{"id":19305,"date":"2021-02-26T17:36:02","date_gmt":"2021-02-26T12:06:02","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=19305"},"modified":"2021-02-26T17:36:02","modified_gmt":"2021-02-26T12:06:02","slug":"mcq-questions-for-class-12-maths-chapter-11","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/mcq-questions-for-class-12-maths-chapter-11\/","title":{"rendered":"MCQ Questions for Class 12 Maths Chapter 11 Three Dimensional Geometry with Answers"},"content":{"rendered":"

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

Three Dimensional Geometry Class 12 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Three Dimensional Geometry Multiple Choice Questions of Class 12 Maths to know different concepts. Practicing the MCQ Questions on Three Dimensional Geometry Class 12 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Three Dimensional Geometry Class 12 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nDistance between two planes:
\n2x + 3y + 4z = 5 and 4x + 6y + 8z = 12 is
\n(a) 2 units
\n(b) 4 units
\n(c) 8 units
\n(d) \\(\\frac { 1 }{\\sqrt{29}}\\) units.<\/p>\n

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

Answer: (d) \\(\\frac { 1 }{\\sqrt{29}}\\) units.<\/p>\n<\/details>\n

\n

Question 2.
\nThe planes 2x – y + 4z = 3 and 5x – 2.5y +10 z = 6 are
\n(a) perpendicular
\n(b) parallel
\n(c) intersect along y-axis
\n(d) passes through (0, 0, \\(\\frac { 5 }{4}\\))<\/p>\n

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

Answer: (b) parallel<\/p>\n<\/details>\n

\n

Question 3.
\nThe co-ordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by:
\n(a) (2, 0, 0)
\n(b) (0, 5, 0)
\n(c) (0, 0, 7)
\n(d) (0, 5, 7).<\/p>\n

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

Answer: (a) (2, 0, 0)<\/p>\n<\/details>\n

\n

Question 4.
\nIf \u03b1, \u00df, \u03b3 are the angles that a line makes with the positive direction of x, y, z axis, respectively, then the direction-cosines of the line are:
\n(a) < sin \u03b1, sin \u00df, sin \u03b3 >
\n(b) < cos \u03b1, cos \u00df, cos \u03b3 >
\n(c) < tan \u03b1, tan \u00df, tan \u03b3 >
\n(d) < cos\u00b2 \u03b1, cos\u00b2 \u00df, cos\u00b2 \u03b3 >.<\/p>\n

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

Answer: (b) < cos \u03b1, cos \u00df, cos \u03b3 ><\/p>\n<\/details>\n

\n

Question 5.
\nThe distance of a point P(a, b, c) from x-axis is
\n(a) \\(\\sqrt { a^2+c^2}\\)
\n(b) \\(\\sqrt { a^2+b^2}\\)
\n(c) \\(\\sqrt { b^2+c^2}\\)
\n(d) b\u00b2 + c\u00b2.<\/p>\n

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

Answer: (c) \\(\\sqrt { b^2+c^2}\\)<\/p>\n<\/details>\n

\n

Question 6.
\nIf the direction-cosines of a line are < k, k, k >, then
\n(a) k > 0
\n(b) 0 < k < 1
\n(c) k = 1
\n(d) k = \\(\\frac { 1 }{\u221a3}\\) or –\\(\\frac { 1 }{\u221a3}\\)<\/p>\n

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

Answer: (c) k = 1<\/p>\n<\/details>\n

\n

Question 7.
\nThe reflection of the point (\u03b1, \u00df, \u03b3) in the xy-plane is:
\n(a) (\u03b1, \u00df, 0)
\n(b) (0, 0, \u03b3)
\n(c) (-\u03b1, -\u00df, \u03b3)
\n(d) (\u03b1, \u00df, -\u03b3).<\/p>\n

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

Answer: (d) (\u03b1, \u00df, -\u03b3).<\/p>\n<\/details>\n

\n

Question 8.
\nWhat is the distance (in units) between two planes:
\n3x + 5y + 7z = 3 and 9x + 15y + 21z = 9?
\n(a) 0
\n(b) 3
\n(c) \\(\\frac { 6 }{\\sqrt{83}}\\)
\n(d) 6.<\/p>\n

