CBSE Class 12

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise

Question 1.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and.atmost 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maxi¬mum amount of vitamin A in the diet?
Solution:
Let x and y be the number of packets of food P and Q. We have the data as

Calcium constraint: 12x + 3y ≥ 240
i.e., 4x + y ≥ 80
Iron constraint: 4x + 20y ≥ 460
i.e., x + 5y ≥ 115
Cholesterol constraint: 6x + 4y ≤ 300
i.e., 3x + 2y ≤ 150
Quantity of vitamin A in the diet Z = 6x + 3y
The LPP is Minimise Z = 6x + 3y
subject to the constraints
4x + y ≥ 80
x + 5y ≥ 115
3x + 2y ≤ 150
x ≥ 0, y ≥ 0

The feasible region is shaded in the figure. We use corner point method to find the minimum value of Z.

Minimum value of Z is at B (15,20). Hence, the amount of vitamin A under constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum quantity of vitamin A will be 15 0 units.
∴ Maximum value of Z is 285 at C(40, 15).
Hence 40 packets of food P and 15 packets of food Q should be used to maximise the amount of vitamin A. The maximum amount of vitamin A in the diet is 285 units.

Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹ 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹ 200 per bag, contains 1.5 units of nutritional element A, 1.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution:
Let x and y denote the number of bags of brand P and brand Q respectively.

Element A constraint: 3x +1, 5y ≥ 18
Element B constraint: 2.5x + 11.25y ≥ 45
Element C constraint: 2x + 3y ≥ 24
Cost function : Z = 250x + 200y
The L.P.P is minimise Z = 250x + 200y
subject to the constraints 3x + 1.5y ≥ 18, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24, x, y ≥ 0.

The feasible region is shaded in the figure.

From the table the minimum value of Z is 1950. Since the region is unbounded 1950 may or may not be the minimum value of Z. Consider the inequality 250x + 200y < 1950. This half plane has no point common with the feasible region The minimum value of Z is 1950 at C(3, 6). To produce a mixture having minimum cost, the farmer should mix 3 bags’ of brand P and 6 bags of brand Q.
The minimum cost of the mixture is ₹ 1950

Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs Rs. 16 and one kg of food Y costs Rs.20. Find the least cost of the mixture which will produce the required diet.
Solution:
Let x kg of food X and y kg of food Y be mixed in the diet. We have the data as

Vitamin A constraint: x + 2y ≥ 10
Vitamin B constraint: 2x + 2y ≥ 12
Vitamin C constraint: 3x + y ≥ 18
Cost function Z = 16x + 20y
The L.P.P. is
Minimise Z = 16x + 20y
subject to the constraints
x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x, y ≥ 0

From the table, the minimum value of Z is 112. Since the region is unbounded 112 may or may not be the minimum value of Z. Consider the inequality 16x + 20y =112. This half plane has no point common with the feasible region.
∴ Minimum value of Z is 112 at B(2, 4). Hence 2kg of food X and 4kg of food Y should be mixed. The least cost of the mix-ture is ₹ 112

Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given be-low:

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution:
Let x be the no. of toys of type A and y be the no. of toys of type B.
We have the data as

Machine I constraint: 12x + 6y ≤ 360
Machine II constraint: 18x ≤ 360
Machine II constraint: 6x + 9y ≤ 360
Profit function : Z = 7.5 x + 5y
The LPP is maximise Z = 7.5 x + 5y
subject to the constraints 12x + 6y ≤ 360, 18x + 0y ≤ 360, 6x + 9y ≤ 360, x, y ≥ 0.

∴ Maximum value of Z is 262.50 at C(15, 30).
Hence 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.

Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution:
Let x tickets of executive class and y tickets of economy class be sold.
We have the data as

Total ticket constraint: x + y ≤ 200
Executive ticket constraint: x ≥ 20
Economy ticket constraint: y ≥ Ax
Profit function : Z = 1000x + 600y
The L.P.P. is maximise Z = 1000x + 600y
subject to the constraints x + y ≤ 200, x ≥ 20, y ≥ 4x, x, y ≥ 0.

The feasible region is shaded in the figure.

Hence 40 tickets of executive class and 160 tickets of economy class are to be sold to get the maximum profit. The maximum profit is ₹ 136000

Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60,50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x quintals of grains are transported from godown Ato ration shop D andy quintals be transported from godown Ato ration shop E.
We have the data as

The transported quantity of rice is non negative x ≥ 0, y ≥ 0, 100 – (x + y) ≥ 0, 60 – x ≥ 0
50 – y ≥ 0 , x + y – 60 ≥ 0
i.e., x + y ≤ 100, x + y ≥ 60, x ≤ 60, y ≤ 50, x ≥ 0, y ≥ 0
The total transportation cost
Z = 6x + 3y + 2.5 (100 – (x + y) + 4(60 – x) + 2(50 – y) + 3 (x + y – 60)
Z = 2.5x + 1.5y + 410
The L.P.P. is Minimise Z = 2.5x + 1.5y + 410
subject to the constraints x ≤ 60, y ≤ 50, x + y ≥ 60, x + y ≤ 100, x, y ≥ 0

The minimum value of Z is 510 at D( 10,50)
Hence the supply from godown A are 10,50,40 quintals and from godown B are 50,0,0 quintals to shops D, E and F respectively and the minimum cost is ₹ 510

Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

Assuming that the transportation cost of 10 litres of oil is ₹ 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x litre of oil be supplied from depot A to pump D and y litre of oil from depot B to pumb E. Then 7000 – (x + y) litre of oil will be transported to pump F.
The oil supplied from B to D, E and F are 4500 – x, 3000 – y, x + y – 3500

