Class 11 Maths Chapter 12 Introduction to Three-dimensional Geometry MCQs with solutions are provided at BYJU’S for students. The multiple-choice questions for Class 11 Maths Chapter 12 are prepared as per the latest exam pattern by our subject teachers. The objective type questions are given in accordance with the CBSE syllabus (2022 – 2023) and NCERT curriculum. By solving the MCQs on Introduction to Three-dimensional geometry, students can score good marks in exam.
Also check: Class 11 Maths Chapter-wise MCQs
MCQs for Class 11 Maths Chapter 12 Introduction to Three-dimensional Geometry
Class 12 Maths MCQs of chapter 12 Introduction to three-dimensional geometry are provided here with the correct option. These multiple-choice questions for 3d geometry are given here with proper explanations. The questions are based on the three-dimensional geometry concept and related formulas. Students are advised to solve the MCQs questions by themselves and then verify with the provided answers.
Download PDF – Chapter 12 Introduction to Three Dimensional Geometry MCQs
Q.1: Which octant do the point (−5,4,3) lie?
A. Octant I
B. Octant II
C. Octant III
D. Octant IV
Answer: B. Octant II
Explanation: Given (−5,4,3) is the point.
Here, the x-coordinate is negative but y and z coordinates are positive. Therefore, (−5,4,3) lie in octant II.
Q.2: A point is on the x-axis. Which of the following represent the point?
A. (0, x, 0)
B. (0, 0, x)
C. (x, 0, 0)
D. None of the above
Answer: C. (x, 0, 0)
Explanation: At x-axis, y and z coordinates are zero.
Q.3: Coordinate planes divide the space into ______ octants.
A. 4
B. 6
C. 8
D. 10
Answer: C. 8
Explanation: The coordinate planes divide the three dimensional space into eight octants.
Q.4: What is the distance between the points (2, –1, 3) and (–2, 1, 3)?
A. 2√5 units
B. 25 units
C. 4√5 units
D. √5 units
Answer: A. 2√5 units
Explanation: Let the points be P (2, – 1, 3) and Q (– 2, 1, 3)
By using the distance formula,
PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
Therefore, the required distance is 2√5 units.
Q.5: The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is:
(a) 3 units
(b) 4 units
(c) 5 units
(d) Cannot be determined
Answer: (c) 5
Explanation: Let the two points be P (3sin θ, 0, 0) and Q (4cos θ, 0, 0)
Now by distance formula,
PQ = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
PQ = √{(4cos θ – 3sin θ)²}
PQ = 4cos θ – 3sin θ
Now, maximum value of 4cos θ – 3sin θ;
= √{(4² + (-3)²}
= √(16 + 9)
= √25
= 5
Thus, PQ = 5 units
So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5.
Q.6: The locus represented by xy + yz = 0 is:
(a) A pair of perpendicular lines
(b) A pair of parallel lines
(c) A pair of parallel planes
(d) A pair of perpendicular planes
Answer: (d) A pair of perpendicular planes
Q.7: Find the image of (−2,3,4) in the y z plane.
A. (-2, 3, 4)
B. (2, 3, 4)
C. (-2, -3, 4)
D. (-2, -3, -4)
Answer: B. (2, 3, 4)
Q.8: The perpendicular distance of the point P(6, 7, 8) from the XY – Plane is:
(a)8
(b)7
(c)6
(d) None of the above
Answer: A. 8
Explanation: Let Q be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane.
Thus, the distance of this foot Q from P is z-coordinate of P, i.e., 8 units
Q.9: The image of the point P(1,3,4) in the plane 2x – y + z = 0 is:
(a) (-3, 5, 2)
(b) (3, 5, 2)
(c) (3, -5, 2)
(d) (3, 5, -2)
Answer: Answer: (a) (-3, 5, 2)
Explanation:
Let the image of the point P(1, 3, 4) is Q.
The equation of the line through P and normal to the given plane is:
(x – 1)/2 = (y – 3)/-1 = (z – 4)/1
Since the line passes through Q, so let the coordinate of Q are (2k + 1, -k + 3, k + 4)
Now, the coordinate of the mid-point of PQ is:
(k + 1, -k/2 + 3, k/2 + 4)
Now, this midpoint lies in the given plane.
2(k + 1) – (-k/2 + 3) + (k/2 + 4) + 3 = 0
⇒ 2k + 2 + k/2 – 3 + k/2 + 4 + 3 = 0
⇒ 3k + 6 = 0
⇒ k = -2
Hence, the coordinate of Q is (2k + 1, -k + 3, k + 4) = (-4 + 1, 2 + 3, -2 + 4)
= (-3, 5, 2)
Q.10: The distance of the point P(a, b, c) from the x-axis is:
(a) √(a² + c²)
(b) √(a² + b²)
(c) √(b² + c²)
(d) None of these
Answer: (c) √(b² + c²)
Explanation:
The coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).
So, the required distance = √{(a – a)² + (b – 0)² + (c – 0)²}
= √(b² + c²)
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