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Chapter 11 · Class 12 Mathematics

Three Dimensional Geometry — Questions & Answers

Board-pattern questions from Three Dimensional Geometry, each with the correct answer and the reasoning behind it. 279 questions from this chapter are on TestSaathi; a few of them are below so you can see what the practice looks like before signing up.

Sample questions from Three Dimensional Geometry

  1. Q1. Two lines through the origin have direction cosines (1/3, 2/3, 2/3) and (2/3, -2/3, 1/3). The angle between them is:

    • A.0 degrees
    • B.60 degrees
    • C.90 degrees✓
    • D.45 degrees
    Solution

    The dot product is (1/3)(2/3) + (2/3)(-2/3) + (2/3)(1/3) = 2/9 - 4/9 + 2/9 = 0, so the two lines are perpendicular.

  2. Q2. The plane passing through the origin and containing the line (x-2)/1 = (y-1)/(-2) = (z+3)/1 is:

    • A.2x + y - 3z = 0
    • B.x - 2y + z = 0
    • C.x - y + z = 0
    • D.x + y + z = 0✓
    Solution

    The plane contains the origin and the point (2,1,-3), so its normal is perpendicular to both (2,1,-3) and the direction (1,-2,1): (2,1,-3) x (1,-2,1) = ((1)(1) - (-3)(-2), (-3)(1) - (2)(1), (2)(-2) - (1)(1)) = (-5, -5, -5), which reduces to (1, 1, 1). The plane is x + y + z = 0.

  3. Q3. The distance of the point (1,2,3) from the plane x + y + z = 9, measured along the line through it with direction ratios (2,3,6), is:

    • A.sqrt(3) units
    • B.7/11 units
    • C.21/11 units✓
    • D.3/sqrt(3) units
    Solution

    The line meets the plane where (1 + 2t) + (2 + 3t) + (3 + 6t) = 6 + 11t = 9, so t = 3/11. The measured distance is |t| * |(2,3,6)| = (3/11)(7) = 21/11 units.

  4. Q4. The area of the triangle with vertices A(2,1,0), B(3,3,2) and C(1,2,3) is:

    • A.sqrt(59) square units
    • B.sqrt(35)/2 square units
    • C.5*sqrt(2)/2 square units✓
    • D.5*sqrt(2) square units
    Solution

    AB = (1,2,2) and AC = (-1,1,3). AB x AC = ((2)(3) - (2)(1), (2)(-1) - (1)(3), (1)(1) - (2)(-1)) = (4, -5, 3), whose magnitude is sqrt(16 + 25 + 9) = sqrt(50) = 5 sqrt(2). Area = (1/2)(5 sqrt(2)) = 5 sqrt(2)/2 square units.

  5. Q5. The image of the point (2,-3,4) in the line x/2 = (y - 4)/(-1) = (z + 4)/2 is:

    • A.(6, 1, 2)
    • B.(8, 4, 1)
    • C.(4, -1, 6)
    • D.(10, 5, 0)✓
    Solution

    The foot of the perpendicular is F(6,1,2), obtained at t = 3. The image is 2F - P = (12 - 2, 2 + 3, 4 - 4) = (10, 5, 0).

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