The transportation cost of 10 litres of oil for 1 km is ₹ 1.
∴ Transportation cost

Z = 0.3 x + 0.1 y + 3950
The transported quantity of oil is non negative.
i.e., x ≥ 0, y ≥ 0, 7000 – (x + y) ≥ 0
4500 – x ≥ 0, 3000 – y ≥0, x + y – 3500 ≥ 0
i.e., x, y ≥ 0
x + y ≤ 7000
x + y ≥ 3500
x ≤ 4500
y ≤ 3000
Hence the L.P.P. is
Minimise Z = 0.3x + 0. 1 y + 3950 subject to the constraints x ≤ 4500, y ≤ 3000, x + y ≤ 7000, x + y ≥ 3500, x, y ≥ 0

The feasible region is shaded in the figure. We use comer point method to find the minimum value of Z.

The minimum of Z is 4400 at E (500, 3000)
When x = 500 and y = 3000 500 l, 3000 l, 3500 l of oil should be trans¬ported from depot A to pumps D, E and F and 4000 l, 0 l, 0 l are transported from B to pumps D, E and F

Question 8.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phosphoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Solution:
Let x bags of brand P and y bags of brand Q be used in the garden.
We have the data as

Phosphoric acid constraint: x + 2y ≥ 240
Potash constraint: 3x + 1.5y ≥ 270
Chlorine constraint: 1.5x + 2y ≤ 310
Quantity of nitrogen : Z = 3x + 3.5y
The L.P.P. is Minimise Z = 3x + 3.5y
subject to the constraints x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x, y ≥ 0.

The feasible region is shaded in the figure.

∴ Minimum value of Z is 470 at C(40, 100)
Hence 40 bags of brand P and 100 bags of brand Q are used.
The minimum amount of nitrogen is 470.

Question 9.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phos-phoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution:
Let x bags of brand P and y bags of brand Q be used in the garden.
We have the data as

Phosphoric acid constraint: x + 2y ≥ 240
Potash constraint: 3x + 1.5y ≥ 270
Chlorine constraint: 1.5x + 2y ≤ 310
Quantity of nitrogen : Z = 3x + 3.5y
The L.P.P. is Minimise Z = 3x + 3.5y
subject to the constraints x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x, y ≥ 0.

The feasible region is shaded in the figure.

The maximum value of Z is 595 obtained at A(140,50)
Hence 140 bags of brand P and 50 bags of brand Q are used.
The maximum amount of nitrogen is 595.

Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is utmost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by utmost 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution:
Let the number of dolls of type A be x and that of type B be y.
We have the data as

Total dolls constraint: x + y ≤ 1200
Doll A constraint: x ≤ 3y + 600 or x – 3y ≤ 600
Profit function : Z = 12x +16y
The L.P.P is maximise Z = 12x + 16y
subject to the constraints
x + y ≤ 1200, x – 3y ≤ 600, y ≤ \(\frac { x }{ 2 }\), x ≥ 0, y ≥ 0

The maximum value of Z is 16000 at C(800,400) Hence 800 dolls of type A and 400 dolls of type B should be manufactured to get maximum profit.
The maximum profit is ₹ 16000.

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-2/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

Class 12 Maths Ex 13.2 NCERT Solutions  Question 1.
If P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\), find P(A∩B) if A and B are independent events.
Solution:
A and B are independent if
P (A ∩ B) = P(A) P(B) = \(\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 } \)

Ex 13.2 Class 12 NCERT Solutions Question 2.
Two cards are drawn at random and without replacement from a pack of 52 playing cards.Find the probability that both the cards are black.
Solution:
Let E : first card drawn is black
F : second card drawn is black
P(E ∩ F) = P(E) P(F|E)
P(E) = Probability of getting a black card \(\frac { 26 }{ 52 }\)
P(F|E) = Probability of getting a black card in the second draw given that the first card is black = \(\frac { 21 }{ 51 }\)
P(E ∩ F) = \(\frac { 26 }{ 52 }\) x \(\frac { 21 }{ 51 }\) = \(\frac { 25 }{ 102 }\)

Question 3.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
S = {12 good oranges, 3 bad oranges),
n(S) = 15
P (a box is approved) = \(\frac { C(12,3) }{ C(15,3) } =\frac { 12\times 11\times 10 }{ 15\times 14\times 13 } =\frac { 44 }{ 91 }\)

Question 4.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not
Solution:

Question 5.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?
Solution:

Question 6.
Let E and F be the events with P(E) = \(\frac { 3 }{ 5 }\), P (F) = \(\frac { 3 }{ 10 }\) and P (E ∩ F) = \(\frac { 1 }{ 5 }\). Are E and F independent?
Solution:
P(E) = \(\frac { 3 }{ 5 }\), P(F) = \(\frac { 3 }{ 10 }\),
∴ P (E) x P (F) = \(\frac { 3 }{ 5 } \times \frac { 3 }{ 10 } =\frac { 9 }{ 50 }\)
P(E ∩ F) ≠ P(E) x P(F)
∴ The event A and B are not independent.

Question 7.
Given that the events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(A∪B) = \(\frac { 3 }{ 5 }\) and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.
Solution:

Question 8.
Let A and B independent events P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A∩B)
(ii) P(A∪B)
(iii) P (A | B)
(iv) P(B | A)
Solution:
P (A) = 0.3,
P (B) = 0.4
A and B are independent events
(i) ∴ P (A ∩ B) = P (A). P (B) = 0.3 x 0.4 = 0.12.

