Question

Class 11 Maths MCQ Questions of Introduction to Three Dimensional Geometry with Answers?

Answer 1

The important Class 11 Maths MCQ Questions of Three Dimensional Geometry with Answers are covered here. The important MCQ questions are taken from the earlier year question papers and test papers, which will assist you with accomplishing more marks in yearly exams. 

Introduction to 3-D Geometry fuses the accompanying significant concepts, such as

• Coordinate Points in the three-dimensional space
• Distance between two points
• Section Formula

Practice the following important MCQ Questions for class 11 Maths, should help you to solve the problems faster in the final examination and get amazing marks in class 11 Maths final examination.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The cartesian equation of the line is 3x + 1 = 6y – 2 = 1 – z then its direction ratio are

(a) 1/3, 1/6, 1
(b) -1/3, 1/6, 1
(c) 1/3, -1/6, 1
(d) 1/3, 1/6, -1

2. 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)

3. Three planes x + y = 0, y + z = 0, and x + z = 0

(a) none of these
(b) meet in a line
(c) meet in a unique point
(d) meet taken two at a time in parallel lines

4. The coordinate of foot of perpendicular drawn from the point A(1, 0, 3) to the join of the point B(4, 7, 1) and C(3, 5, 3) are

(a) (5/3, 7/3, 17/3)
(b) (5, 7, 17)
(c) (5/3, -7/3, 17/3)
(d) (5/7, -7/3, -17/3)

5. The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a

(a) Straight line
(b) Plane
(c) Sphere
(d) None of these

6. The plane 2x – (1 + a)y + 3az = 0 passes through the intersection of the planes

(a) 2x – y = 0 and y + 3z = 0
(b) 2x – y = 0 and y – 3z = 0
(c) 2x + 3z = 0 and y = 0
(d) 2x – 3z = 0 and y = 0

7. If the end points of a diagonal of a square are (1, -2, 3) and (2, -3, 5) then the length of the side of square is

(a) \(\sqrt3\) units
(b) \(2\sqrt3\) units
(c) \(\sqrt5\) units
(d) \(2\sqrt4\) units

8. The coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ plane is

(a) (0, 17/2, 13/2)
(b) (0, -17/2, -13/2)
(c) (0, 17/2, -13/2)
(d) None of these

9. The equation of plane passing through the point i + j + k and parallel to the plane r . (2i – j + 2k) = 5 is

(a) r . (2i – j + 2k) = 2
(b) r . (2i – j + 2k) = 3
(c) r . (2i – j + 2k) = 4
(d) r . (2i – j + 2k) = 5

10. The points on the y- axis which are at a distance of 3 units from the point (2, 3, -1) is

(a) either (0, -1, 0) or (0, -7, 0)
(b) either (0, 1, 0) or (0, 7, 0)
(c) either (0, 1, 0) or (0, -7, 0)
(d) either (0, -1, 0) or (0, 7, 0)

11. If α, β, γ are the angles made by a half ray of a line respectively with positive directions of X-axis Y-axis and Z-axis, then sin² α + sin² β + sin² γ =

(a) 1
(b) 0
(c) -1
(d) None of these

12. If P(x, y, z) is a point on the line segment joining Q(2, 2, 4) and R(3, 5, 6) such that the projections of OP on the axes are 13/5, 19/5, 26/5 respectively, then P divides QR in the ration

(a) 1 : 2
(b) 3 : 2
(c) 2 : 3
(d) 1 : 3

13. A  plane is parallel to yz-plane so it is perpendicular to

(a) z-axis
(b) y-axis
(c) x-axis
(d) None of these

14.  The locus of a point for which y = 0, z = 0 is   

(a) equation of x-axis
(b) equation of y-axis
(c) equation of z-axis
(d) None of these

15. The distance of point P(3, 4, 5) from the yz-plane is 

(a) 3 units
(b) 4 units
(c) 5 units
(d) 550 units

16. The points A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a 

(a) square
(b) rhombus
(c) rectangle
(d) None of these

17. Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a) 8
(b) 7
(c) 6
(d) None of the above

18. What is the ratio in which the line joining the points (2,4, 5) and (3, 5, - 4) is internally divided by the xy-plane?

(a) 5:4
(b) 3:4
(c) 1:2
(d) 7:5

19. In a three dimensional space, the equation 6y – 2z = 0 represents

(a) a plane containing X axis
(b) none of these
(c) a plane containing Z axis
(d) a plane containing X axis

20. The angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is

(a) π/2
(b) π/3
(c) π/4
(d) π/6

Answer 2

1. Answer: (a) 1/3, 1/6, 1

Explanation: Given 3x + 1 = 6y – 2 = 1 – z

= (3x + 1)/1 = (6y – 2)/1 = (1 – z)/1

= (x + 1/3)/(1/3) = (y – 2/6)/(1/6) = (1 – z)/1

= (x + 1/3)/(1/3) = (y – 1/3)/(1/6) = (1 – z)/1

Now, the direction ratios are: 1/3, 1/6, 1

2. Answer: (a) (-3, 5, 2)

Explanation: Let image of the point P(1, 3, 4) is Q in the given plane.

