n

\n

Question 13.
\nhe equation of plane passing through the point i + j + k and parallel to the plane r . (2i – j + 2k) = 5 is
\n(a) r . (2i – j + 2k) = 2
\n(b) r . (2i – j + 2k) = 3
\n(c) r . (2i – j + 2k) = 4
\n(d) r . (2i – j + 2k) = 5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) r . (2i – j + 2k) = 3
\nThe equation of plane parallel to the plane r . (2i – j + 2k) = 5 is
\nr . (2i – j + 2k) = d
\nSince it passes through the point i + j + k, therefore
\n(i + j + k) . (2i – j + 2k) = d
\n\u21d2 d = 2 – 1 + 2
\n\u21d2 d = 3
\nSo, the required equation of the plane is
\nr . (2i – j + 2k) = 3<\/p>\n<\/details>\n

\n

Question 14.
\nThe cartesian equation of the line is 3x + 1 = 6y – 2 = 1 – z then its direction ratio are
\n(a) 1\/3, 1\/6, 1
\n(b) -1\/3, 1\/6, 1
\n(c) 1\/3, -1\/6, 1
\n(d) 1\/3, 1\/6, -1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1\/3, 1\/6, 1
\nGive 3x + 1 = 6y – 2 = 1 – z
\n= (3x + 1)\/1 = (6y – 2)\/1 = (1 – z)\/1
\n= (x + 1\/3)\/(1\/3) = (y – 2\/6)\/(1\/6) = (1 – z)\/1
\n= (x + 1\/3)\/(1\/3) = (y – 1\/3)\/(1\/6) = (1 – z)\/1
\nNow, the direction ratios are: 1\/3, 1\/6, 1<\/p>\n<\/details>\n

\n

Question 15.
\nUnder what condition does the equation x\u00b2 + y\u00b2 + z\u00b2 + 2ux + 2vy + 2wz + d represent a real sphere
\n(a) u\u00b2 + v\u00b2 + w\u00b2 = d\u00b2
\n(b) u\u00b2 + v\u00b2 + w\u00b2 > d
\n(c) u\u00b2 + v\u00b2 + w\u00b2 < d
\n(d) u\u00b2 + v\u00b2 + w\u00b2 < d\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) u\u00b2 + v\u00b2 + w\u00b2 > d
\nEquation x\u00b2 + y\u00b2 + z\u00b2 + 2ux + 2vy + 2wz + d represent a real sphere if
\nu\u00b2 + v\u00b2 + w\u00b2 – d > 0
\n\u21d2 u\u00b2 + v\u00b2 + w\u00b2 > d<\/p>\n<\/details>\n

\n

Question 16.
\nThe locus of a first-degree equation in x, y, z is a
\n(a) sphere
\n(b) straight line
\n(c) plane
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) plane
\nIn an x-y-z cartesian coordinate system, the general form of the equation of a plane is
\nax + by + cz + d = 0
\nIt is an equation of the first degree in three variables.<\/p>\n<\/details>\n

\n

Question 17.
\nThe image of the point P(1,3,4) in the plane 2x – y + z = 0 is
\n(a) (-3, 5, 2)
\n(b) (3, 5, 2)
\n(c) (3, -5, 2)
\n(d) (3, 5, -2)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (-3, 5, 2)
\nLet image of the point P(1, 3, 4) is Q in the given plane.
\nThe equation of the line through P and normal to the given plane is
\n(x – 1)\/2 = (y – 3)\/-1 = (z – 4)\/1
\nSince the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
\nNow, the coordinate of the mid-point of PQ is
\n(r + 1, -r\/2 + 3, r\/2 + 4)
\nNow, this point lies in the given plane.
\n2(r + 1) – (-r\/2 + 3) + (r\/2 + 4) + 3 = 0
\n\u21d2 2r + 2 + r\/2 – 3 + r\/2 + 4 + 3 = 0
\n\u21d2 3r + 6 = 0
\n\u21d2 r = -2
\nHence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
\n= (-3, 5, 2)<\/p>\n<\/details>\n

\n

Question 18.
\nThe distance of the point P(a, b, c) from the x-axis is
\n(a) \u221a(a\u00b2 + c\u00b2)
\n(b) \u221a(a\u00b2 + b\u00b2)
\n(c) \u221a(b\u00b2 + c\u00b2)
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) \u221a(b\u00b2 + c\u00b2)
\nThe coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).
\nSo, the required distance = \u221a{(a – a)\u00b2 + (b – 0)\u00b2 + (c – 0)\u00b2}
\n= \u221a(b\u00b2 + c\u00b2)<\/p>\n<\/details>\n

\n

Question 19.
\nThe vector equation of a sphere having centre at origin and radius 5 is
\n(a) |r| = 5
\n(b) |r| = 25
\n(c) |r| = \u221a5
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) |r| = 5
\nWe know that the vector equation of a sphere having center at the origin and radius R
\n= |r| = R
\nHere R = 5
\nHence, the equation of the required sphere is |r| = 5<\/p>\n<\/details>\n

\n

Question 20.
\nThe ratio in which the line joining the points(1, 2, 3) and (-3, 4, -5) is divided by the xy-plane is
\n(a) 2 : 5
\n(b) 3 : 5
\n(c) 5 : 2
\n(d) 5 : 3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 3 : 5
\nLet the points are P(1, 2, 3) and Q(-3, 4, -5)
\nLet the line joining the points P(1, 2, 3) and Q(-3, 4, -5) is divided by the xy-plane at point R in the ratio k : 1
\nNow, the coordinate of R is
\n{(-3k + 1)\/(k + 1), (4k + 2)\/(k + 1), (-5k + 3)\/(k + 1)}
\nSince R lies on the xy-plane.
\nSo, z-coordinate is zero
\n\u21d2 (-5k + 3)\/(k + 1) = 0
\n\u21d2 k = 3\/5
\nSo, the ratio = 3\/5 : 1 = 3 : 5<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. …<\/p>\n

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