(ii) P(A∪B) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58.

(iii) P (A | B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.12}{0.4}=0.3\)

(iv) P(B | A) = \(\frac{P(A \cap B)}{P(A)}=\frac{0.12}{0.3}=0.4\)

Question 9.
If A and B are two events, such that P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\). Find P (not A and not B)
Solution:
P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\)
P(A) P(B) = \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(A).P(B) = P(A ∩ B)
∴ A and B are independent events
Since A and B are independent events, A’ and B’ are also independent events.
∴ P (A’ ∩ B’) = P (A’). P (B’)
= [1 – P(A)] [1 – P(B)]
= \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{2}\right)=\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}\)

Question 10
Events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(B) = \(\frac { 7 }{ 12 }\) and P (not A or not B) = \(\frac { 1 }{ 4 }\). State whether A and Bare independent
Solution:

Hence A and B are not independent events.

Question 11
Given two independent events A and B such that P (A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Solution:
P(A) = 0.3, P(B) = 0.6
A and B are independent events.
i. P(A and B) = P(AB) = P(A).P(B)
= 0.3 x 0.6 = 0.18

ii. P(A and not B) = P(A B’ ) = P(A) P( B’)
= P(A) [1 – P(B)] = 0.3 (1 – 0.6)
= 0.3 x 0.4 = 0.12

iii P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(AB)
= 0.3 + 0.6 – 0.18 from (i)
= 0.72

iv. P(neither A nor B) = P( A’ B’)
= P(A’)P(B’)
= [1 – P(A)] [1 – P(B)]
= [ 1 – 0.3] [1 – 0.6]
= 0.7 x 0.4 = 0.28

Another Method:
P(neither A nor B) = P(A’ B’)
= P(A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B) = 1 – 0.72 from (iii) = 0.28

Question 12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
P(getting an odd number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
P(getting an even number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
Let A : getting even number in all the 3 throws
P(A) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(getting an odd number atleast once) = P(not getting even number in the 3 throws)
= 1 – P(A) = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\)

Question 13
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution:
Let R : red ball is drawn
B : black ball is drawn
P(R) = \(\frac { 8 }{ 18 }\), P(B) = \(\frac { 10 }{ 18 }\)
i. P(both are red) = P(1^st ball is red and 2^nd ball is red)
= P(RR) = P(R). P(R),
since the events are independent
= \(\frac { 8 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 6 }{ 81 }\)

ii. P(1^st ball is black and 2^nd ball is red)
= P(BR) = P(B) . P(R)
since the events are independent
\(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 20 }{ 81 }\)

iii. P(one of them is black and the other is red) = P(1^st ball is black & 2^nd ball is red or 1^st ball is red & 2^nd ball is black) = P(BR or RB)
= P(BR) + P(RB)
= P(B) P(R) + P(R) P(B)
= 2 x \(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 40 }{ 81 }\)

Question 14
Probability of solving specific problem independently by A and B are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\) respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
Solution:

Question 15
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace ’

(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen ’
F: ‘the card drawn is a queen or jack ’.
Solution:
(i) E : card drawn is a spade
F : card drawn is an ace
E ∩ F : card drawn is an ace of spade

Hence E, F are independent events

(ii) E : card drawn is black
F : card drawn is a king
E ∩ F : card drawn is a black king

Hence E and F are independent events.

(iii) E : card drawn is king or queen
F : card drawn is queen or jack
E ∩ F : card drawn is queen

Hence E and F are not independent events.

Question 16
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspapers.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H : student reads Hindi newspaper
E : student reads English newspaper

Question 17
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
(a) 0
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 1 }{ 12 }\)
(d) \(\frac { 1 }{ 36 }\)
Solution:
(d) n(S) = 36
The sample space contains 36 simple events
Let E: getting even prime number on both dice
∴ E = {(2, 2)} ∴ P(E) = \(\frac { 1 }{ 36 }\)

Question 18
Two events A and B are said to be independent, if
(a) A and B are mutually exclusive
(b) P(A’B’) = [1 – P(A)] [1 – P(B)]
(c) P(A) = P(B)
(d) P (A) + P (B) = 1
Solution:
A and B are independent events if P(A ∩ B) = P(A) P(B)
P( A’B’) = P( A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= [1 – P(A)] – P(B)[1 – P(A)]
= [1 – P(A)][1 – P(B)]
= P(A’)P(B’)
If A and B are mutually exclusive, then A and B are not independent events.
If P(A) = P(B), then A and B need not be independent events.
If P(A) + P(B) = 1, then A and B are complements of each other. Then also A and B are not independent events.

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1/

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1

Question 1
Given that E and Fare events such that P (E) = 0.6, P (F) = 0.3 and P(E∩F) = 0.2 find P(E|F) and P (F|E).
Solution:
Given: P (E)=0.6, P (F)=0.3, P (E ∩ F)=0.2
P(E|F) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 }\)
P(F|E) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }\)

Question 2
Compute P(A|B) if P(B)=0.5 and P (A∩B) = 0.32.
Solution:
Given: P(B) = 0.5, P(A∩B) = 0.32
P(A|B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.50 } =\frac { 32 }{ 50 }\) = 0.64

Question 3.
If P (A) = 0.8, P (B)=0.5 and P(B/A)=0.4, find
(i) P(A∩B)
(ii) P(A/B)
(iii)P(A∪B)
Solution:
(i) P(B/A) = \(\frac { P(A\cap B) }{ P(A) } \Rightarrow 0.4=\frac { P(A\cap B) }{ 0.8 }\)
∴ P(A∩B) = 0.4 x 0.8 = 0.32

