MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers

Answer: (c) 12/13, 4/13, 3/13
Let AB be the given line and the DCs of AB be l, m, n. Then
Projection on x-axis = AB . l = 12 (Given)
Projection on y-axis = AB . m = 4 (Given)
Projection on z-axis = AB . n = 3 (Given)
⇒ (AB²) (l² + m² + n²) = 144 + 16 + 9
⇒ (AB²) = 169 {since l² + m² + n² = 1}
⇒ AB = 13
Hence, DCs of AB are 12/13, 4/13, 3/13

Answer: (c) cos θ = (n₁ . n₂)/{|n₁| × |n₂|}
The angle between the planes r . n₁ = d₁ and r . n₂ = d₂ is defined as
cos θ = (n₁ . n₂)/{|n₁| × |n₂|}

Answer: (c) z = 0
The perpendicular distance of P(x, y, z) from xy-plane is zero.

Answer: (b) Line parallel to x-axis
Since x = 0 and y = 0 together represent x-axis, therefore x = a and y = b represent a line parallel to x-axis.

Answer: (d) x + 3y + 6z – 7 = 0
Let the equation of the plane is
(2x – 5y + z – 3) + λ(x + y + 4z – 5) = 0
⇒ (2 + λ)x + (λ – 5)y + (4λ + 1)z – (3 + 5λ) = 0
Since the plane is parallel to x + 3y + 6z – 1 = 0
⇒ (2 + λ)/1 = (λ – 5)/3 = (1 + 4λ)/6
⇒ 6 + 3λ = λ – 5
⇒ 2λ = -11
⇒ λ = -11/2
Again,
6λ – 30 = 3 + 12λ
⇒ -6λ = -33
⇒ λ = -33/6
⇒ λ = -11/2
So, the required equation of plane is
(2x – 5y + z – 3) + (-11/2) × (x + y + 4z – 5) = 0
⇒ 2(2x – 5y + z – 3) + (-11) × (x + y + 4z – 5) = 0
⇒ 4x – 10y + 2z – 6 – 11x – 11y – 44z + 55 = 0
⇒ -7x – 21y – 42z + 49 = 0
⇒ x + 3y + 6z – 7 = 0

Answer: (a) (5/3, 7/3, 17/3)
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: (c) (0, 17/2, -13/2)
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: (c) 6√2i + 6j ± 6k
Let l, m, n be the DCs of OP.
Then it is given that l = cos 45 = 1/√2
m = cos 60 = 1/2
Now, l² + m² + n² = 1
⇒ 1/2 + 1/4 + n² = 1
⇒ n² = 1/4
⇒ n = ±1/2
Now, r = |r|(li + mj + nk)
⇒ r = 12(i/√2 + j/2 ± k/√2)
⇒ r = 6√2i + 6j ± 6k

Answer: (a) (-3, 5, 2)
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: (a) 4 points not in the same plane
Sphere is referred to its center and it follows a quadratic equation with 2 roots. The mid-point of chords of a sphere and parallel to fixed direction lies in the normal diametrical plane.
Now, general equation of the plane depends on 4 constants. So, one sphere passes through 4 points and they need not be in the same plane.

Answer: (d) either (0, -1, 0) or (0, 7, 0)
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: (c) (0, 17/2, -13/2)
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) r . (2i – j + 2k) = 3
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: (a) 1/3, 1/6, 1
Give 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: (b) u² + v² + w² > d
Equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere if
u² + v² + w² – d > 0
⇒ u² + v² + w² > d

Answer: (c) plane
In an x-y-z cartesian coordinate system, the general form of the equation of a plane is
ax + by + cz + d = 0
It is an equation of the first degree in three variables.

Answer: (a) (-3, 5, 2)
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) √(b² + c²)
The coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).
So, the required distance = √{(a – a)² + (b – 0)² + (c – 0)²}
= √(b² + c²)

Answer: (a) |r| = 5
We know that the vector equation of a sphere having center at the origin and radius R
= |r| = R
Here R = 5
Hence, the equation of the required sphere is |r| = 5

Answer: (b) 3 : 5
Let the points are P(1, 2, 3) and Q(-3, 4, -5)
Let the line joining the points P(1, 2, 3) and Q(-3, 4, -5) is divided by the xy-plane at point R in the ratio k : 1
Now, the coordinate of R is
{(-3k + 1)/(k + 1), (4k + 2)/(k + 1), (-5k + 3)/(k + 1)}
Since R lies on the xy-plane.
So, z-coordinate is zero
⇒ (-5k + 3)/(k + 1) = 0
⇒ k = 3/5
So, the ratio = 3/5 : 1 = 3 : 5