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Main reasons for increase of plastic pollution : 

1. Rising human population and their needs, people depend on plastic. 

2. The food industry which packs everything on plastic. 

3. Increase in usage of plastic bottles and container caps. 

4. Indiscriminate usage of plastic bags and carriers. 

5. Over usage of plastic straws and stirrers. 

6. Lack of proper management of plastic waste. 

7. Unawareness among the public about the hazardous affects of plastic wastes. 

8. People are not following the 3’Rs (Recycle – Reduce – Reuse) method to reduce plastic wastes.

1. Take some water in a test tube and add urea to it. Heat the test tube till all the urea dissolves. 

2. Add more urea to it. Keep on adding to it until no more urea can be dissolved in it. 

3. Let the solution cool down for sometime. 

4. Observe the test tube after about half an hour. 

5. We get large size crystals of urea.

The correct option is B) galvanization.

1. When iron nails, iron gates, iron benches or pieces of iron are left is the open ground for a long time, we find a brown layer on the surface of iron articles. 

2. This is called rust’ and process of forming this layer is called rusting. 

3. When iron is exposed to air for a long time, the Oxygen present in air reacts with it in the presence of moist air and forms a new substance called iron oxide as rust on iron articles. 

This process is known as rusting. Iron + Oxygen (from air) + Water → rust (Iron oxide) 

4. As a new substance is formed in this change, we call it a chemical change.

No, because they are coated with zinc or chromium.

1. When coal and oils are burnt, they release acidic gases like NO₂ and SO₂ . 

2. They mix up with the water vapour and come down as acid rains.

The correct option is D) growing of nails.

(d) Change of day and night

The correct option is B) fast change.

1. Metals like gold, silver, zinc do not rust even though they are exposed to moist air. 

2. Zinc is used in Galvanization process to prevent rusting of iron articles.

The correct option is C) making of a cake.

1. Firing crackers during festivals and celebrations increase the concentration of dust and pollutants in the air. 

2. The dust particles get settled on the surrounding surfaces which are packed with chemicals like copper, zinc, sodium and magnesium. They cause damage to paintings. 

3. The quality of ambient air will be decreased drastically. 

4. Firing of crackers causes lot of sound pollution. 

5. The oxides of sulphur pollute the air and causes lung diseases to human beings. 

6. The increase in levels of air pollutants cause acid rains, corrosion of objects and decrease in visibility.

1. Swarming of honeybees 

2. Rainbow formation 

3. Clouds passing across the mountains. 

4. Rows of birds flying in the sky.

The correct option is D) all.

1. A cow giving birth to a calf 

2. Plants producing vegetables 

3. A new born baby trying to search for milk. 

4. These are all wonderful changes we feel happy to observe in the nature.

1. Calcium carbide is more commonly known as “masala” is used for the artificial ripening of fruits. 

2. It is very harmful to health. 

3. Calcium carbide is a carcinogenic agent. 

4. The most important precaution to avoid eating such artificially ripened fruits is to go in for fruits and vegetables which are not unseasonal. 

5. Always wash the vegetables and fruits properly before consuming them.

Cotyledons and scutellum.

 Micelles: When molecular ions in soaps and detergents aggregate, they form micelles. It is formed because soap has hydrophobic part. Water can attract hydrophilic part but not hydrophobic part. 

No, micelle will not be formed in ethanol, as soap will dissolve in ethanol. Micelles trap (attract) dirt, grease, oily spot, etc. which is washed away by water.

(i) Soap molecules have two ends- one end is the hydrocarbon chain which is water repellent, whereas the other end is the ionic part which it water-soluble end. When soap is dissolved in water it forms a group of many molecules, known as micelle

(ii) These micelles are formed because their hydrocarbon chains come together and the polar ends are projected outwards.

(iii) Micelle formation in ethanol will not occur because the hydrocarbon chain end of the soap will dissolve in ethanol.

(iv) Soaps in the form of a micelle are able to clean dirty clothes having oily spots, as the oily dirt is collected in the centre of the micelle, which forms an emulsion in water and on rinsing, the water washes away the micelles with dirt attached to them

When soap is added to water, micelle formation takes place, this is because the hydrocarbon chains of soap molecules are hydrophobic while the ionic ends are hydrophilic and hence soluble in water.