(ii) P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.5 } =\frac { 32 }{ 50 } =\frac { 16 }{ 25 }\)

(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98

Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = \(\frac { 5 }{ 13 }\) and P(A|B) = \(\frac { 2 }{ 5 }\).
Solution:

Question 5.
If P(A) = \(\frac { 6 }{ 11 }\),P(B) =\(\frac { 5 }{ 11 }\) and P(A∪B) = \(\frac { 7 }{ 11 }\),
Find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Solution:

Question 6.
A coin is tossed three times, where
(i) E: head on third toss F: heads on first two tosses.
(ii) E: at least two heads F : at most two heads
(iii) E: at most two tails F: at least one tail
Solution:
The sample space S = {HHH, HHT, THH, TTH, THH, THT, TTT}

Question 7.
Two coins are tossed once
(i) E: tail appears on one coin F: one coin shows head
(ii) E: no tail appears F: no head appears.
Solution:
The Sample space S = {HH, TH, HT, TT}

Question 8.
A die is thrown three times.
E: 4 appears on the third toss
F: 6 and 5 appears respectively on first two tosses.
Solution:

Question 9.
Mother, father and son line up at random for a family picture:
E: son on one end, F: father in middle
Solution:
Mother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
F: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
E∩F = {(m, f, s), (s, f, m)}

Question 10.
A Mack and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
Let S be the sample space. Then
n(S) = 36
a. E : event of getting a sum greater than 9
E = {(4, 6), (5, 5), (5, 6), (6,4), (6, 5), (6, 6)}
F: event of getting 5 on black die.
F = {(5, 1), (5, 2), (5, 3), (5,4), (5, 5), (5, 6))
E ∩ F = {(5, 5), (5, 6))
\(\mathrm{P}(\mathrm{F})=\frac{6}{36}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}\)
\(P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{6}{36}\right)}=\frac{2}{6}=\frac{1}{3}\)

b. E : event of getting a sum 8
E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F : event of getting a number less than 4 on red die
F= {(1, 1),(2, 1),(3, 1),(4, 1),(5, 1),(6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1,3), (2,3), (3, 3), (4, 3), (5, 3), (6, 3)}
E ∩ F = {(5,3),(6,2)}
\(P(E \cap F)=\frac{2}{36}, P(F)=\frac{18}{36}\)
∴ P(E|F) = \(\frac{P(E ∩ F)}{P(F)}\) = \(\frac{(\frac{2}{36})}{(\frac{18}{36})}\) = \(\frac{1}{9}\)

Question 11
A fair die is rolled. Consider events E = {1,3,5} F = {2,3} and G = {2,3,4,5}, Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F) |G) and P (E ∩ F)|G)
Solution:

Question 12.
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is girl?
Solution:
Let B : The elder boy
G : The elder girl
b : The younger boy
g : The younger girl
The sample space S = {(B, h), (B, g), (G, b),(G, g)}

i. Let E : both children are girls
F : the youngest is a girl

ii. Let E : both children are girls
F : atleast on child is a girl

Question 13.
An instructor has a question bank consisting of 300 easy True/ False questions, 200 difficult True/ False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution:

Question 14.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
Let S be the sample space.
∴ n(S) =36
Let E : getting different numbers on the dice
∴ n(E) = 30 ∴P(E) = \(\frac { 30 }{ 36 }\)
Let F : getting the sum of numbers on the dice is 4
∴ F= {(1, 3), (2, 2), (3, 1))
n(F) = 3 ∴ P(F) = \(\frac { 3 }{ 36 }\)
E ∩ F = {(1,3),(3, 1)}
n(E ∩ F) = 2 ∴ P(E ∩ F) = \(\frac{2}{36}\)
\(\mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{30}{36}\right)}=\frac{1}{15}\)

Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’ given that ‘at least one die shows a 3’.
Solution:
The sample space S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, H). (1, T), (2, H). (2, T), (4, H). (4, T), (5, H), (5, T), (6, 1),(6, 2),(6, 3), (6,4),(6, 5),(6, 6))
Let E: coin shows tail
F : atleast one die shows 3
Now E ∩ F = Φ or P(E ∩ F) = 0
P(E|F) = 0 since P(E ∩ F) = 0

Question 16.
If P(A) = \(\frac { 1 }{ 2 }\), P (B) = 0 then P (A | B) is
(a) 0
(b) \(\frac { 1 }{ 2 }\)
(c) not defined
(d) 1
Solution:
P(A) = P(B) = 0
∴ P(A∩B) = 0
∴ P(A|B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0 }{ 0 }\)
Thus option C is correct.

Question 17.
If A and B are events such that P(A | B) = P(B | A) then
(a) A⊂B but A≠B
(b) A = B
(c) A∩B = φ
(d) P(A) = P(B)
Solution:
P(A | B) = P(B | A)
Thus option (d) is correct.
\(\frac { P(A\cap B) }{ P(B) } =\frac { P(B\cap A) }{ P(A) }\)
⇒ P(A) = P(B)

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-5/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.5

Question 1.
Find [a, b, c] if \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-2 \hat{k}\)
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ 2 & -3 & 1 \\ 3 & 1 & -2 \end{array}\right|\) = 1(6 – 1)+2(- 4 – 3) + 3(2 + 9) = 1(5) + 2(- 7) + 3(11) = 24.

Question 2.
Show that the vectors \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=-2 \hat{i}+3 \hat{j}-4\hat{k}\) and \(\vec{c}=\hat{i}-3\hat{j}+5 \hat{k}\) are coplanar.
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|\)
= 1(15 – 12)+2(- 10 + 4) + 3(6 – 3)
= 1(3) + 2(- 6) + 3(3) = 0.

