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MCQ Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers - Set - 1
Question 1:
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
Correct Answer – (A)
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
Question 2 :
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)
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)
Question 3 :
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
Correct Answer – (B)
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
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)
Correct Answer – (A)
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)
Question 5 :
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)
MCQ Class 11 Mathematics Introduction to Three Dimensional Geometry with Answers
Question 6 :
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
Correct Answer – (D)
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
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
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 8 :
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
Correct Answer – (B)
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.
Question 9 :
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
Correct Answer – (C)
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.
Question 10 :
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
Correct Answer – (A)
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
- 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
