Introduction to Three Dimensional Geometry Class 11 MCQ Online Test With Answers Questions

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

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)

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.

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)

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.

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

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

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)

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

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

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)

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

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

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)

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

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)

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)

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

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

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.