Question 3.
Find λ, if the vectors \(\hat{i}-\hat{j}+\hat{k}, 3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\hat{i}+\lambda \hat{j}-3 \hat{k}\)are coplanar.
Solution:
Let \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\), \(\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{c}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) be the given vectors.
Since the given vectors are coplanar \([\vec{a}, \vec{b}, \vec{c}]\) = 0
i.e., \(\left|\begin{array}{ccc} 1 & -1 & 1 \\ 3 & 1 & 2 \\ 1 & \lambda & -3 \end{array}\right|\) = 0
1(- 3 – 2λ) + 1(- 9 – 2) + 1(3λ – 1) = 0
– 3 – 2λ – 11 + 3λ – 1 = 0 ⇒ λ = 15

Question 4.
i. If c₁ = 1 and c₂ = 2, find c₃ which makes \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
ii. Then if c₂ = – 1 and c₃ = 1, show that no value of c₁ can make \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
Solution:
i. Given that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ \([\vec{a}, \vec{b}, \vec{c}]\) = 0

Question 5.
Consider the 4 points A, B, C and D with position vectors \(4 \hat{i}+8 \hat{j}+12 \hat{k}, 2 \hat{i}+4 \hat{j}+6 \hat{k}\) \(3 \hat{i}+5 \hat{j}+4 \hat{k} \text { and } 5 \hat{i}+8 \hat{j}+5 \hat{k}\).
i. Find \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\).
ii. Show that the 4 points are coplanar.
Solution:

Question 6.
Find x such that the four points A(3, 2, 1), B(4, x, 5), C(4, 2, – 2) and D(6, 5, – 1) are coplanar
Solution:

ii. Since the 4 points A, B, C and D are coplanar \([\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}]\) = 0
\(\left|\begin{array}{ccc}
1 & (x-2) & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{array}\right|\) = 0
1(0 + 9) – (x – 2)(- 2+9) + 4(3 – 0) = 0
9 – 7(x – 2) + 12 = 0
9 – 7x + 14 + 12 = 0
⇒ 7x = 35
∴ x = 5

Question 7.
Show that the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar if \(\vec{a}\) + \(\vec{b}\), \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\) are coplanar.
Solution:

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Show that the line joining the origin to the point (2,1,1) is perpendicular to the line de-termined by the points (3, 5, – 1), (4, 3, -1).
Solution:
The direction ratios of the line joining (0, 0, 0) and (2, 1, 1) are 2, 1,1 The direction ratios Of the line joining (3, 5, -1) and (4, 3, -1) are 1, -2, 0
∴ a₁a₂ + b₁b₁ + C₁C₂
= 2(1) + 1(-2) + 1 (0) = 0
The lines are perpendicular.

Question 2.
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ – m₂ n₁, n₁ l₂ – n₂ l₁, l₁ m₂ – l₂ m₁
Solution:.
Let \(\hat{a}\) and \(\hat{b}\) be unit vectors in the direction of the lines
∴ \(\hat{a}=l_{1} \hat{i}+m_{1} \hat{j}+n_{1} \hat{k}\)
\(\hat{b}=l_{2} \hat{i}+m_{2} \hat{j}+n_{2} \hat{k}\)
Since a and b are mutually perpendicular, then \(\hat{a}\) x \(\hat{b}\) is a unit vector perpendicular to \(\hat{a}\) and \(\hat{b}\)

∴ The direction cosines are m₁n₂ – m₂ n₁, n₁ l₂ – n₂ l₁, l₁ m₂ – l₂ m₁

Question 3.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Solution:
a₁ = a, b₁ = b, c₁ = c
a₂ = b – c, b₂ = c – a, c₂ = a – b
Let θ be the angle between the lines

Question 4.
Find the equation of a line parallel to x- axis and passing the origin.
Solution:
The direction ratios of any line parallel to x-axis is 1, 0, 0. Hence the equation of the line passing through the origin having direction ratios 1,0,0 is
\(\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\)
i.e., \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\)

Question 5.
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (- 4, 3, -6) and (2, 9, 2) respectively, then find the angle between the line AB and CD.
Solution:
The direction ratios of AB
= 4 – 1, 5 – 2, 7 – 3 = 3, 3, 4
The direction ratios of CD = 6, 6, 8
Let θ be the angle between AB and CD

Question 6.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution:
The equation of the lines are
\(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and
\(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\)
The direction ratios of 1^st line = – 3, 2k, 2
The direction ratios of II^nd line = 3k, 1, – 5
Since the two lines are perpendicular,
a₁a₂ + b₁b₂ + c₁c₂ = 0
– 3(3k) + 2k(1) + 2(- 5) = 0
– 9k + 2k – 10 = 0
– 7k = 10
k = \(\frac { -10 }{ 7 }\)

Question 7.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9\) = 0.
Solution:
Let \(\hat{a}\) be the position vector of the point (1, 2, 3)
a = \(\hat{i}+2 \hat{j}+3 \hat{k}\)
The direction ratios of the normal to the plane are 1, 2, – 5.
Let b = \(\hat{i}+2 \hat{j}-5 \hat{k}\)
Let \(\hat{r}\) be the position vector of any point on the line.
The vector equation is \(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e., \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j}-5 \hat{k})\)

