Check the below NCERT MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Introduction to Three Dimensional Geometry Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

## Introduction to Three Dimensional Geometry Class 11 MCQs Questions with Answers

**MCQ On Three Dimensional Geometry Class 11 Question 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

## Answer

Answer: (a) 1/3, 1/6, 1

Hint:

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

**MCQ On Three Dimensional Geometry Question 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)

## Answer

Answer: (a) (-3, 5, 2)

Hint:

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)

**Introduction To 3d Geometry Class 11 Extra Questions Question 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

## Answer

Answer: (c) meet in a unique point

Hint:

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.

**Important Questions Of 3d Geometry Class 11 Question 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)

## Answer

Answer: (a) (5/3, 7/3, 17/3)

Hint:

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)

**MCQ On Introduction To Three Dimensional Geometry Class 11 Question 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

## Answer

Answer: (b) Plane

Hint:

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.

**3d Geometry Class 11 Questions Question 6.**

The equation of the set of point P, the sum of whose distance from A(4, 0, 0) and B(-4, 0, 0) is equal to 10 is

(a) 9x² + 25y² + 25z² + 225 = 0

(b) 9x² + 25y² + 25z² – 225 = 0

(c) 9x² + 25y² – 25z² – 225 = 0

(d) 9x² – 25y² – 25z² – 225 = 0

## Answer

Answer: (b) 9x² + 25y² + 25z² – 225 = 0

Hint:

Let the point P is (x, y, z)

Now given that

PA + PB = 10

⇒ √{(x-4)² + y² + z²} + √{(x+4)² + y² + z²} = 10

⇒ √{(x-4)² + y² + z²} = 10 – √{(x+4)² + y² + z²}

Now square both side

[√{(x-4)² + y² + z²}]² = (10)² + [{(x+4)² + y² + z²}]² – 2 ×10×√{(x+4)² + y² + z²}

⇒ {(x-4)² + y² + z²} = 100 + {(x+4)² + y² + z²} – 20×√{(x+4)² + y² + z²}

⇒ x² + 16 – 8x + y² + z² = 100 + x² + 16 + 8x + y² + z² – 20×√{(x+4)² + y² + z²}

⇒ – 8x = 100 + 8x – 20×√{(x+4)² + y² + z²}

⇒ -8x -8x – 100 = – 20×√{(x+4)² + y² + z²}

⇒ -16x -100 = – 20×√{(x+4)² + y² + z²}

⇒ 4x + 25 = 5×√{(x+4)² + y² + z²}

Again square both side,

(4x + 25)² = 25 ×[√{(x+4)² + y² + z²}]²

⇒ 16x² + 625 + 200x = 25×{(x+4)² + y² + z²}

⇒ 16x² + 625 + 200x = 25×(x² + 16 + 8x + y² + z²)

⇒ 16x² + 625 + 200x = 25x² + 400 + 200x + 25y² + 25z²

⇒ 25x² + 400 + 200x + 25y² + 25z² – 16x² – 625 – 200x = 0

⇒ 9x² + 25y² + 25z² – 225 = 0

**Important Questions Of Three Dimensional Geometry Class 11 Question 7.**

The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is

(a) 3

(b) 4

(c) 5

(d) Can not be find

## Answer

Answer: (c) 5

Hint:

Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)

Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}

⇒ distance = √{(4cos θ – 3sin θ)²}

⇒ distance = 4cos θ – 3sin θ ……………. 1

Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}

= √(16 + 9)

= √25

= 5

From equation 1, we get

distance = 5

So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

**MCQ On 3d Geometry Class 11 Question 8.**

A vector r is equally inclined with the coordinate axes. If the tip of r is in the positive octant and |r| = 6, then r is

(a) 2√3(i – j + k)

(b) 2√3(-i + j + k)

(c) 2√3(i + j – k)

(d) 2√3(i + j + k)

## Answer

Answer: (d) 2√3(i + j + k)

Hint:

Let l, m, n are DCs of r.

