CBSE Class 12 Mathematics Three Dimensional Geometry MCQs Set E

Question. If a line makes an angle θ₁, θ₂, θ₃ with the axes respectively, then the value of cos 2θ₁ + cos 2θ₂ + cos 2θ₃ is
(a) 1
(b) –1
(c) 4
(d) 3

Answer: B

Question. If the equation of a line AB is x − 3/1 = y + 2/−2 = z − 5/4 , find the direction ratios of a line parallel to AB.
(a) 1, 2, 4
(b) 1, 2, –4
(c) 1, –2, –4
(d) 1, –2, 4

Answer: D

Question. If α, β, γ, are the angles made by a line with the co-ordinate axes. Then sin²α + sin²β + sin²γ is
(a) 0
(b) –1
(c) 2
(d) 1

Answer: C

Question. If α, β, γ, are the direction angles of a vector and cos α = 14/15, cos β = 1/3, then cos γ =
(a) ± (2/15)
(b) ± (1/5)
(c) ± (1/15)
(d) ± (4/15)

Answer: A

Question. If a line makes angles 90°, 60° and 30° with the positive directions of x, y and z-axis respectively, then its direction cosines are
(a) (1/2, 0, √3/2)
(b) (√3/2, 1/2, 0)
(c) (√3/2, 0, 1/2)
(d) (0, 1/2, √3/2)

Answer: D

Question. Find the direction cosines of the line that makes equal angles with the three axes in space.
(a) ± (1/√2)
(b) ±1
(c) ± (1/√3)
(d) √3

Answer: C

Question. Find the equation of a line passing through a point (2, –1, 3) and parallel to the line r = (i + j) + (2i + j − 2k ).
(a) r = (i + j) + μ (2i − j + 3k )
(b) r = (2i − j + 3k ) + μ (2i + j − 2k )
(c) r = (i − j) + μ (2i − j + 3k )
(d) r = (2i + j + 3k ) + μ (2i + j − 2k )

Answer: B

Question. If (1/2, 1/3, n) are the direction cosines of a line, then the value of n is
(a)√ 23/6
(b) 23/6
(c) 2/3
(d) 3/2

Answer: A

Question. The distance of the plane 3x – 6y + 2z + 11 = 0 from the origin is
(a) (11/7) units
(b) (1/7) unit
(c) (7/11) unit
(d) (13/7) units

Answer: A

Question. Find the distance of the point (2, 3, 4) from the plane r (3i − 6j + 2k ) + 11 = 0.
(a) 1 unit
(b) 4 units
(c) 2 units
(d) 3 units

Answer: B

Question. Write the direction cosines of a line parallel to the line 3 - x/3 = y + 2/−2 = z + 2/6
(a) 1/7, 2/7, 3/7
(b) −3/7, 2/7, 6/7
(c) 3/7, 2/7, 6/7
(d) 3/7, -2/7, 6/7 

Answer: B

Question. The equation of a plane with intercepts 2, 3 and 4 on the X, Y and Z-axes respectively is A. Here, A refers to
(a) 2x + 3y + 4z = 12
(b) 6x + 4y + 3z = 12
(c) 2x + 3y + 4z = 1
(d) 6x + 4y + 3z = 1

Answer: B

Question. What is the distance between the planes 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0 ?
(a) 2/6 unit
(b) 3/2 units
(c) 1/6 unit
(d) 1/4 unit

Answer: C

Question. The equation of a line is given by 4 - x/2 = y + 3/3 = z + 2/6 , the direction cosines of line parallel to the given line is
(a) −2/7, 3/7, 6/7
(b) 2/7, -3/7, 6/7
(c) 2/7, 3/7, 6/7
(d) −2/7, 3/7, 6/7

Answer: D

Question. A line makes angles a, b and g with the co-ordinate axes. If a + b = 90°, then the value of angle g is
(a) 60°
(b) 90°
(c) 45°
(d) 30°

Answer: B

Question. The vector equation of the plane passing through a point having position vector 2i + 3j + 4k and perpendicular to the vector 2i + j − 2k is
(a) r .( i j k) 2 + − 2 = −1
(b) r .( i j k) 2 + − 2 = 8
(c) r .( i j k) 2 + − 2 = 9
(d) r .( i j k) 2 + − 2 = 15

Answer: A

Question. The cartesian equation of a line is x + 3/2 = y − 5/4 = z + 6/2. The vector equation for the line is
(a) 2i + 3 j − 6k + λ(2i − 3 j + 2k)
(b) − 3i + 5 j − 6k + λ(2i + 4 j + 2k)
(c) − 3i − 5 j + 6k + λ(2i − 3 j − 2k)
(d) 3i + 5 j + 6k + λ(2i − 4 j − 2k)