Micelle will not form in all types of solvents. It will form in such type of solvent where soap is insoluble in that particular solvent.

The unattended trolley was commandeered by a crew of two boys. One served as the driver and the other as a mechanic.

In the later Vedic period, Vajpeya sacrifice was performed by a new king at the time of his coronation. The Rajasuya sacrifice was performed to appease Gods in order to ensure the material well-being of the kingdom. The Ashvamedha sacrifice was performed for a powerful king to proclaim that he was the ‘King of Kings’.

The Aryans propitiated their Gods by performing sacrifices. The common items of offerings for the sacrifice included ghee, milk, grain, flesh and soma juice. Every householder performed the sacrifice by kindling the sacred fire and reciting hymns. All the members of the family participated in these ceremonies.

Answer is (b) cos2x/2 + k

Students are encouraged to solve the Three Dimensional Geometry MCQ Questions of Class 12 to know various ideas. Every one of the ideas of Three Dimensional Geometry of class 12 is significant for Students according to the exam perspective to get more marks.  Practicing the Multiple choice Questions on Three Dimensional Calculation Class 12 with answers will support your certainty consequently assisting you with scoring high in the exam.

Class 12 Maths MCQ Questions of Three Dimensional Geometry with Answers are accessible here to help the understudies who are appearing up for the CBSE board exams. Understudies can refer to all the significant MCQ Questions for various ideas gave at Sarthaks eConnect to better prepare for exams. Explore MCQ Questions for Class 12 Maths with answers furnished with definite solutions.

Practice MCQ Question for Class 12 Maths chapter-wise

1. The distance of point (2, 5, 7) from the x-axis is

(a) 2
(b) \(\sqrt{74}\)
(c) \(\sqrt{29}\)
(d) \(\sqrt{53}\)

2. P is a point on the line segment joining the points (3, 2, -1) and (6, 2, -2). If y -coordinate of point P is 2, then its x-coordinate will be

(a) 2
(b) 1
(c) -1
(d) -2

3. P is a point on the line segment joining the points (3, 5, -1) and (6, 3, -2). If y -coordinate of point P is 2, then its x-coordinate will be

(a) 2
(b) 17/3
(c) 15/2
(d) -5

4. Direction ratios of a line are 2, 3, -6. Then direction cosines of a line making obtuse angle with the y-axis are

(a) 2/7,3/7,-6/7
(b) -2/7,3/7,-6/7
(c) -2/7,-3/7,6/7
(d) -2/7,-3/7,-6/7

5.  A line makes angle α, β, γ with x-axis, y-axis and z-axis respectively then cos 2α + cos 2β + cos 2γ is equal to

(a) 2
(b) 1
(c) -2
(d) -1

6. The equations of y-axis in space are

(a) x = 0, y = 0
(b) x = 0, z = 0
(c) y = 0, z = 0
(d) y = 0.

7. If the direction cosines of a line are k/3,k/3,k/3, then value of k is

(a) k > 0
(b) 0 < k < 1.
(c) k = 1/3
(d) k = ±3

8. The line joining the points (0, 5, 4) and (1, 3, 6) meets XY-plane at the point 

(a) -2, 9, 0 
(b) -2, 9, 0 
(c) -2, 8, 0 
(d) -2, 9, 6

9. A line makes angles π/4,3π/4 with x-axis and y-axis respectively. Then the angle which it makes with z-axis can be _

(a) π, 0
(b) π , π
(c) π , 1
(d) none of these

10. The direction cosines of the y-axis are

(a) (6, 0, 0)
(b) (1, 0, 0)
(c) (0, 1, 0)
(d) (0, 0, 1)

11. In three dimensional space, the equation x²−x−2 = 0 represents

(a) a pair of straight lines
(b) a set containing two distinct points
(c) a pair of parallel planes
(d) none of these