Given, l = m = n

⇒ l² + m² + n² = 1

⇒ 3l² = 1

⇒ l² = 1/3

⇒ l = m = n = 1/√3

So, r = |r|(li + mj + nk)

⇒ r = 6(i/√3 + j/√3 + k/√3)

⇒ r = 2√3(i + j + k)

**Class 11 Maths Chapter 12 Important Questions Question 9.**

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

2x – y = 0 and y + 3z = 0

2x – y = 0 and y – 3z = 0

2x + 3z = 0 and y = 0

2x – 3z = 0 and y = 0

## Answer

Answer: (d) A

Hint:

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

**Introduction To 3d Geometry Class 11 Questions Question 10.**

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) √3 unit

(b) 2√3 unit

(c) 3√3 unit

(d) 4√3 unit

## Answer

Answer: (a) √3 unit

Hint:

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 = √{(1 – 2)² + (-2 + 3)² + (3 – 5)²}

= √{1 + 1 + 4}

= √6

Again length of the diagonal of square is √2 times the length of side of the square.

⇒ a√2 = √6

⇒ a√2 = √3×√2

⇒ a = √3

So, the length of side of square is √3 unit

**MCQ Questions For Class 11 Maths With Answers Pdf Download Question 11.**

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

## Answer

Answer: (c) (0, 17/2, -13/2)

Hint:

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-plane

So, 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)

**Class 11 Maths MCQ Questions Question 12.**

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

Answer: (b) π/3

Hint:

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)/{√(16 + 9 + 25)×√(9 + 16 + 25)}

⇒ cos θ = 25/{√(50)×√(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

**Chapter 12 Class 11 Maths Question 13.**

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

## Answer

Answer: (b) r . (2i – j + 2k) = 3

Hint:

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

**Questions On 3d Geometry Question 14.**

A vector r is equally inclined with the coordinate axes. If the tip of r is in the positive octant and |r| = 6, then r is

(a) 2√3(i – j + k)

(b) 2√3(-i + j + k)

(c) 2√3(i + j – k)

(d) 2√3(i + j + k)

## Answer

Answer: (d) 2√3(i + j + k)

Hint:

Let l, m, n are DCs of r.

Given, l = m = n

⇒ l² + m² + n² = 1

⇒ 3l² = 1

⇒ l² = 1/3

⇒ l = m = n = 1/√3

So, r = |r|(li + mj + nk)

⇒ r = 6(i/√3 + j/√3 + k/√3)

⇒ r = 2√3(i + j + k)

**Chapter 12 Maths Class 11 Question 15.**

The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is

(a) 3

(b) 4

(c) 5

(d) Can not be find

## Answer

Answer: (c) 5

Hint:

Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)

Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}

⇒ distance = √{(4cos θ – 3sin θ)²}

⇒ distance = 4cos θ – 3sin θ …………….1

Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}

= √(16 + 9)

= √25

= 5

From equation 1, we get

distance = 5

So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

**Introduction To 3d Geometry Class 11 Formulas Question 16.**

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)

Hint:

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)

**Introduction To Three Dimensional Geometry Ncert Solutions Question 17.**

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)

## Answer

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

Hint:

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 = √4

⇒ 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)

Question 18.

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

## Answer

Answer: (d) None of these

Hint:

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 – sin² α – sin² β – sin² γ = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

⇒ sin² α + sin² β + sin² γ = 2

Question 19.

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

## Answer

Answer: (b) 3 : 2

Hint:

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

Question 20.

In a three dimensional space, the equation 3x – 4y = 0 represents

(a) a plane containing Y axis

(b) none of these

(c) a plane containing Z axis

(d) a plane containing X axis

## Answer

Answer: (c) a plane containing Z axis

Hint:

Given, equation is 3x – 4y = 0

Here z = 0

So, the given equation 3x – 4y = 0 represents a plane containing Z axis.

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