Question 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
Solution:
The given plane is \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
i.e., x + y + z – 2
The equation of any plane parallel to the given plane is x +y + z = k … (1)
Since the required plane passes through the point (a, b, c), we get a + b + c = k
Thus (1) gives the required plane as x + y + z = a + b + c

Question 9.
Find the shortest distance between the lines \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
Solution:

Question 10.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution:
The equation of the line through the points (5,1, 6) and (3,4, 1) is
\(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}=\lambda\)
i.e., \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ. + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the YZ – plane, x coordinate is zero
i.e., – 2λ + 5 = 0 ∴ λ = \(\frac { 5 }{ 2 }\)
\(y=3\left(\frac{5}{2}\right)+1=\frac{17}{2}\)
\(z=-5\left(\frac{5}{2}\right)+6=\frac{-13}{2}\)
∴ The required point is \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\)

Question 11.
Find the coordinates of the point where the line through (5, 1,6) and (3,4, 1) crosses the ZX- plane.
Solution:
The equation of the line through (5, 1, 6) and (3, 4, 1) is
\(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the ZX – plane,
y coordinate is zero.

Question 12.
Find the coordinates of the point where the line through (3, – 4, -5) and (2, -3, 1) crosses the plane 2x + y + z = 7.
Solution:
The equation of the line through (3, – 4, – 5) and (2, – 3, 1) is

The coordinate of any point P on the line is (- λ + 3, λ – 4, 6λ – 5)
P is a point on the plane 2x + y + z = 7
i.e., 2(- λ + 3) + (λ – 4) + (6λ – 5) = 7
– 2λ + 6 + λ – 4 + 6λ – 5 = 7
5λ – 3 = 7 ∴ λ = 2
∴ P is (- 2+ 3, 2 – 4, 6(2) – 5) = (1, – 2, 7)
∴ The coordinates of the point which crosses the plane is (1, -2, 7)

Question 13.
Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y +3z = 5 and 3x + 3y + z = 0.
Solution:
The plane passes through (- 1, 3, 2)
∴ x₁ = – 1, y₁ = 3, z₁ = 2
The direction ratios of the normal to the planes are 1, 2, 3 and 3, 3, 1.
Hence the equation of the plane is
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|\) = 0
\(\left|\begin{array}{ccc} x+1 & y-3 & z-2 \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{array}\right|\) = 0
– 7(x + 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x – 7 + 8y – 24 – 3z + 6 = 0
– 7x + 8y – 3z – 25 = 0
7x – 8y + 3z + 250 = 0

Another method:
The equation of a plane passing through the point(- 1, 3, 2) is
a(x + 1) + b(y – 3) + c(z – 2) = 0 … (1)
where a, b and c are the direction ratios of the normal to the plane.
Since (1) is perpendicular to the planes
x + 2y + 3z = 5 and 3x + 3y + z = 0, we get
a + 2b + 3c = 0 … (2) and
3a + 3b + c = 0 … (3)
∴ By the rule of cross multiplication, we get
\(\frac{a}{\left|\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right|}=\frac{0}{\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|}=\frac{c}{\left|\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right|}\)
i.e., \(\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}\)
Since a, b, c are the direction ratios of the normal to the plane, we can take a = – 7, b = 8, c = – 3.
∴ (1) → 7(x+ 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x + 8y – 3z – 25 = 0
or 7x – 8y + 3z + 25 = 0

Question 14.
If the points (1, 1, p) and (-3, 0, 1) are equidistant from the plane \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13\) = 0, then find the value of p.
Solution:
The equation of the plane is,

Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1 and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4\) = 0 and parallel to x- axis.
Solution:
Equation of the given planes are
x + y + z – 1 = 0 and 2x + 3y – z + 4 = 0
The equation of the plane passing through the line of intersection of the planes is
(x + y + z – 1) + (2x + 3y – z + 4) = 0
(1 + 2λ)x + (1 + 3λ)y + (1 – λ )z + (4 – 1) = 0 … (1)
The direction ratios of the normal to the plane are 1 + 2λ , 1 + 3λ , 1 – λ
The direction ratios of x-axis are 1, 0, 0
Since the required plane is parallel to the x-axis, the normal to the plane is perpendicular to x-axis.

Question 16.
If O is the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
Solution:
\(\overrightarrow{\mathrm{OP}}=\hat{i}+2 \hat{j}-3 \hat{k}\)
Direction ratios of OP are 1, 2, -3
OP is perpendicular to the plane
∴ Direction ratios of the normal to the plane
are a = 1, b = 2, c = -3
P( 1, 2, – 3) is a point on the plane.
Equation of the plane is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
i.e, 1(x – 1) + 2(y – 2) – 3(z + 3) = 0
x – 1 + 2y – 4 – 3z – 9 = 0
x + 2y – 3z – 14 = 0
The equation of the plane is x + 2y – 3z – 14 = 0

Question 17.
Find the equation of the plane which contains the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4\) = 0, \(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5\) = 0 and which is perpendicular to the plane \(\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8\) = 0.
Solution:
The equation of the given planes are x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The equation of the plane containing the line of inÍersection of the above plane is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y (3 – λ)z + (5λ – 4) = 0 … (1)
Since (1) is perpendicular to the plane
5x + 3y – 6z + 8 = 0, we get
5(1 + 2λ) + 3(2 + λ) – 6(3 – λ) = 0
5 + 10λ + 6 + 3λ – 18 + 6λ = 0
19λ – 7 = 0
∴ λ = \(\frac { 7 }{ 19 }\)