12. The vector equation for the line passing through the points (–1, 0, 2) and (3, 4, 6) is:

(a) i +2k + λ(4i + 4j + 4k)
(b) i –2k + λ(4i + 4j + 4k)
(c) -i+2k+ λ(4i + 4j + 4k)
(d) -i+2k+ λ(4i – 4j – 4k)

13. Number of lines making equal angles with the co-ordinate axes are is 

(a) 1
(b) 4
(c) 8
(d) 2

14. Distance of the point (3, 4, 5) from the origin (0, 0, 0) is

(a) \(\sqrt{50}\)
(b) 3
(c) 4
(d) 5

15. The distance between the planes 2x−3y + 6z+12=0 and 2x−3y+6z−2=0 is

(a) 10/7
(b) 2/7
(c) 2
(d) 24/7

16. The locus represented by xy+ yz =0 is

(a) a pair of perpendicular lines
(b)a pair of parallel lines
(c) a pair of parallel planes
(d) a pair of perpendicular planes

17. Which of the following is false?

(a) Addition is commutative in N. 
(b) Multiplication is associative in N.
(c) If a * b =a^b for all a,b ∈ N then * is commutative in N.
(d) Addition is associative in N

18. Which of the following is false?

(a) meet in a unique point
(b) meet in a line
(c) meet taken two at a time in parallel lines
(d) None of these

19. The coordinates of the mid-point of the line segment joining the points (2,3) and (4,7).

(a) (3,5) 
(b) (2,5)
(c) (4,5)
(d) None of these

20. The coordinates of the mid-point of the line segment joining the points (3,4) and (5,6).

(a) (3,5) 
(b) (2,5)
(c) (4,5)
(d) None of these

Answer:

1. Answer: (b) 

Explanation: \(\sqrt{5^2+7^2}\)

\(=\sqrt{25+49}\)

\(=\sqrt{74}\)

2. Answer: (a) 2

Explanation: Let P divides the line segment in the ratio of λ: 1, x - coordinate of the point P may be expressed as

\(x=\frac{6\lambda+3}{\lambda+1}\) giving \(\frac{6\lambda+3}{\lambda+1}=5\)

so that λ = 2. Thus y-coordinate of P is

\(\frac{2\lambda+2}{\lambda+1}=2\)

3. Answer: (c) 15/2

Explanation: Let P divides the line segment in the ratio of λ: 1, x - coordinate of the point P may be expressed as

\(\therefore=\frac{3k+5}{k+1}=2\)  

 \(\frac{6k+3}{ka+1}=\frac{-18+3}{-3+1}\)

= 15/2

4. Answer: (c) -2/7,-3/7,6/7

Explanation:  As direction cosines of a line whose direction ratio are 2,3, -6 are 2/7,3/7,−6/7.
As angle with the y-axis is obtuse, ∴ cos β < 0, Therefore direction ratios are -2/7,-3/7,6/7.

5. Answer: (d) -1

Explanation: cos²α + cos²β + cos²\(\gamma\)

= 2(cos² + cos²+ cos²\(\gamma\))⁻³

= 2−3

= −1

= (cos²α + cos²β + cos²\(\gamma\)= 1)

6. Answer: (b) x = 0, z = 0

Explanation: As on the y-axis, x-coordinate and z-coordinate are zeroes.

7. Answer: (d) k = ± 3

Explanation: \(3\times\frac{k^2}{9}=1\)  

k = ±3  

8. Answer: (a) -2, 9, 0 

Explanation: (a) -2, 9, 0 as lone in \(\frac{x-1} 1=\frac{y-3}{-2} =\frac{z-6}{2}=\lambda\)

General point on line is (λ + 1, -2λ + 3, 2λ + 6). If it meets AT-plane, then 

2λ + 6 = 0

λ = – 3

∴ Point is (-2, 9, 0)

9. Answer: (a) 

Explanation: 0,π, as \(cos^2\frac{\pi}{4}+cos^2\frac{\pi}{4}+cos^2\gamma=1\)

= 1/2 + 1/2 + \(cos^\gamma=1\)

\(cos\gamma=0\)

\(\gamma=0,\pi\)

10. Answer: (c) (0, 1, 0)

Explanation: The Y-axis makes angles 90°, 0° and 90° with the positive directions of X-axis, Y-axis and Z-axis, respectively, then cos 90° = 0, cos 0° = 1, cos 90° = 0 are the directions cosines of the line.