Answer: B

Question. Find the equation of plane passing through the point (1, 2, 3) and the direction cosines of the normal as l, m, n.
(a) lx + my + nz = l + 2m + 3n
(b) lx + my + nz + (l + 2m + 3n) = 0
(c) lx + my + nz = 1/2 (l + 2m + 3n)
(d) None of the options

Answer: A

Question. The lines x − 1/2 = y + 1/-3 = z + 10/8 and x - y/1 = y + 3/k = z + 1/7 are coplanar if k =
(a) 4
(b) –4
(c) 2
(d) –2

Answer: B

Question. Find the equation of the plane passing through (2, 3, –1) and is perpendicular to the vector 3i − 4j + 7k .
(a) 3x – 4y + 7z + 13 = 0
(b) 3x + 4y – 7z – 13 = 0
(c) 3x + 4y + 7z – 13 = 0
(d) 3x – 4y – 7z + 13 = 0

Answer: A

Question. The equation of a line passing through the point (– 3, 2, –4) and equally inclined to the axes are
(a) x – 3 = y + 2 = z – 4
(b) x + 3 = y – 2 = z + 4
(c) x + 3/1 = y − 2/2 = z + 4/3
(d) None of these

Answer: B

Question. Find the direction cosines of the line x − 2/2 = 2y − 5/-3 = z + 1/0
(a) –2, –5, 1
(b) 2, –3, 0
(c) 2, − 3/2, 0
(d) 4/5, - 3/5, 0

Answer: D

Question. An equation of the plane passing through the points (3, 2, –1), (3, 4, 2) and (7, 0, 6) is 5x + 3y – 2z = l, where l is
(a) 23
(b) 21
(c) 19
(d) 27

Answer: A

Question. The distance of the plane 2x – 3y + 4z – 6 = 0 from the origin is A. Here, A refers to
(a) 6
(b) –6
(c) − 6/√29
(d) 6/√29

Answer: D

Question. What is the distance (in units) between the two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9?
(a) 0
(b) 3
(c) 6/√83
(d) 6

Answer: A

Question. Distance of the point (a, b, g) from y-axis is
(a) b
(b) |b|
(c) |b| + |g|
(d) 2 + 2

Answer: D

Question. The reflection of the point (a, b, g) in the xyplane is
(a) (a, b, 0)
(b) (0, 0, g)
(c) (–a, –b, g)
(d) (a, b, –g)

Answer: D

Question. P is a point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If x co-ordinate of P is 5, then its y co-ordinate is
(a) 2
(b) 1
(c) –1
(d) –2

Answer: A

Question. The vector equation of the line through the points A(3, 4, –7) and B(1, –1, 6) is
(a) r = (3i − 4j − 7k ) + λ(i − j + 6k )
(b) r = (i − j + 6k ) + λ(3i − 4j − 7k )
(c) r = (3i + 4j − 7k ) + λ(−2i − 5j + 13k )
(d) r = (i − j + 6k ) + λ(4i + 3j − k )

Answer: C

Question. If the line joining (2, 3, –1) and (3, 5, –3) is perpendicular to the line joining (1, 2, 3) and (3, 5, l), then l =
(a) –3
(b) 2
(c) 5
(d) 7

Answer: D

Question. The value of p, so that the lines 1 - x/3 = 7y - 14/2p = z − 3/2 and 7 - 7x/3p = y - 5/1 = 6 - z/5 intersect at right angle, is
(a) 10/11
(b) 70/11
(c) 10/7
(d) 70/9

Answer: B

Question. The lines x + 3/-3 = y - 1/1 = z - 5/5 and x + 1/-1 = y - 2/2 = z - 5/5 are
(a) coplanar
(b) non-coplanar
(c) perpendicular
(d) none of these

Answer: A

Question. If the lines x − 2/1 = y − 9/2 = z − 13/3 and x − a/1 = y − 1/-2 = z + 2/3 are coplanar, then a =
(a) 2
(b) –2
(c) 3
(d) –3

Answer: D

Case Based MCQs

Case-I : Read the following passage and answer the questions.

In a diamond exhibition, a diamond is covered in cubical glass box having coordinates
O(0, 0, 0), A(1, 0, 0), B(1, 2, 0), C(0, 2, 0), O′(0, 0, 3), A′(1, 0, 3), B′(1, 2, 3) and C′(0, 2, 3).

Question. Direction ratios of OA are
(a) < 0, 1, 0 >
(b) < 1, 0, 0 >
(c) < 0, 0, 1 >
(d) none of these

Answer: B

Question. Equation of diagonal OB′ is
(a) x/1 = y/2 = z/3
(b) x/0 = y/1 = z/2
(c) x/1 = y/0= z/2
(d) none of these

Answer: A

Question. Equation of plane OABC is
(a) x = 0
(b) y = 0
(c) z = 0
(d) none of these

Answer: C

Question. Equation of plane O′A′B′C′ is
(a) x = 3
(b) y = 3
(c) z = 3
(d) z = 2

Answer: C

Question. Equation of plane ABB′A′ is
(a) x = 1
(b) y = 1
(c) z = 2
(d) x = 3

Answer: A

Case-II : Read the following passage and answer the questions.