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

Answer: (a) 0
\nHint:
\nThe given planes are
\n3x + 5y + 7z = 3 ……(1)
\nand 9x + 15y + 21z = 9 …….(2)
\nDividing (2) by 3, 3x + 5y + 7z = 3, which is (1)
\nThus, the given planes are coincident and as such the distance between them = 0 (units).<\/p>\n<\/details>\n

\n

Question 9.
\nThe equation of the line in vector form passing through the point (-1, 3, 5) and parallel to line \\(\\frac { x-3 }{2}\\) = \\(\\frac { y-4 }{3}\\), z = 2 is
\n(a) \\(\\vec r\\) = (-\\(\\hat i\\) + 3\\(\\hat j\\) + 5\\(\\hat k\\)) + \u03bb(2\\(\\hat i\\) +3\\(\\hat j\\) + \\(\\hat k\\))
\n(b) \\(\\vec r\\) = (-\\(\\hat i\\)+ 3\\(\\hat j\\) + 5\\(\\hat k\\)) + \u03bb(2\\(\\hat i\\) + 3\\(\\hat j\\))
\n(c) \\(\\vec r\\) = (2\\(\\hat i\\)+ 3\\(\\hat j\\) – 2\\(\\hat k\\)) + \u03bb(-\\(\\hat i\\) + 3\\(\\hat j\\) + 5\\(\\hat k\\))
\n(d) \\(\\vec r\\) = (2\\(\\hat i\\) + 3\\(\\hat j\\)]) + \u03bb(-\\(\\hat i\\) + 3\\(\\hat j\\) + 5\\(\\hat k\\)).<\/p>\n

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

Answer: (b) \\(\\vec r\\) = (-\\(\\hat i\\)+ 3\\(\\hat j\\) + 5\\(\\hat k\\)) + \u03bb(2\\(\\hat i\\) + 3\\(\\hat j\\))
\nHint:
\nThe given line is
\n\\(\\frac { x-3 }{2}\\) = \\(\\frac { y-4 }{3}\\) = \\(\\frac { z-2 }{0}\\)
\n\u2234 Reqd. equation of the line is:
\n\\(\\vec r\\) = (-\\(\\hat i\\)+ 3\\(\\hat j\\) + 5\\(\\hat k\\)) + \u03bb(2\\(\\hat i\\) + 3\\(\\hat j\\))<\/p>\n<\/details>\n

\n

Question 10.
\nLet the line \\(\\frac { x-2 }{3}\\) = \\(\\frac { y-1 }{-5}\\) = \\(\\frac { z-2 }{2}\\) lie in the plane x + 3y – \u03b1z + \u00df = 0. Then (\u03b1, \u00df) equals:
\n(a) (-6, -17)
\n(b) (5, -15)
\n(c) (-5, 5)
\n(d) (6, -17).<\/p>\n

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

Answer: (a) (-6, -17)
\nHint:
\nThe line = \\(\\frac { x-2 }{3}\\) = \\(\\frac { y-1 }{-5}\\) = \\(\\frac { z-2 }{2}\\) lies in the plane x + 3y – \u03b1z + \u00df = 0,
\n\u2234 2 + 3(1) – \u03b1(2) + \u00df = 0
\n\u21d2 2\u03b1 – \u00df = 5
\nand (1)(3) + (-5)(3) + (2) (-\u03b1) = 0
\n\u21d2 3 – 15 – 2\u03b1 = 0
\n\u21d2 2\u03b1 = -12\u03b1
\n\u21d2 \u03b1 = -6.
\nPutting in (1),
\n2(-6) – \u00df = 5
\n\u21d2 \u00df = -12 – 5 = -17.
\nHence, (\u03b1, \u00df) is (-6, -17).<\/p>\n<\/details>\n

\n

Question 11.
\nThe projections of a vector on the three co-ordinate axes are 6, -3, 2 respectively. The direction-cosines of the vector are:
\n(a) \\(\\frac { 6 }{5}\\), –\\(\\frac { 3 }{5}\\), \\(\\frac { 2 }{5}\\)
\n(b) \\(\\frac { 6 }{7}\\), –\\(\\frac { 3}{7}\\), \\(\\frac { 2 }{7}\\)
\n(c) \\(\\frac { -6 }{7}\\), \\(\\frac { -3 }{7}\\), \\(\\frac { 1 }{7}\\)
\n(d) 6, -3, 2.<\/p>\n

