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MCQ Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers - Set - 2
Question 1:
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
Correct Answer – (C)
Given, equation is 3x – 4y = 0
Here z = 0
So, the given equation 3x – 4y = 0 represents a plane containing Z axis.
Question 2 :
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
Correct Answer – (D)
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 3 :
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)
Correct Answer – (A)
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)
Question 4 :
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)
Correct Answer – (D)
Question 5 :
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
Correct Answer – (B)
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
MCQ Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers
Question 6 :
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
Correct Answer – (B)
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 7 :
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)
Correct Answer – (D)
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 8 :
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
Correct Answer – (C)
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
Question 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
Correct Answer – (B)
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
Question 10 :
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
Correct Answer – (C)
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)
- NCERT Solutions Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers : Exercise 12.1
- NCERT Solutions Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers : Exercise 12.2
- NCERT Solutions Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers : Exercise 12.3
- NCERT Solutions Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers : Exercise 12 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
