MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers
Introduction to Three Dimensional Geometry Class 11 Maths MCQs Questions with Answers
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.
Class 11 Maths Chapter 12 Quiz
Class 11 Maths Chapter 12 MCQ Online Test
You can refer to NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
Introduction to Three Dimensional Geometry Class 11 Maths MCQ online test
| Q1. | The projections of a directed line segment on the coordinate axes are 12, 4, 3. The DCS of the line are |
A.12/13, -4/13, 3/13 |
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B.-12/13, -4/13, 3/13 |
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C.12/13, 4/13, 3/13 |
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D.None of these |
Ans: 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
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
| Q2. | The angle between the planes r . n1 = d1 and r . n1 = d2 is |
A.cos θ ={|n1| × |n2|}/ (n1. n2) |
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B.cos θ = (n1 . n2)/{|n1| × |n2|}² |
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C.cos θ = (n1 . n2)/{|n1| × |n2|} |
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D.cos θ = (n1 . n2)² /{|n1| × |n2|} |
Ans: 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|}
The angle between the planes r . n1 = d1 and r . n2 = d2 is defined as
cos θ = (n1 . n2)/{|n1| × |n2|}
| Q3. | For every point P(x, y, z) on the xy-plane |
A.x = 0 |
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B.y = 0 |
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C.z = 0 |
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D.None of these |
Ans: z = 0
The perpendicular distance of P(x, y, z) from xy-plane is zero.
The perpendicular distance of P(x, y, z) from xy-plane is zero.
| Q4. | The locus of a point P(x, y, z) which moves in such a way that x = a and y = b, is a |
A.Plane parallel to xy-plane |
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B.Line parallel to x-axis |
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C.Line parallel to y-axis |
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D.Line parallel to z-axis |
Ans: 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.
Since x = 0 and y = 0 together represent x-axis, therefore x = a and y = b represent a line parallel to x-axis.
| Q5. | The equation of the plane containing the line 2x – 5y + z = 3, x + y + 4z = 5 and parallel to the plane x + 3y + 6z = 1 is |
A.x + 3y + 6z + 7 = 0 |
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B.x + 3y – 6z – 7 = 0 |
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C.x – 3y + 6z – 7 = 0 |
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D.x + 3y + 6z – 7 = 0 |
Ans: 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
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
| Q6. | 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) |
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B.(5, 7, 17) |
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C.(5/3, -7/3, 17/3) |
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D.(5/7, -7/3, -17/3) |
Ans: (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)
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)
| Q7. | 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) |
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B.(0, -17/2, -13/2) |
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C.(0, 17/2, -13/2) |
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D.None of these |
Ans: (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)
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)
| Q8. | If P is a point in space such that OP = 12 and OP inclined at angles 45 and 60 degrees with OX and OY respectively, then the position vector of P is |
A.6i + 6j ± 6√2k |
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B.6i + 6√2j ± 6k |
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C.6√2i + 6j ± 6k |
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D.None of these |
Ans: 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
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
| Q9. | The image of the point P(1,3,4) in the plane 2x – y + z = 0 is |
A.(-3, 5, 2) |
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B.(3, 5, 2) |
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C.(3, -5, 2) |
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D.(3, 5, -2) |
Ans: (-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)
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)
| Q10. | There is one and only one sphere through |
A.4 points not in the same plane |
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B.4 points not lie in the same straight line |
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C.none of these |
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D.3 points not lie in the same line |
Ans: 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.
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.
| Q11. | 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) |
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B.either (0, 1, 0) or (0, 7, 0) |
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C.either (0, 1, 0) or (0, -7, 0) |
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D.either (0, -1, 0) or (0, 7, 0) |
Ans: 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)
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)
| Q12. | 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) |
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B.(0, -17/2, -13/2) |
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C.(0, 17/2, -13/2) |
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D.None of these |
Ans: (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)
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)
| Q13. | he 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 |
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B.r . (2i – j + 2k) = 3 |
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C.r . (2i – j + 2k) = 4 |
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D.r . (2i – j + 2k) = 5 |
Ans: 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
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
| Q14. | The cartesian equation of the line is 3x + 1 = 6y – 2 = 1 – z then its direction ratio are |
A.1/3, 1/6, 1 |
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B.-1/3, 1/6, 1 |
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C.1/3, -1/6, 1 |
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D.1/3, 1/6, -1 |
Ans: 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
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
| Q15. | Under what condition does the equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere |
A.u² + v² + w² = d² |
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B.u² + v² + w² > d |
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C.u² + v² + w² < d |
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D.u² + v² + w² < d² |
Ans: 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
Equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere if
u² + v² + w² – d > 0
⇒ u² + v² + w² > d
| Q16. | The locus of a first-degree equation in x, y, z is a |
A.sphere |
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B.straight line |
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C.plane |
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D.none of these |
Ans: 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.
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.
| Q17. | The image of the point P(1,3,4) in the plane 2x – y + z = 0 is |
A.(-3, 5, 2) |
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B.(3, 5, 2) |
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C.(3, -5, 2) |
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D.(3, 5, -2) |
Ans: (-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)
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)
| Q18. | The distance of the point P(a, b, c) from the x-axis is |
A.√(a² + c²) |
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B.√(a² + b²) |
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C.√(b² + c²) |
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D.None of these |
Ans: √(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)
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)
| Q19. | The vector equation of a sphere having centre at origin and radius 5 is |
A.|r| = 5 |
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B.|r| = 25 |
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C.|r| = √5 |
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D.none of these |
Ans: |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
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
| Q20. | The ratio in which the line joining the points(1, 2, 3) and (-3, 4, -5) is divided by the xy-plane is |
A.2 : 5 |
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B.3 : 5 |
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C.5 : 2 |
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D.5 : 3 |
Ans: 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
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
MCQ Questions for Class 11 Maths
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MCQ Questions for Class 11 Maths Chapter 1 Sets
MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions
MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities
MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations
MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem
MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series
MCQ Questions for Class 11 Maths Chapter 10 Straight Lines
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections
MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry
MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives
MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning
MCQ Questions for Class 11 Maths Chapter 15 Statistics
MCQ Questions for Class 11 Maths Chapter 16 Probability
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