Question 18.
Find the distance of the point (- 1, – 5, – 10) from the point of intersection of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) = 5
Solution:
The cartesian equation of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) is
\(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}\) = λ … (1)
The cartesian equation of the plane
\(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) is x – y + z = 5 … (2)
The coordinate of any point on this line is P(2 + 3λ, – 1 + 4λ, 2 + 2λ).
Let the line intersects the plane at P, then P is a point on the plane.
∴ (2) → (2 + 3λ) – (-1 + 4λ) + (2 + 2λ) = 5
2 + 3λ + 1 – 4λ + 2 + 2λ = 5
∴ λ = 0
∴ P is (2, – 1,2) and the given point is Q(- 1, – 5, – 10)
∴ PQ = \(\sqrt{(-1-2)^{2}+(-5–1)^{2}+(-10-2)^{2}}\)
= \(\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}\)
= \(\sqrt{9+16+144}=\sqrt{169}\)
= 13

Question 19.
Find the vector equation of the line passing through (1,2,3) and parallel to the planes \(\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \text { and } \vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6\)
Solution:
The planes are \(\hat{r}\). \(\hat{n}\)₂ = d₁
and \(\hat{r}\). \(\hat{n}\)₂ = d₂
where \(\hat{n}\)₁ = \(\hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{n}_{2}=3 \hat{i}+\hat{j}+\hat{k}\)
Since the required line is parallel to the planes, it is perpendicular to both \(\hat{n}\)₁ and \(\hat{n}\)₂.
\(\hat{n}\)₁ x \(\hat{n}\)₂ is perpendicular to both \(\hat{n}\)₁ and \(\hat{n}\)₂
\(\vec{n}_{1} \times \vec{n}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 2 \\
3 & 1 & 1
\end{array}\right|=-3 \hat{i}+5 \hat{j}+4 \hat{k}\)
The required line passes through the point (1, 2, 3) and is parallel to \(\hat{n}\)₁ x \(\hat{n}\)₂.
∴ The equation of the line is
\(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})\)

Question 20.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\)
\(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)
Solution:
Let \(\vec{b}_{1}\) and \(\vec{b}_{2}\) be the direction of the lines where \(\vec{b}_{1}=3 \hat{i}-16 \hat{j}+7 \hat{k}\) and \(\vec{b}_{2}=3 \hat{i}+8 \hat{j}-5 \hat{k}\).
The required line is perpendicular to both \(\vec{b}_{1}\) and \(\vec{b}_{2}\). Hence it is parallel to \(\vec{b}_{1}\) x \(\vec{b}_{2}\)
\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|=24 \hat{i}+36 \hat{j}+72 \hat{k}\)
The direction ratios of \(\vec{b}_{1}\) x \(\vec{b}_{2}\) = 24, 36, 72
= 2, 3, 6
The required line passes through the point (1, 2, – 4 ) and has direction ratios 2, 3, 6.
Hence the equation is
\(\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Question 21.
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}\).
Solution:
The equation of a plane in the intercept from is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1.

Question 22.
Distance between the two planes:
2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
a. 2 units
b. 4 units
c. 8 units
d. \(\frac{2}{\sqrt{29}}\) units
Solution:
d. \(\frac{2}{\sqrt{29}}\) units

Hence the planes are parallel.
The distance between the parallel planes is the difference between their distance from the origin.
The distance of the plane 2x + 3y + 4z = 4

Question 23.
The planes 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
a. perpendicular
b. parallel
c. intersect y-axis
d. passes through (0, 0, \(\frac { 5 }{ 4 }\))
Solution:
b. parallel

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise

Question 1.
Write down a unit vector in XY – plane, making an angle of 30° with the positive direction of x – axis.
Solution:
Let P(x, y) be any point on the line making 30° with x-axis. Let \(\vec{a}\) = x\(\hat{i}\) + y\(\hat{i}\) be a unit vector.

Another method:
The line in the above figure lies in X Y plane and makes 30° with positive direction of x – axis. Hence the line makes 60° with y- axis and 90° with z – axis.
The direction cosines of the line
= cos 30°, cos 60°, cos 90° = \(\frac{\sqrt{3}}{2}\), \(\frac { 1 }{ 2 }\), 0
Hence the unit vector is \(\frac{\sqrt{3}}{2}\)\(\hat{i}\) + \(\frac { 1 }{ 2 }\)\(\hat{j}\) + 0\(\hat{k}\).

Question 2.
Find the scalar components and magnitude of the vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).
Solution:

Question 3.
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution:

Let the girl starts from O and stops at P. Let the coordinates of P be (x, y).
\(\vec{r}\) = \(\overrightarrow{OP}\)
∴ \(\vec{r}\) = – x\(\hat{i}\) + y \(\hat{j}\) … (1)
In ∆ QMP, sin60° = \(\frac { y }{ 3 }\)

Question 4.
If \(\vec{a}\) = \(\vec{b}\) + \(\vec{c}\), then is it true that |\(\vec{a}\)| = |\(\vec{b}\)| + |\(\vec{c}\)|? Justify your answer.
Solution:

Question 5.
Find the value of x for which x\((\hat{i}+\hat{j}+\hat{k})\) is a unit vector.
Solution:

Question 6.
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\).
Solution:
\(\vec{a}+\vec{b}\) be the resultant of \(\vec{a}\) and \(\vec{b}\).

Question 7.
If \(\vec{a}\) = \(\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{c}\) = \(\hat{i}-2 \hat{j}+\hat{k}\), find a unit vector parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:

Question 8.
Show that the points A (1, – 2, – 8), B (5, 0, – 2)and C (11, 3, 7) are collinear and find the ratio in which B divides AC.
Solution:

\(\overrightarrow{AB}\) is a scalar multiple of \(\overrightarrow{BC}\)
∴ \(\overrightarrow{AB}\) || \(\overrightarrow{BC}\) and B is the common point.
Hence A, B, C are collinear.
Also \(\frac{|\overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{BC}}|}=\frac{2}{3}\)
∴ The point B divides AC in the ratio 2 : 3.