A football match is organised between students of class XII of two schools, say school A and school B. For which a team from each school is chosen. Remaining students of class XII of school A and B are respectively sitting on the plane represented by the equation r (i + j + 2k ) = 5 and r (2i − j + k ) = 6, to cheer up the team of their respective schools.

Question. The cartesian equation of the plane on which students of school A are seated is
(a) 2x – y + z = 8
(b) 2x + y + z = 8
(c) x + y + 2z = 5
(d) x + y + z = 5

Answer: C

Question. The magnitude of the normal to the plane on which students of school B are seated, is
(a) √5
(b) √6
(c) √3
(d) √2

Answer: B

Question. The intercept form of the equation of the plane on which students of school B are seated, is
(a) x/6 + y/6 + z/6 = 1
(b) x/3 + y/(-6) + z/6 = 1
(c) x/3 + y/6 + z/6 = 1
(d) x/3 + y/6 + z/3 = 1

Answer: B

Question. Which of the following is a student of school B?
(a) Mohit sitting at (1, 2, 1)
(b) Ravi sitting at (0, 1, 2)
(c) Khushi sitting at (3, 1, 1)
(d) Shewta sitting at (2, –1, 2)

Answer: C

Question. The distance of the plane, on which students of school B are seated, from the origin is
(a) 6 units
(b) (1/√6) units
(c) (5/√6) units
(d) √6 units

Answer: D

Assertion & Reasoning Based MCQs

Directions (Q.-51 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose
the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

Question. Assertion : The points (1, 2, 3), (–2, 3, 4) and (7, 0, 1) are collinear.
Reason : If a line makes angles π/2, 3π/4, and  π/4 with X, Y, and Z-axes respectively, then its direction cosines are 0, −1/√2 and 1/√2.

Answer: B

Question. Assertion : If the cartesian equation of a line is x − 5/3 = y + 4/7 = z − 6/2, then its vector form is r = 5i − 4j + 6k + λ(3i + 7j + 2k ) .
Reason : The cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by x + 3/3 = y − 4/5 = z + 8/6 is x + 3/-2 = y - 4/4 = z + 8/-5

Answer: C

Question. Assertion : The three lines with direction cosines 12/13, -3/13, 4/13, 4/13, 12/13, 3/13, 3/13, -4/13, 12/13, are mutually perpendicular.
Reason : The line through the points (1, –1, 2) and (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Answer: B

Question. Assertion : The pair of lines given by r = i − j + λ(2i + k ) and r = 2i − k + μ(i + j − k ) intersect.
Reason : Two lines intersect each other, if they are not parallel and shortest distance = 0.

Answer: A

Question. Assertion : There exists only one plane that is perpendicular to the given vector.
Reason : Through a given point perpendicular to the given vector only one plane exists.

Answer: D

Question. Assertion : If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, then the small angle δq between the two positions is given by δq² = δl² + δm² + δn².
Reason : If O is the origin and A is (a, b, c), then the equation of plane through A at right angle to OA is given by ax + by + cz = a² + b² + c².

Answer: B

Question. Consider the lines L₁ : x + 1/3 = y + 2/1 = z + 1/2, L₂ : x - 2 = y + 2/2 = z - 3/3
Assertion : The distance of point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L₁ and L₂ is 13/5√3
Reason : The unit vector perpendicular to both the lines L₁ and L₂ is −i − 7j + 5k/ 5√3

Answer: A

Question. Assertion : The equation of a plane which passes through (2, –3, 1) and normal to the line joining the points (3, 4, –1) and (2, –1, 5) is given by x + 5y – 6z + 19 = 0.
Reason : The length of perpendicular from the point (7, 14, 5) to the plane 2x + 4y – z = 2 is 2√21.

Answer: C

Question. Assertion : Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′ respectively from the origin, then 1/a2 + 1/b2 + 1/c2 = a1/a2 + 1/b2 + 1/c2
Reason : The points (i − j + 3k ) and 3(i + j + k ) are equidistant from the plane r (5i + 2j − 7k ) + 9 = 0.

Answer: B

Question. Assertion : The straight line x − y / − 4 = y − 4 /− 7 = z + 3 / 13 lies in the plane 5x – y + z = 8.
Reason : The straight line x - x1/ l = y - y1/m = z - z1/c lies in the plane ax + by + cz + d = 0 iff normal to the plane is perpendicular to the line & every point of the line satisfies the equation of the plane.