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

Answer: (b) \\(\\frac { 6 }{7}\\), –\\(\\frac { 3}{7}\\), \\(\\frac { 2 }{7}\\)
\nHint:
\nDirection-cosines are:
\n< \\(\\frac { 6 }{\\sqrt{36+9+4}}\\), \\(\\frac {-3}{\\sqrt{36+9+4}}\\), \\(\\frac { 2 }{\\sqrt{36+9+4}}\\) >
\ni.e., < \\(\\frac { 6 }{7}\\), –\\(\\frac { 3}{7}\\), \\(\\frac { 2 }{7}\\) >.<\/p>\n<\/details>\n

\n

Question 12.
\nA line AB in three-dimensional space makes angles 45\u00b0 and 120\u00b0 with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle \u03b8 with the positive z-axis, then \u03b8 equals:
\n(a) 30\u00b0
\n(b) 45\u00b0
\n(c) 60\u00b0
\n(d) 15\u00b0.<\/p>\n

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

Answer: (c) 60\u00b0
\nHint:
\ncos\u00b2 \u03b1 + cos\u00b2 \u00df + cos\u00b2 \u03b3 = 1
\nHere \u03b1 = 45\u00b0, \u00df = 120\u00b0, \u03b3 = 0.
\n\u2234 cos\u00b2 45\u00b0 + cos\u00b2 120\u00b0 + cos\u00b2 \u03b8 = 1
\n\u21d2 \\(\\frac { 1 }{2}\\) + \\(\\frac { 1 }{4}\\) + cos\u00b2 \u03b8 = 1
\n\u21d2 1 – cos\u00b2 \u03b8 = \\(\\frac { 3 }{4}\\)
\n\u21d2sin\u00b2 \u03b8 = \\(\\frac { 3 }{4}\\) = sin\u00b2 60\u00b0
\n\u21d2 \u03b8 = 60\u00b0<\/p>\n<\/details>\n

\n

Question 13.
\nIf the angle between the line x = \\(\\frac { y-1 }{2}\\) = \\(\\frac { z-3 }{\u03bb}\\) and the plane x + 2y + 3z = 4is cos-1<\/sup> (\\(\\sqrt{\\frac { 5}{14}}\\)) then \u03bb, equals:
\n(a) \\(\\frac { 2 }{3}\\)
\n(b) \\(\\frac { 3 }{2}\\)
\n(c) \\(\\frac { 2 }{5}\\)
\n(d) \\(\\frac { 5 }{3}\\)<\/p>\n

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

Answer: (a) \\(\\frac { 2 }{3}\\)
\nHint:
\nThe given line is \\(\\frac { x-0 }{1}[\/latex = [latex]\\frac { y-1 }{2}\\) = \\(\\frac { z-3 }{\u03bb}\\) and the plane is x + 2y + 3z = 4.
\n\u2234 Angle between the line and the plane is:
\n\"MCQ
\n\u21d2 14 (5 + \u03bb\u00b2) – (25 + 9\u03bb\u00b2 + 30\u03bb) = 5(5 + \u03bb\u00b2)
\n\u21d2 45 + 5\u03bb\u00b2 – 30\u03bb = 25 + 5\u03bb\u00b2
\n\u21d2 30\u03bb = 20
\n\u21d2 \u03bb = \\(\\frac { 2 }{3}\\)<\/p>\n<\/details>\n

\n

Question 14.
\nThe length of the perpendicular drawn from the point (3, -1, 11) to the line \\(\\frac { x }{2}\\) = \\(\\frac { y-2 }{3}\\) = \\(\\frac { z-3 }{4}\\) is
\n(a) \\(\\sqrt { 29}\\)
\n(b) \\(\\sqrt { 33}\\)
\n(c) \\(\\sqrt { 53}\\)
\n(d) \\(\\sqrt { 65}\\)<\/p>\n

