Class 12 Maths Chapter 11 Three Dimensional Geometry MCQs are provided here with solutions. The multiple-choice questions for Three-dimensional geometry are given as per the CBSE syllabus (2022-2023) and NCERT curriculum. Class 12 Maths MCQs for all chapters are prepared as per the latest exam pattern. Practising these questions will help students to score good marks in the board exam.
MCQs for Class 12 Maths Three Dimensional Geometry
Find the mcqs for class 12 Maths chapter 11 three dimensional geometry with solutions here.
Download PDF – Chapter 11 Three Dimensional Geometry MCQs
Q.1: The direction cosines of the y-axis are:
A. (9, 0, 0)
B. (1, 0, 0)
C. (0, 1, 0)
D. (0, 0, 1)
Answer: C. (0, 1, 0)
Q.2: Find the equation of the plane passing through the points P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4).
A. x + 2y = 0
B. x – y – 2 = 0
C. -x + 2y – 2 = 0
D. x + y – 2 = 0
Answer: D. x + y – 2 = 0
Explanation: Given three points, P(1, 1, 1), Q(3, -1, 2) and R(-3, 5, -4).
On solving we get;
⇒ x + y – 2 = 0
Q.3: The equation x² – x – 2 = 0 in three-dimensional space is represented by:
A. A pair of parallel planes
B. A pair of straight lines
C. A pair of the perpendicular plane
D. None of these
Answer: A. A pair of parallel planes
Q.4: The direction ratios of the normal to the plane 7x + 4y – 2z + 5 = 0 are:
A. 7, 4,-2
B. 7, 4, 5
C. 7, 4, 2
D. 4, -2, 5
Answer: A. 7, 4,-2
Q.5: If l, m, n are the direction cosines of a line, then;
A. l2+ m2+ 2n2 = 1
B. l2+ 2m2+ n2 = 1
C. 2l2+ m2+ n2 = 1
D. l2+ m2+ n2 = 1
Answer: D. l2+ m2+ n2 = 1
Q.6: Direction ratio of line joining (2, 3, 4) and (−1, −2, 1), are:
A. (−3, −5, −3)
B. (−3, 1, −3)
C. (−1, −5, −3)
D. (−3, −5, 5)
Answer: A. (−3, −5, −3)
Explanation: The direction ratio of the line joining A (2, 3, 4) and B (−1, −2, 1), are:
(−1−2), (−2−3), (1−4)
= (−3, −5, −3)
Q.7: If a line has direction ratios 2, – 1, – 2, determine its direction cosines:
A. ⅓, ⅔, -⅓
B. ⅔, -⅓, -⅔
C. -⅔, ⅓, ⅔
D. None of the above
Answer: B. ⅔, -⅓, -⅔
Explanation: Direction cosines are:
= ⅔, -⅓, -⅔
Q.8: The vector equation for the line passing through the points (–1, 0, 2) and (3, 4, 6) is:
A. i + 2k + λ(4i + 4j + 4k)
B. i – 2k + λ(4i + 4j + 4k)
C. -i+2k+ λ(4i + 4j + 4k)
D. -i+2k+ λ(4i – 4j – 4k)
Answer: C. -i+2k+ λ(4i + 4j + 4k)
Explanation: The vector equation of the line is given by:
r = a + λ (b – a), λ ∈ R
Let a = -i + 2k
And b = 3i + 4j + 6k
b – a = 4i + 4j + 4k
Let the vector equation be r, then;
r = -i + 2k + λ (4i + 4j + 4k)
Q.9: If the lines x-2/1 =y-2/1 =z-4/k and x-1/k = y-4/2 = z-5/1 are coplanar, then k can have:
A. Exactly two values
B. Exactly three values
C. Exactly one value
D. Any value
Answer: A. Exactly two values
Q.10: What are the direction cosines of the equation of the plane 2x + 3y – z = 5?
A. 1/√14, 3/√14, -2/√14
B. 2/√14, 3/√14, -1/√14
C. 2/√14, 1/√14, -1/√14
D. 2/√14, -2/√14, -3/√14
Answer: B. 2/√14, 3/√14, -1/√14
Explanation: The equation of the plane, 2x + 3y – z = 5…. (1)
Direction ratio of the normal (2, 3, -1)
By using the formula,
√[(2)2 + (3)2 + (-1)2] = √14
Now,
Divide both the sides of equation (1) by √14, we get
2x/(√14) + 3y/(√14) – z/(√14) = 5/√14
So this is of the form lx + my + nz = d
Where, l, m, n are the direction cosines and d is the distance
Therefore, the direction cosines are 2/√14, 3/√14, -1/√14
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