11. Answer: (c) a pair of parallel planes

Explanation: x² −x −2 = 0

⇔(x−2) (x+1) = 0

x = 2, x= −1 . which are the planes (both parallel to YOZ plane

12. Answer: (c) 

Explanation: The vector equation of the line is given by:

r = a + λ (b – a), λ ∈ R

Let a = -i + 2k

And b = 3i + 4j + 6k

b – a = 4i + 4j + 4k

Let the vector equation be r, then;

r = -i + 2k + λ (4i + 4j + 4k)

13.  Answer: (d) 2

Explanation: There are two different lines that make equal angles with the co-ordinate axes.They are y = x line and y = - x line.

14. Answer: (a) 

Explanation: Let P be the point whose coordinate is (3, 4, 5) and Q represents the origin

\(PQ=\sqrt{(3-0)^2+(4-0)^2+(5-0)^2}\)

\(=\sqrt{9+16+25}\)

\(=\sqrt{50}\)

∴ Distance of the point (3, 4, 5) from the origin (0, 0, 0) is \(\sqrt{50}\)

15. Answer: (c) 

Explanation: The given equations of the plane are 2x − 3y + 6z +12=0 and 2x −3y + 6z + −2 = 0.

\(=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}\)

\(=\frac{|12-(-2)}{\sqrt{2^2+3^2+6^2}}\)

= 14/7

= 2.

16. Answer: (d) a pair of perpendicular planes

Explanation: Locus represented by xy+ yz =0

⇒ y(x+z) =0

The planes y=0 and x+z =0 are perpendicular.

17. Answer: (c) 

Explanation: Since, for the set of natural numbers with respect to addition, it is commutative and associative.  Also, for the set of natural numbers with respect to multiplication, it is associative.

18. Answer: (a) meet in a unique point

Explanation: The planes x + y = 0 i.e. x = - y and y + z = 0 i.e. z = -y meet in the line

x/1 = y/−1 = z/1 .

Any point on this line is (t, -t, t). This point lies in the plane x + z = 0 if t + t = 0 ⇒ t = 0 .

So the three planes meet in a unique point (0, 0, 0).

19. Answer: (a) (3,5) 

Explanation: \((\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\)

\(=(\frac{2+4}{2},\frac{3+7}{2})\)

\(=(\frac{6}{2},\frac{10}{2})\)

= (3,5).

20. Answer: (c) (4,5)

Explanation: \((\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\)

\(=(\frac{3+5}{2},\frac{4+6}{2})\)

\(=(\frac{8}{2},\frac{10}{2})\)

= (4,5)

Click here to practice more MCQ Question for Three Dimensional Geometry Class 12

The important Class 11 Maths MCQ Questions of Three Dimensional Geometry with Answers are covered here. The important MCQ questions are taken from the earlier year question papers and test papers, which will assist you with accomplishing more marks in yearly exams. 

Introduction to 3-D Geometry fuses the accompanying significant concepts, such as

• Coordinate Points in the three-dimensional space
• Distance between two points
• Section Formula

Practice the following important MCQ Questions for class 11 Maths, should help you to solve the problems faster in the final examination and get amazing marks in class 11 Maths final examination.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. 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

2. 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)

3. 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

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)

5. 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

6. The plane 2x – (1 + a)y + 3az = 0 passes through the intersection of the planes

(a) 2x – y = 0 and y + 3z = 0
(b) 2x – y = 0 and y – 3z = 0
(c) 2x + 3z = 0 and y = 0
(d) 2x – 3z = 0 and y = 0

7. 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) \(\sqrt3\) units
(b) \(2\sqrt3\) units
(c) \(\sqrt5\) units
(d) \(2\sqrt4\) units

8. 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

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

10. 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)

11. 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

12. 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

13. A  plane is parallel to yz-plane so it is perpendicular to

(a) z-axis
(b) y-axis
(c) x-axis
(d) None of these

14.  The locus of a point for which y = 0, z = 0 is   

(a) equation of x-axis
(b) equation of y-axis
(c) equation of z-axis
(d) None of these

15. The distance of point P(3, 4, 5) from the yz-plane is 

(a) 3 units
(b) 4 units
(c) 5 units
(d) 550 units

16. The points A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a 

(a) square
(b) rhombus
(c) rectangle
(d) None of these

17. Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a) 8
(b) 7
(c) 6
(d) None of the above

18. What is the ratio in which the line joining the points (2,4, 5) and (3, 5, - 4) is internally divided by the xy-plane?