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 (2r + 1, -r + 3, r + 4) . Now, the coordinate of the mid-point of PQ is

(r + 1, -r/2 + 3, r/2 + 4)

Now, this point lies in the given plane.

2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0

⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0

⇒ 3r + 6 = 0

⇒ r = -2

Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)

= (-3, 5, 2)

3. Answer: (c) meet in a unique point

Explanation: Given, three planes are

x + y = 0 …….. 1

y + z = 0 …….. 2

and x + z = 0 ……… 3

add these planes, we get

2(x + y + z) = 0

⇒ x + y + z = 0 ……… 4

From equation 1

0 + z = 0

⇒ z = 0

From equation 2

x + 0 = 0

⇒ x = 0

From equation 3

y + 0 = 0

⇒ y = 0

So, (x, y, z) = (0, 0, 0)

Hence, the three planes meet in a unique point.

4. Answer: (a) (5/3, 7/3, 17/3)

Explanation: Let D be the foot of perpendicular and let it divide BC in the ration m : 1

Then the coordinates of D are {(3m + 4)/(m + 1), (5m + 7)/(m + 1), (3m + 1)/(m + 1)}

Now, AD ⊥ BC

⇒ AD . BC = 0

⇒ -(2m + 3) – 2(5m + 7) – 4 = 0

⇒ m = -7/4

So, the coordinate of D are (5/3, 7/3, 17/3)

5. Answer: (b) Plane

Explanation: Let the position vectors of the given points A and B be a and b respectively and that of

the variable point be r.

Now, given that

PA² – PB² = k (constant)

⇒ |AP|² – |BP|² = k

⇒ |r – a|² – |r – b|² = k

⇒ (|r|² + |a|²– 2r.a) – (|r|² + |b|² – 2r.b) = k

⇒ 2r.(b – a) = k + |b|² – |a|²

⇒ r.(b – a) = (k + |b|² – |a|²)/2

⇒ r.(b – a) = C where C = (k + |b|² – |a|²)/2 = constant

So, it represents the equation of a plane.

6. Answer: (a) 2x – y = 0 and y + 3z = 0

Explanation: Given, equation of plane is:

2x – (1 + a)y + 3az = 0

=> (2x – y) + a(-y + 3z) = 0

which is passing through the intersection of the planes

2x – y = 0 and -y + 3z = 0

2x – y = 0 and y – 3z = 0

7. Answer: \(\sqrt3\) unit

Explanation: Let a is the length of the side of a square.

Given, the diagonal of a square are (1,–2,3) and (2, -3, 5)

Now, length of the diagonal of square = \(\sqrt{{(1 – 2)2 + (-2 + 3)2 + (3 – 5)2}}\)

= \(\sqrt{{1 + 1 + 4}}\)

= \(\sqrt6\)

Again length of the diagonal of square is \(\sqrt2\) times the length of side of the square.

⇒ \(a\sqrt2\) = \(\sqrt6\)

⇒ \(a\sqrt2=\sqrt3\times\sqrt2\)

⇒ a = \(\sqrt3\)

So, the length of side of square is \(\sqrt3\) unit.

8. Answer: (c) (0, 17/2, -13/2)

Explanation: The line passing through the points (5, 1, 6) and (3, 4, 1) is given as

(x-5)/(3-5) = (y-1)/(4-1) = (z-6)/(1-6)

⇒ (x-5)/(-2) = (y-1)/3 = (z-6)/(-5) = k(say)

⇒ (x-5)/(-2) = k

⇒ x – 5 = -2k

⇒ x = 5 – 2k

(y-1)/3 = k

⇒ y – 1 = 3k

⇒ y = 3k + 1

and (z-6)/(-5) = k

⇒ z – 6 = -5k

⇒ z = 6 – 5k

Now, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)

The equation of YZ-plane is x = 0

Since the line passes through YZ-planeSo, 5 – 2k = 0

⇒ k = 5/2

Now, 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 

= 17/2

and 6 – 5k = 6 – 5×5/2 = 6 – 25/2 

= -13/2

Hence, the required point is (0, 17/2, -13/2)

9. Answer: (b) r . (2i – j + 2k) = 3

Explanation: The equation of plane parallel to the plane r . (2i – j + 2k) = 5 is

r . (2i – j + 2k) = d

Since it passes through the point i + j + k, therefore

(i + j + k) . (2i – j + 2k) = d

⇒ d = 2 – 1 + 2

⇒ d = 3

So, the required equation of the plane is

r . (2i – j + 2k) = 3

10. Answer: (d) either (0, -1, 0) or (0, 7, 0)

Explanation: Let the point on y-axis is O(0, y, 0)