Another method:
Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be the position vectors of A, B and C.
∴ \(\vec{a}\) = \(\hat{i}-2 \hat{j}-8 \hat{k}\), \(\vec{b}\) = \(5 \hat{i}+0 \hat{j}-2 \hat{k}\), \(\vec{c}\) = \(11 \hat{i}+3 \hat{j}+7 \hat{k}\)
Let B divide AC in the ratio λ : 1
By section formula, the position vector of

Equating components of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get

∴ λ = \(\frac{2}{3}\) satisfies (1) and (3)
Hence A, B and C are collinear and B divides AC in the ratio \(\frac{2}{3}\) : 1 That is, in the ratio 2 : 3.

Question 9.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2\(\vec{a}\) + \(\vec{b}\)) and (\(\vec{a}\)-3\(\vec{b}\)) externally in the ratio 1 : 2. Also, show that P is the midpoint of the line, segment RQ.
Solution:
\(\overrightarrow{\mathrm{OP}}=2 \vec{a}+\vec{b}, \overrightarrow{\mathrm{OQ}}=\vec{a}-3 \vec{b}\),
R divides PQ in the ratio 1 : 2 externally.

∴ P is the midpoint of the line segment RQ.

Question 10.
The two adjacent sides of a parallelogram are \(2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to its diagonal. Also, find its area.
Solution:
Let \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\) be any two adjacent sides of a parallelogram.
\(\vec{a}\) + \(\vec{b}\) is a vector along the diagonal

Question 11.
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
Solution:
Let a, P and y be the angles made by the vector with the axes OX, OY and OZ.
Let l, m and n be the direction cosines.
Since the vector is equally inclined to the
axes, α = ß = γ
∴ l = cosα, m = cos α, n = cos α
We have l² + m² + n² = 1
i.e, cos² α + cos² α + cos² α = 1
⇒ 3 cos² α = 1
cos² α = \(\frac { 1 }{ 3 }\)
cos α = \(\frac{1}{\sqrt{3}}\)
∴ l = m = n = \(\frac{1}{\sqrt{3}}\)
Hence the direction cosines of the vector equally inclined to the axes are \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\).

Question 12.
Let \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\) Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{c}\). \(\vec{d}\) = 15.
Solution:
Let \(\vec{d}\) = \(x \hat{i}+y \hat{j}+z \hat{k}\)
Since \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\)
\(\vec{d}\).\(\vec{a}\) = 0 and \(\vec{d}\).\(\vec{b}\) = 0.
x + 4y + 2z = 0 … (1) and
3x – 2y + 7z = 0 … (2)
Given \(\vec{c}\).\(\vec{d}\) = 15 … (3)
2x – y + 4z = 15 … (3)
Solving (1), (2) and (3), we get x = \(\frac { 160 }{ 3 }\)
y = \(\frac { – 5 }{ 3 }\) and z = \(\frac { -70 }{ 3 }\)
∴ \(\vec{d}=\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{3} \hat{k}\)
\(\vec{d}=\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})\)

Question 13.
The scalar product of the vector \(\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k} \text { and } \lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equal to one. Find the value of λ.
Solution:

Question 14.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular vectors of equal magnitudes, show that the vectors \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution:
Given that \(|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda\) … (1)
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular

Hence \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).

Question 15.
Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\), if and only if \(\vec{a}\), \(\vec{b}\) are perpendicular, given a ≠ \(\vec{0}\), b ≠ \(\vec{0}\)
Solution:
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b}) \ldots \ldots\) … (1)
Case 1
Let
From (1), we get \(\vec{a}\), \(\vec{b}\) = 0
Hence \(\vec{a}\) ⊥ \(\vec{b}\) since \(\vec{a}\) and \(\vec{b}\) are nonzero vectors.

Case 2
Let \(\vec{a}\) ⊥ \(\vec{b}\). Then \(\vec{a}\).\(\vec{b}\) = 0
(1) → \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\) + 0
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\)

Question 16.
If θ is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a}\).\(\vec{b}\) ≥ 0 only when
a. 0 < θ < \(\frac { π }{ 2 }\)
b. 0 ≤ θ ≤ \(\frac { π }{ 2 }\)
c. 0 < θ < π
d. 0 ≤ θ ≤ π
Solution:

Question 17.
Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and θ is the angle between them. Then \(\vec{a}\) + \(\vec{b}\) is a unit vector if
a. θ = \(\frac { π }{ 4 }\)
b. θ = \(\frac { π }{ 3 }\)
c. θ = \(\frac { π }{ 2 }\)
d. θ = \(\frac { 2π }{ 3 }\)
Solution:

Question 18.
The value of
\(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \text { is }\)
a. 0
b. -1
c. 1
d. 3
Solution:
c. 1
\(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \text { is }\)
= \(\hat{i} \cdot \hat{i}-\hat{j} \cdot \hat{j}+\hat{k} \hat{k}\)
= 1 – 1 + 1 = 1

Question 19.
If θ is the angle between any two vectors \(\vec{a}\) and \(\vec{b}\), then \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\), when θ is equal to
a. 0
b. \(\frac { π }{ 4 }\)
c. \(\frac { π }{ 2 }\)
d. π
Solution:
b. \(\frac { π }{ 4 }\)