(a) 5:4
(b) 3:4
(c) 1:2
(d) 7:5

19. In a three dimensional space, the equation 6y – 2z = 0 represents

(a) a plane containing X axis
(b) none of these
(c) a plane containing Z axis
(d) a plane containing X axis

20. 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

Answer:

1. Answer: (a) 1/3, 1/6, 1

Explanation: 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

2. Answer: (a) (-3, 5, 2)

Explanation: 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)

3. Answer: (c) meet in a unique point

Explanation: 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.

4. Answer: (a) (5/3, 7/3, 17/3)

Explanation: 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)

5. Answer: (b) Plane

Explanation: 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.

6. Answer: (a) 2x – y = 0 and y + 3z = 0

Explanation: 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

7. Answer: \(\sqrt3\) unit

Explanation: 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 = \(\sqrt{{(1 – 2)2 + (-2 + 3)2 + (3 – 5)2}}\)

= \(\sqrt{{1 + 1 + 4}}\)

= \(\sqrt6\)

Again length of the diagonal of square is \(\sqrt2\) times the length of side of the square.

⇒ \(a\sqrt2\) = \(\sqrt6\)

⇒ \(a\sqrt2=\sqrt3\times\sqrt2\)

⇒ a = \(\sqrt3\)

So, the length of side of square is \(\sqrt3\) unit.

8. Answer: (c) (0, 17/2, -13/2)

Explanation: 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-planeSo, 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)

9. Answer: (b) r . (2i – j + 2k) = 3

Explanation: 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

10. Answer: (d) either (0, -1, 0) or (0, 7, 0)

Explanation: 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 = \(\sqrt4\)

⇒ 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)

11. Answer: (d) None of these

Explanation: 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 – sin²α – sin² β – sin² γ = 1

⇒ 3 – 1 = sin²α + sin² β + sin²γ

⇒ sin2α + sin2β + sin2γ = 2

12. Answer: (b) 3 : 2

Explanation: 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

13. Answer: (c) x-axis

Explanation: A plane is parallel to yz-plane which is always perpendicular to x-axis.

14. Answer: (a) equation of x-axis

Explanation: Locus of the point y=0, z=0 is x-axis, since on x-axis both y=0 and z=0.

15. Answer: (a) 3 units

Explanation: From basic ides of three-dimensional geometry, we know that x-coordinate of a point is its distance from yz plane.

∴ Distance of Point P (3, 4, 5) from y z plane is given by its x coordinate.

∵ x-coordinate of point P = 3

∴ Distance of (3, 4, 5) from y z plane is 3 units.

16. Answer: (b) rhombus

Explanation: Since mid-point of [AC] coincides with that of [BD], therefore, ABCD is a parallelogram.

Also |AB|= \(\sqrt{2^2+3^2+6^2}=7\)

and |AD|= \(\sqrt{ 6^2+2^2+3^2}=7\)

therefore, ABCD is a rhombus.

Further, |AC|= \(\sqrt{4^2+5^2+9^2} =\sqrt{122}\)

and |BD|= \(\sqrt{8^2+1^2+3^2}=\sqrt{74}\)

⇒|AC| ≠ |BD|,

 therefore, ABCD is a neither a rectangle nor a square.

17. Answer: (a) 8

Explanation: Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units.

18. Answer: (a) 5:4

Explanation: Let the line joining the points (2, 4, 5) and (3, 5, -4) is internally divided by the xy - plane in the ratio k: 1.