Given point is A(2, 3, -1)

Given OA = 3

⇒ OA² = 9

⇒ (2 – 0)² + (3 – y)² + (-1 – 0)² = 9

⇒ 4 + (3 – y)² + 1 = 9

⇒ 5 + (3 – y)² = 9

⇒ (3 – y)² = 9 – 5

⇒ (3 – y)² = 4

⇒ 3 – y = \(\sqrt4\)

⇒ 3 – y = ±4

⇒ 3 – y = 4 and 3 – y = -4

⇒ y = -1, 7

So, the point is either (0, -1, 0) or (0, 7, 0)

Answer 3

11. Answer: (d) None of these

Explanation: Let l, m, n be the direction cosines of the given vector.

Then, α, β, γ

l = cos α

m = cos β

n = cos γ

Now, l² + m² + n² = 1

⇒ cos² α + cos² β + cos² γ = 1

⇒ 1 – sin² α + 1 – sin² β + 1 – sin² γ = 1

⇒ 3 – sin²α – sin² β – sin² γ = 1

⇒ 3 – 1 = sin²α + sin² β + sin²γ

⇒ sin2α + sin2β + sin2γ = 2

12. Answer: (b) 3 : 2

Explanation: Since OP has projections 13/5, 19/5 and 26/5 on the coordinate axes, therefore

OP = 13i/5 + 19j/5 + 26/5k

Let P divides the join of Q(2, 2, 4) and R(3, 5, 6) in the ratio m : 1

Then the position vector of P is

{(3m + 2)/(m + 1), (5m + 2)/(m + 1), (6m + 4)/(m + 1)}

So, 13i/5 + 19j/5 + 26/5k = (3m + 2)/(m + 1)+ (5m + 2)/(m + 1)+ (6m + 4)/(m + 1)

⇒ (3m + 2)/(m + 1) = 13/5

⇒ 2m = 3

⇒ m = 3/2

⇒ m : 1 = 3 : 2

Hence, P divides QR in the ration 3 : 2

13. Answer: (c) x-axis

Explanation: A plane is parallel to yz-plane which is always perpendicular to x-axis.

14. Answer: (a) equation of x-axis

Explanation: Locus of the point y=0, z=0 is x-axis, since on x-axis both y=0 and z=0.

15. Answer: (a) 3 units

Explanation: From basic ides of three-dimensional geometry, we know that x-coordinate of a point is its distance from yz plane.

∴ Distance of Point P (3, 4, 5) from y z plane is given by its x coordinate.

∵ x-coordinate of point P = 3

∴ Distance of (3, 4, 5) from y z plane is 3 units.

16. Answer: (b) rhombus

Explanation: Since mid-point of [AC] coincides with that of [BD], therefore, ABCD is a parallelogram.

Also |AB|= \(\sqrt{2^2+3^2+6^2}=7\)

and |AD|= \(\sqrt{ 6^2+2^2+3^2}=7\)

therefore, ABCD is a rhombus.

Further, |AC|= \(\sqrt{4^2+5^2+9^2} =\sqrt{122}\)

and |BD|= \(\sqrt{8^2+1^2+3^2}=\sqrt{74}\)

⇒|AC| ≠ |BD|,

 therefore, ABCD is a neither a rectangle nor a square.

17. Answer: (a) 8

Explanation: Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units.

18. Answer: (a) 5:4

Explanation: Let the line joining the points (2, 4, 5) and (3, 5, -4) is internally divided by the xy - plane in the ratio k: 1.

∴ For xy plane, z = 0

\(0=\frac{-k\times 4+5}{k+1}\) = 4k = 5

⇒k = 5/4

So, ratio is 5 : 4

19. Answer: (a) a plane containing X axis

Explanation: Given, equation is 6y – 2z = 0

Here x = 0

So, the given equation 6y- 2z  represents a plane containing X-axis.

20. Answer: (b) π/3

Explanation: Let a is a vector parallel to the vector having direction ratio is 4, -3, 5

⇒ a = 4i – 3j + 5k

Let b is a vector parallel to the vector having direction ratio is 3 ,4, 5

⇒ b = 3i + 4j + 5k

Let θ be the angle between the given vectors.

Now, cos θ = (a . b)/(|a|×|b|)

⇒ cos θ = (12 – 12 + 25)/\(\sqrt{16 + 9 + 25)}\)×\(\sqrt{9+16+25}\)

⇒ cos θ = 25/\(\sqrt{50}\times\sqrt{50}\)

⇒ cos θ = 25/50

⇒ cos θ = 1/2

⇒ cos θ = π/3

⇒ θ = π/3

So, the angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is π/3

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