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

Answer: (c) \\(\\sqrt { 53}\\)
\nHint:
\nLet any point on the line
\n\\(\\frac { x }{2}\\) = \\(\\frac { y-2 }{3}\\) = \\(\\frac { z-3 }{4}\\) be P (2k, 2 + 3k, 3 + 4k).
\nIf P be the foot of perpendicular,then direction ratios of the perpendicular are
\n< 2k – 3, 2 + 3k + 1, 3 + 4k – 11 >
\ni.e.< 2k -3, 3k + 3, 4k – 8 >.
\nAnd direction-ratios of the line are < 2, 3, 4 >.
\n\u2234 2 (2k – 3) + 3 (3k + 3) + 4(4k – 8) = 0
\n\u21d2 29k – 29 = 0
\n\u21d2 k = 1.
\n\u2234 P is (2, 2 + 3, 3 + 4) i.e. (2, 5, 7).
\nAlso Q is (3, -1, 11).
\n\u2234 Length of perpendicular
\n= \\(\\sqrt {(2 – 3)^2 + (5 + 1)^2 + (7 – 11)^2}\\)
\n= \\(\\sqrt {1 + 36 + 16}\\)
\n= \\(\\sqrt {53}\\)<\/p>\n<\/details>\n

\n

Question 15.
\nThe distance of the point (1, -5, 9) from the plane x – y + z = 5, measured along a straight line x = y = z is:
\n(a) 10\u221a3
\n(b) 5\u221a3
\n(c) 3\\(\\sqrt {10}\\)
\n(d) 3\u221a5<\/p>\n

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

Answer: (a) 10\u221a3
\nHint:
\nThe line through P (1, -5, 9) parallel to x = y = z is
\n\\(\\frac { x-1 }{1}\\) = \\(\\frac { y+5 }{1}\\) = \\(\\frac { z-9 }{1}\\) …….. (1)
\nAny point on (1) is Q (1 + \u03bb, -5 + \u03bb, 9 + \u03bb).
\nThis lies on x – y + z = 5
\n\u21d2 1 + \u03bb + 5 – \u03bb + 9 + \u03bb = 5
\n\u21d2 \u03bb = -10.
\nQ is (-9, -15, -1).
\n\u2234 |PQ| = \\(\\sqrt {(-9 – 1)^2 + (-15 + 5)^2 + (-1 – 9)^2}\\)
\n= \\(\\sqrt {100+100+100}\\)
\n= 10\u221a3<\/p>\n<\/details>\n

\n

Question 16.
\nAn equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is:
\n(a) x – 2y + 2z – 3 = 0
\n(b) x – 2y + 2z + 1 = 0
\n(c) x – 2y + 2z – 1 = 0
\n(d) x – 2y + 2z + 5 = 0.<\/p>\n

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

Answer: (a) x – 2y + 2z – 3 = 0
\nHint:
\nAny plane parallel to x – 2y + 2z – 5 = 0 is
\nx – 2y + 2z + k = 0 …….(1)
\nIts distance from the origin = 1
\n\u21d2 \\(\\frac { |0-0+0+k| }{\\sqrt{1+4+4}}\\) = 1
\n\u21d2 |k| = 3
\n\u21d2 k = \u00b13
\nPutting in (1), x – 2y + 2z – 3 = 0.<\/p>\n<\/details>\n

\n

Question 17.
\nIf the lines: \\(\\frac { x-2 }{1}\\) = \\(\\frac { y-3 }{1}\\) = \\(\\frac { z-4 }{-k}\\) and \\(\\frac { x-1 }{k}\\) = \\(\\frac { y-4 }{2}\\) = \\(\\frac { z-5 }{1}\\) are coplanar, then k can have:
\n(a) exactly one value
\n(b) exactly two values
\n(c) exactly three values
\n(d) any value.<\/p>\n

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

Answer: (b) exactly two values
\nHint:
\nThe given parallel planes are:
\n\\(\\left|\\begin{array}{rrr}
\n1 & -1 & -1 \\\\
\n1 & 1 & -k \\\\
\nk & 2 & 1
\n\\end{array}\\right|\\) = 0
\nif (1)(1 + 2k) + (1)(1 + k\u00b2) + (-1)(2 – k) = 0
\nif 1 + 2k + 1 + k\u00b2 – 2 + k = 0
\nif k\u00b2 + 3k = 0 if k(k + 3) = 0
\nif k = 0, -3.
\nHence, k can have exactly two values.<\/p>\n<\/details>\n

