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Chapter-11 Introduction to Three-Dimensional Geometry — Online MCQ Test

MATHS · CLASS 11 FIRST PUC · Karnataka State Board

Practice Chapter-11 Introduction to Three-Dimensional Geometry with a free chapter-wise online MCQ test for Karnataka State Board CLASS 11 FIRST PUC MATHS. This chapter covers: Coordinate axes - Coordinate planes - Octants - Coordinates of a point in space - Distance formula - Section formula. AI-generated questions from basic to board-exam level, with instant results and explanations.

10
Questions
20m
Time Limit
3
Attempts Left
  • 10 random questions from this chapter (mixed difficulty)
  • Questions you've seen before won't repeat until the pool resets
  • You have 20 minutes — exam auto-submits when time is up
  • Maximum 3 attempts per chapter
  • Results and explanations shown immediately after submission
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Chapter-11 Introduction to Three-Dimensional Geometry — Important Questions & Answers (FAQ)

Frequently asked questions from Karnataka State Board CLASS 11 FIRST PUC MATHS — Chapter-11 Introduction to Three-Dimensional Geometry, with answers and explanations. These are sample questions; the exam has its own separate question set.

In three-dimensional geometry, how many coordinate axes are there?
  • A. 2
  • B. 3 ✓
  • C. 4
  • D. 5
Answer: B. 3
In 3D geometry, there are three mutually perpendicular coordinate axes: X-axis, Y-axis, and Z-axis.
The three coordinate planes in 3D geometry are XY-plane, YZ-plane, and _____?
  • A. XZ-plane
  • B. ZX-plane
  • C. Both A and B ✓
  • D. None of these
Answer: C. Both A and B
The three coordinate planes are XY-plane, YZ-plane, and ZX-plane (or XZ-plane), which are the same plane with different notations.
Find the distance between points A(1, 2, 3) and B(4, 5, 6).
  • A. √27 ✓
  • B. √36
  • C. 9
  • D. 6
Answer: A. √27
Distance = √[(4-1)² + (5-2)² + (6-3)²] = √[9 + 9 + 9] = √27.
The coordinates of a point equidistant from A(1, 2, 3) and B(3, 2, 1) lie on:
  • A. A circle
  • B. A line perpendicular to AB
  • C. The perpendicular bisector plane of AB ✓
  • D. A sphere
Answer: C. The perpendicular bisector plane of AB
Points equidistant from two given points lie on the perpendicular bisector plane of the line segment joining them.
A point P(x, y, z) is at distance 13 from the origin and also from the point (2, 3, 6). Which equation satisfies this condition?
  • A. 2x + 3y + 6z = 49 ✓
  • B. 4x + 6y + 12z = 49
  • C. x + y + z = 13
  • D. x² + y² + z² = 169
Answer: A. 2x + 3y + 6z = 49
Setting PA = PB where A is origin and B(2,3,6): x² + y² + z² = (x-2)² + (y-3)² + (z-6)² simplifies to 2x + 3y + 6z = 49.

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