∴ For xy plane, z = 0

\(0=\frac{-k\times 4+5}{k+1}\) = 4k = 5

⇒k = 5/4

So, ratio is 5 : 4

19. Answer: (a) a plane containing X axis

Explanation: Given, equation is 6y – 2z = 0

Here x = 0

So, the given equation 6y- 2z  represents a plane containing X-axis.

20. Answer: (b) π/3

Explanation: 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)/\(\sqrt{16 + 9 + 25)}\)×\(\sqrt{9+16+25}\)

⇒ cos θ = 25/\(\sqrt{50}\times\sqrt{50}\)

⇒ 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

Click here to practice MCQ Questions for Three Dimensional Geometry class 11

Important Class 11 Maths MCQ Questions of Sequences and Series with Answers are given here. Sequences and series have a few applications in our everyday life. The significant MCQ Questions for class 11 Maths cover every one of the topics and subtopics in the NCERT.

Go through the significant MCQ Questions gave at “Sarthaks eConnect” and accomplish astounding marks in the yearly assessment.  It’s based on latest syllabus and exam pattern. The answers of the questions are given in a detailed explanation. so, students can understand them in a superior manner.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. If a, b, c are in AP then

(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

2. Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is

(a) 2 + \(\sqrt3\)
(b) 2 – \(\sqrt3\)
(c) 2 ± \(\sqrt3\)
(d) None of these

3. The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

4.  The sum of the series 1/2! − 1/3! + 1/4! −... up to infinity is

(a) e^-2
(b) e^-1
(c) e^-1/2
(d) e^1/2

5. The third term of a geometric progression is 4. The product of the first five terms is

(a) 4³
(b) 4⁵
(c) 4⁴
(d) none of these

6. Let Tr be the r^th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have T_m = 1/n and T_n = 1/m, then T_mn equals

(a) 1/mn
(b) 1/m + 1/n
(c) 1
(d) 0

7. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is

(a) 2
(b) 4
(c) 6
(d) 8

8. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in

(a) A.P
(b) G.P
(c) H.P
(d) A.G.P

9. If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

10. The 35^th partial sum of the arithmetic sequence with terms an = n/2 + 1

(a) 240
(b) 280
(c) 330
(d) 350

11. The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is

(a) 1
(b) 2
(c) 3
(d) 4

12. If 2/3, k, 5/8 are in AP then the value of k is

(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31

13. If the third term of an A.P. is 7 and its 7^th term is 2 more than three times of its third term, then the sum of its first 20 terms is

(a) 228
(b) 74
(c) 740
(d) 1090

14. If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals

(a) 10
(b) 12
(c) 11
(d) 13

15. If a is the A.M. of b and c and G1 and G2 are two GM between them then the sum of their cubes is

(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

16. 3, 5, 7, 9, …….. is an example of

(a) Geometric Series
(b) Arithmetic Series
(c) Rational Exponent
(d) Logarithm

17.  2, 3, 5, 7, 11, ?, 17

(a) 12
(b) 13
(c) 14
(d) 15

18. If the positive numbers a,b,c,d are in AP. Then, abc,abd,acd,bcd are

(a) not in AP/GP/HP
(b) in AP
(c) in GP
(d) in HP

19. The minimum value of expression 3^x+3^1−x,x ∈ R, is

(a) 0
(b) 1/3
(c) 3
(d) \(2\sqrt3\)

20. The sum of odd integers from 1 to 2001 is

(a) 10201
(b) 102001
(c) 100201
(d) 1002001

Answer:

1. Answer: (b) 2b = a + c

Explanation: Given, a, b, c are in AP

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

2. Answer: (a) 2+ \(\sqrt3\)

Explanation: Let the three numbers be a/r, a, ar

Since the numbers form an increasing GP, So r > 1

Now, it is given that a/r, 2a, ar are in AP

⇒ 4a = a/r + ar

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± \(\sqrt3\)

⇒ r = 2 + \(\sqrt3\) {Since r > 1}

3. Answer: (a) n/(n+1)

Explanation: Given series is:

S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

4. Answer: (b) e^-1

Explanation: e^−x = 1−x + x²/ 2! − x³/3! + x⁴/4!−... .....

put x = 1

1/2! − 1/3! + 1/4! .....= e^-1

5. Answer: (b) 4⁵

Explanation: here it is given that T3 = 4.