\n

Question 18.
\nDistance between two parallel planes:
\n2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
\n(a) \\(\\frac { 5 }{2}\\)
\n(b) \\(\\frac { 7 }{2}\\)
\n(c) \\(\\frac { 9 }{2}\\)
\n(d) \\(\\frac { 3 }{2}\\)<\/p>\n

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

Answer: (b) \\(\\frac { 7 }{2}\\)
\nHint:
\nThe given parallel planes are:
\n2x + y + 2z – 8 = 0
\nand 2x + y + 2z + \\(\\frac { 5 }{2}\\) = 0.
\n\u2234 Distance between the planes
\n\\(\\frac { |8+5\/2| }{\\sqrt{4+1+4}}\\) = \\(\\frac { 21 }{6}\\) = \\(\\frac { 7 }{2}\\)<\/p>\n<\/details>\n

\n

Question 19.
\nThe image of the line \\(\\frac { x-1 }{3}\\) = \\(\\frac { y-3 }{1}\\) = \\(\\frac { z-4 }{-5}\\) in the plane:
\n2x – y + z + 3 = 0 is the line:
\n(a) \\(\\frac { x+3 }{-3}\\) = \\(\\frac { y-5 }{-1}\\) = \\(\\frac { z+2 }{5}\\)
\n(b) \\(\\frac { x-3 }{3}\\) = \\(\\frac { y+5 }{1}\\) = \\(\\frac { z-2 }{-5}\\)
\n(c) \\(\\frac { x-3 }{-3}\\) = \\(\\frac { y+5 }{-1}\\) = \\(\\frac { z-2 }{5}\\)
\n(d) \\(\\frac { x+3 }{3}\\) = \\(\\frac { y-5 }{1}\\) = \\(\\frac { z-2 }{-5}\\)<\/p>\n

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

Answer: (d) \\(\\frac { x+3 }{3}\\) = \\(\\frac { y-5 }{1}\\) = \\(\\frac { z-2 }{-5}\\)
\nHint:
\nSince (3) (2) + (1) (-1) + (-5) (1) = 0
\n\u2234 the line is parallel to the plane.
\nImage of (1, 3, 4) is (-3, 5, 2).
\n\u2234 The required, image is
\n\\(\\frac { x+3 }{3}\\) = \\(\\frac { y-5 }{1}\\) = \\(\\frac { z-2 }{-5}\\)<\/p>\n<\/details>\n

\n

Question 20.
\nThe distance of the point (1, 0, 2) from the point of intersection of the line
\n\\(\\frac { x-2 }{3}\\) = \\(\\frac { y+1 }{4}\\) = \\(\\frac { z-2 }{12}\\) and flie plane x – y + z = 16 is
\n(a) 2\\(\\sqrt { 14 }\\)
\n(b) 8
\n(c) 3\\(\\sqrt { 21 }\\)
\n(d) 13.<\/p>\n

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

Answer: (d) 13.
\nHint:
\nAny point on the line is
\n(3k + 2, 4k – 1, 12k + 2).
\nThis lies on the plane
\n\u21d2 3k + 2 – 4k + 1 + 12k + 2 = 16
\n\u21d2 11k = 11
\n\u21d2 k = 1.
\n\u2234 Point of intersection is (5, 3, 14).
\n\u2234 Its distance from (1, 0, 2)
\n\\(\\sqrt {(5 – 1)^2 + (3 – 0)^2 + (14 – 2)^2}\\)
\n= \\(\\sqrt {16+9+144}\\)
\n= \\(\\sqrt {169}\\)
\n= 13<\/p>\n<\/details>\n

\n

Question 21.
\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) 2x + 6y+ 12z = 13
\n(b) x + 3y + 6z = – 7
\n(c) x + 3y + 6z = 7
\n(d) 2x + 6y – 12z = -13.<\/p>\n

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

Answer: (c) x + 3y + 6z = 7
\nHint:
\nPutting z = 0, 2x – 5y = 3 and x + y = 5.
\nSolving, x = 4, y = 1.
\nLet x + 3y + 6z = k be a plane parallel to given plane.
\n\u2234 4 + 3 + 0 = k
\n\u21d2 k = 7
\n\u2234 Required, equation of the plane is x + 3y + 6z = 7<\/p>\n<\/details>\n

\n

Question 22.
\nIf the line \\(\\frac { x-3 }{2}\\) = \\(\\frac { y+2 }{-1}\\) = \\(\\frac { z+4 }{3}\\) lies in the plane lx + my – z = 9,then l\u00b2 + m\u00b2 is equal to
\n(a) 18
\n(b) 5
\n(c) 2
\n(d) 26.<\/p>\n

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

Answer: (c) 2
\nHint:
\nSince the line \\(\\frac { x-3 }{2}\\) = \\(\\frac { y+2 }{-1}\\) = \\(\\frac { z+4 }{3}\\) lies in the plane lx + my – z = 9,
\n\u2234 3l – 2m + 4 = 9 and 2l – m – 3 = 0
\nSolving for l and m, we get:
\nl = 1 and m = -1
\n\u2234 l\u00b2 + m\u00b2 = 1 + 1 = 2.<\/p>\n<\/details>\n

\n

Fill in the blanks<\/span><\/p>\n

Question 1.
\nDirection-cosines of x-axis are ………………<\/p>\n

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

Answer: < 1, 0, 0 ><\/p>\n<\/details>\n

\n

Question 2.
\nDirection-cosines of y-axis are ………………<\/p>\n

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

Answer: < 0, 1, 0 >.<\/p>\n<\/details>\n

\n

Question 3.
\nDirection-cosines of z-axis are ………………..<\/p>\n

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

Answer: < 0, 0, 1 >.<\/p>\n<\/details>\n

\n

Question 4.
\nIf a line makes angles 90\u00b0, 60\u00b0 and \u03b8 with x, y and z-axis respectively, then acute \u03b8 = …………………<\/p>\n

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

Answer: 30\u00b0.<\/p>\n<\/details>\n

\n

Question 5.
\nDirection-cosines of the vector -2\\(\\hat i\\) + \\(\\hat j\\) – 5\\(\\hat k\\) are …………………<\/p>\n

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

Answer: < \\(\\frac { -2 }{\\sqrt{30}}\\), \\(\\frac { 1 }{\\sqrt{30}}\\), \\(\\frac { 5 }{\\sqrt{30}}\\) ><\/p>\n<\/details>\n

\n

Question 6.
\nThe points (1, 2, 7); (2, 6, 3); (3, 10, -1) are ………………..<\/p>\n

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

Answer: collinear.<\/p>\n<\/details>\n

\n

Question 7.
\nThe value of \u2018\u03bb\u2019 so that the lines
\n\\(\\frac { 1-x }{3}\\) = \\(\\frac { 7y-14 }{\u03bb}\\) = \\(\\frac { z-3 }{2}\\) and \\(\\frac { 7-7x }{3\u03bb}\\) = \\(\\frac { y-5 }{1}\\) = \\(\\frac { 6-z }{5}\\) are at right-angles is ………………..<\/p>\n

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

Answer: 7<\/p>\n<\/details>\n

\n

Question 8.
\nThe sum of the intercepts cut off by the plane
\n\\(\\vec r\\)(2\\(\\hat i\\) + \\(\\hat j\\) – \\(\\hat k\\)) – 5 = 0 on the three axes is …………………..<\/p>\n

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

Answer: 5\/2<\/p>\n<\/details>\n

\n

Question 9.
\nIf \u03b1, \u00df, \u03b3 are direction-angles of a line, then:
\ncos 2\u03b1 + cos 2\u00df + cos 2\u03b3 = ……………….<\/p>\n

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

Answer: -1.<\/p>\n<\/details>\n

\n

Question 10.
\nThe equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axis respectively is …………………..<\/p>\n

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

Answer: \\(\\frac { x }{2}\\) + \\(\\frac { y }{3}\\) + \\(\\frac { z }{4}\\) = 1<\/p>\n<\/details>\n

\n

We believe the knowledge shared regarding NCERT MCQ Questions for Class 12 Maths Chapter 11 Three Dimensional Geometry with Answers Pdf free download has been useful to the possible extent. If you have any other queries regarding CBSE Class 12 Maths Three Dimensional Geometry MCQs Multiple Choice Questions with Answers, feel free to reach us via the comment section and we will guide you with the possible solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

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

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