⇒ ar² = 4

Now product of first five terms = a.ar.ar².ar³.ar4

= a⁵r¹⁰

= (ar²)⁵

= 4⁵

6. Answer: (c) 1

Explanation: Let first term is a and the common difference is d of the AP

Now, T_m = 1/n

⇒ a + (m-1)d = 1/n ………… 1

and T_n = 1/m

⇒ a + (n-1)d = 1/m ………. 2

From equation 2 – 1, we get

(m-1)d – (n-1)d = 1/n – 1/m

⇒ (m-n)d = (m-n)/mn

⇒ d = 1/mn

From equation 1, we get

a + (m-1)/mn = 1/n

⇒ a = 1/n – (m-1)/mn

⇒ a = {m – (m-1)}/mn

⇒ a = {m – m + 1)}/mn

⇒ a = 1/mn

Now, T_mn = 1/mn + (mn-1)/mn

⇒ T_mn = 1/mn + 1 – 1/mn

⇒ T_mn = 1

7.  Answer: (c) 6

Explanation: Let a and b are two numbers such that

a + b = 13/6

Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b

Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

8. Answer: (c) H.P

Explanation: (c) H.P

ax² + bx + c = 0

Let p and q are the roots of this equation.

Now p+q = -b/a

and pq = c/a

Given that

p + q = 1/p² + 1/q²

⇒ p + q = (p² + q²)/(p² ×q²)

⇒ p + q = {(p + q)² – 2pq}/(pq)²

⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²

⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}

⇒ -bc²/a³ = {b² – 2ca}/a²

⇒ -bc²/a = b² – 2ca

Divide by bc on both side, we get

⇒ -c /a = b/c – 2a/b

⇒ 2a/b = b/c + c/a

⇒ b/c, a/b, c/a are in AP

⇒ c/a, a/b, b/c are in AP

⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP

⇒ a/c, b/a, c/b are in HP

9. Answer: (b) a², b², c² are in AP

Explanation: Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

10. Answer: (d) 350

Explanation: The 35th partial sum of this sequence is the sum of the first thirty-five terms.The first few terms of the sequence are:

a₁ = 1/2 + 1 = 3/2

a₂ = 2/2 + 1 = 2

a₃ = 3/2 + 1 = 5/2

Here common difference d = 2 – 3/2 = 1/2

Now, a₃₅ = a₁ + (35 – 1)d 

= 3/2 + 34 ×(1/2) 

= 17/2

Now, the sum

= (35/2) × (3/2 + 37/2)

= (35/2) × (40/2)

= (35/2) × 20

= 35 × 10

= 350

11. Answer: (c) 3

Explanation: Let first term of the GP is a and common ratio is r.

3rd term = ar²

5th term = ar⁴

Now

⇒ ar² + ar⁴ = 90

⇒ a(r² + r⁴) = 90

⇒ r² + r⁴ = 90

⇒ r² ×(r² + 1) = 90

⇒ r²(r² + 1) = 3² ×(3² + 1)

⇒ r = 3

So the common ratio is 3

12. Answer: (b) 31/48

Explanation: Given, 2/3, k, 5/8 are in AP

⇒ 2k = 2/3 + 5/8

⇒ 2k = 31/24

⇒ k = 31/48

So, the value of k is 31/48

13. Answer: (c) 740

Explanation: Let a is the first term and d is the common difference of AP

Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term

⇒ a + 2d = 7 ………….. 1

and

3(a + 2d) + 2 = a + 6d

 ⇒ 3×7 + 2 = a + 6d

⇒ 21 + 2 = a + 6d

⇒ a + 6d = 23 ………….. 2

From equation 1 – 2, we get

4d = 16

⇒ d = 16/4

⇒ d = 4

From equation 1, we get

a + 2×4 = 7

⇒ a + 8 = 7

⇒ a = -1

Now, the sum of its first 20 terms

= (20/2)×{2×(-1) + (20-1)×4}

= 10×{-2 + 19×4)}

= 10×{-2 + 76)}

= 10 × 74

= 740

14. Answer: (c) 11

Explanation: the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the

A.P. 57, 59, 61, ….

⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}

⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}

⇒ 6n + 1 = {2n + 112}/2

⇒ 6n + 1 = n + 56

⇒ 6n – n = 56 – 1

⇒ 5n = 55

⇒ n = 55/5

⇒ n = 11

15. Answer: (b) 2abc

Explanation:  Given, a is the A.M. of b and c

⇒ a = (b + c)

⇒ 2a = b + c ………… 1

Again, given G₁ and G₂ are two GM between b and c,

⇒ b, G₁, G₂, c are in the GP having common ration r, then

⇒ r = (c/b)1/(2+1) = (c/b)1/3

Now,

G₁ = br = b × (c/b)1/3

and G₁ = br = b × (c/b)2/3

Now,

(G1)³ + (G₂)³ = b³ ×(c/b) + b³ ×(c/b)²

⇒ (G1)³ + (G₂)³ = b³ ×(c/b)×( 1 + c/b)

⇒ (G1)³ + (G₂)³ = b³ ×(c/b)×( b + c)/b

⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b

⇒ (G₁)³ + (G₂)³ = b² × c×( b + c)/b ………….. 2

From equation 1

2a = b + c

⇒ 2a/b = (b + c)/b

Put value of(b + c)/b in eqaution 2, we get

(G₁)³ + (G₂)³ = b × c × (2a/b)

⇒ (G₁)³ + (G₂)² = b × c × 2a

⇒ (G₁)³ + (G₂)³ = 2abc

16. Answer: (b) Arithmetic Series

Explanation: This is an arithmetic sequence since there is a common difference between each term.

17. Answer: 13

Explanation: Clearly, the given series consists of prime numbers starting from 2. So, the missing term is the prime number after 11, which is 13.

18. Answer: (d) in HP

Explanation: Since, a,b,c,d are in AP.

⇒ a/abcd,b/abcd,c/abcd,d/abcd are in AP

⇒1/bcd,1/cda,1/abd,1/abc are in AP

⇒ bcd,cda,abd,abc are in HP

⇒ abc,abd,cda,bcd are in HP

19. Answer: (d) \(2\sqrt3\)

Explanation: We know that A.M.≥G.M. for positive numbers.

\(\frac{3^x+3^{1-x}}{2}\) \(\geq\sqrt{3^x.3^{1-x}}\)

⇒3^x + 3^1−x ≥ \(2\sqrt3\)

20. Answer: (d) 1002001 

Explanation: The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Here,

a+(n−1)d = 2001

=> 1+(n−1)(2) = 2001

=> 2n−2 = 2000

=> n = 1001

S_n = n/2[2a+(n−1)d]

∴ S_n = 1001/2[2×1+(1001−1)×2]

=1001/2 [2+1000×2]

=1001/2×2002

=1001×1001

=1002001

Click here to practice MCQ Questions for Sequences and Series class 11

The Class 11 Online Mock Test Sequences and Series cover all the topics and subtopics in the NCERT textbook. The solutions for the test are given in a step-by-step procedure so that students can understand them in a better way. Go through the Mock Test provided at Sarthaks and achieve excellent marks in the annual examination. It also helps to check your prepartaion level.

Improve your Class 11 Mock Test Sequences and Series whose online practice test is made under the supervision of experts and that too free of cost. Sequence and Series practice test help the students to focus and solve the general sequencing problems and also other topics that are related to Sequence and series. With the use of this worksheets, students can also have a good revision and a practice of the subject and topics which appear in the examination.

Click here to Practice: -  Class 11 Sequences and Series Mock Test

Since, sum of all probability of a probability distribution is 1

So, P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 7) = 1

= 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k =1

= 10k² + 9k = 0

= (10k - 1)(k + 1) = 0 

∴ k = 1/10, -1 but, k > 0

∴ k = 1/10

Hence, P(x < 2) = P(x = 0) + P(x = 1) = 0 + k 

k = 1/10