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QUESTION: 1

The equation of line of intersection of the planes x + 2y + z = 3 and 6x + 8y + 3z = 13 can be written as

Solution:

Let the d.r’s of a required line be a, b and c. Since, the normal to the given planes x + 2y + z = 3 and 6x + 8y + 3z = 13 are perpendicular to the line.

∴ a +2b+c = 0 and 6a + 8b + 3c = 0

⇒

⇒

or

Also, line passes through (2, -1, 3).

QUESTION: 2

The direction cosines of two lines are such that l + m + n = 0, l^{2} + m^{2} – n^{2} = 0, then angle between them is

Solution:

l + m + n = 0

and l^{2} + m^{2} - n^{2} = 0

l^{2} + m^{2} (- l - m)^{2} = 0

2lm = 0 (l = 0 or m = 0)

If l = 0 then n = - m

l : m : n = 0 : l : -1

If m = 0 then n = - l

l : m : n = l : 0 : -1

⇒ θ = π/3

QUESTION: 3

The point of intersection of the lines and

Solution:

Find general pt. of both line and equate them

⇒ (3λ - 1, 5λ - 3, 7λ - 5 ) .......(i)

⇒ (μ +2, 3μ + 4, 5μ + 6) .......(ii)

on solving λ = 1/2, & μ = -3/2

on putting λ = 1/2, & μ = -3/2 we get

QUESTION: 4

If the straight lines x = 1 + s, y = - 3 - λs, z = 1 + λs and x = t/2, y = 1 + t, z = -t + 2 with parameters s and t respectively are coplanar, then λ equals

Solution:

Given line can be rewritten as

and

since two line are coplanar, then

λ = -2

QUESTION: 5

Shortest distance between the line and

Solution:

Given, lines can be written as

and

∴ Requires SD =

Where

∴ Requires SD

QUESTION: 6

If P(B) = 3/4, = 1/3 = 1/3, then

Solution:

QUESTION: 7

If events are independent and P(A) = 1/3, P(B) = 1/3, P(C) = 1/4 then is equal to

Solution:

QUESTION: 8

A fair die is tossed eight times. The probability that a third six is observed on the eight throw is

Solution:

Required probability

QUESTION: 9

For k = 1, 2, 3, the box B_{k} contains k red balls and (k + 1) white balls. Let P(B_{1}) = 1/2, P(B_{2}) = 1/3, P(B_{3}) = 1/6. A box is selected at random and a ball is drawn from it. If a red ball is drawn, then probability that it has come from box B_{2} is

Solution:

Use Baye’s Theorem formula

A : Red ball is drawn

= 14/39

QUESTION: 10

Let A and B be two event such that stands for complement of event A. Then events A and B are

Solution:

Independent event but not equally likely

QUESTION: 11

Three of the six vertices of a regular hexagon are chosen at random. The probability that triangle with three chosen vertices is an equilateral triangle is

Solution:

Total No. of ways = ^{6}C_{3}

No. of favorable ways = 2

QUESTION: 12

If A and B are two events such that P(A∪B) = 3/4, P(A∩B) 1/4, = 2/3, then

Solution:

P(A∪B) = P(A) = P(A) + P(B) - P(A∩B)

P(B) = 2/3

= P(B) - P(A∩B) =

QUESTION: 13

Three critics review a book. Odds in favour of book are 5 : 2, 4 : 3 and 3 : 4, respectively for the three critics. The probability that majority are in favour of the book is

Solution:

Probability that first critic favour is P(ε_{1}) = 5/7

Similarly, P(ε_{2}) = 4/7 P(ε_{3}) = 3/7

Majority are in favour if at least two favour

QUESTION: 14

Out of a set of integers given by {1, 2, 3, …. 30}, three numbers are selected at random. Find the probability the sum of the number chosen is divisible by 3.

Solution:

number of type 3k → 10

3k + 1 → 10

3k + 2 → 10

Either all the no. are of the same type or one No. from each type

QUESTION: 15

A is a set containing n element. A subset P of A is chosen at random. The set A is reconstructed by replacing the element of P. A set ‘Q’ is again chosen at random. Find probability such that P∩Q = φ.

Solution:

For 1 element not to be present in P∩Q possibilities for n element ⇒ Total

No. of fovourable cases = 3^{n}

QUESTION: 16

If the integers m and n are chosen at random from 1 to 100, then probability that a no. of form 7^{m} + 7^{n} is divisible by 5 equals

Solution:

Consider last digit of power of 7

last digit

7^{4k} → 1

7^{4k+1} → 7

7^{4K+2} → 9

7^{4k+3} → 3

7^{m} + 7^{n} is divided by 5 if

(i) m → 4k + 1, n → 4k + 3

(ii) m→ 4k + 2 and n → 4k

(iii) m→ 4k+3 andn→ 4k+1

(iv) m → 4k and n → 4k + 2

QUESTION: 17

If are the probability of three mutually exclusive and exhaustive events, then the set of all value of p is

Solution:

Since events are mutually exclusive and exhaustive

Hence the set of value satisfying all the above inequalities are

QUESTION: 18

A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is

Solution:

Probability of getting head = 1/2 and probability of throwing 5 or 6 with a dice = 2/6 1/3 . He starts with a coin and alternately tosses the coin and throws the dice and he will if he get a head before he get 5 or 6.

QUESTION: 19

Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is

Solution:

The numbers should be divisible by 6. Thus, the number of favourable ways is ^{16}C_{3} (as there are 16 numbers in first 100 natural numbers, divisible by 6).

Required probability is

QUESTION: 20

A student appears for test I, II and III. The student is successful if he passes either in test I and II or test I and III. The probability of the student passing in test I, II, III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then

Solution:

Let A, B and C be the events that the student is successful in test I, II and III respectively, then P(the student is successful)

P[A∩B∩C')∪(A∩B'∩C)∪(A∩B∩C)]

= P(A∩B∩C') + P(A∩B'∩C) + P(A∩B∩C)

P(A).P(B).P(C') + P(A)P(B')P(C) + P(A)P(B)P(C)

{∴ A, B, C are independent}

This equation has infinitely many values of p and q.

QUESTION: 21

It is given that events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3, then P (B) is

Solution:

P(A/B) = P(A⋂B)/P(B)

P(B/A) = P(B⋂A)/P(A)

2/3 = P(B⋂A)/(1/4)

P(A⋂B) = 2/3*1/4

= 1/6

Therefore, P(A/B) = P(A⋂B)/P(B)

1/2 = 1/6P(B)

P(B) = 1/3

QUESTION: 22

Solution:

number of type 3k → 10

3k + 1 → 10

3k + 2 → 10

Either all the no. are of the same type or one No. from each type

QUESTION: 23

The equation of plane passing through the point (0, 7, -7) and containing the line

Solution:

The equation of plane passing through (0, 7, -7) is a(x – 0) + b(y – 7) + c(z + 7) = 0

Plane contains line and passes through (-1, 3, -2)

∴ a(-1) + b(3 – 7) + c(-2 + 7) = 0

-3a + 2b + c = 0

∴ a : b : c = 1 : 1 : 1

x + y + z = 0

QUESTION: 24

The angle between line and plane 3x – 2y + 6z = 0 is (μ is scalar)

Solution:

D. R’s of line are 2 : 1 : 2

D.R’s of Normal to plane is :3 : -2 : 6

Angle between line and plane

QUESTION: 25

Let P(3, 2, 6) be a point in space and Q be a point on line Then value of m for which the vector is parallel to the plane x-4y + 3z = 1 is

Solution:

= i(- 2 - 3μ) + j(μ - 3) + k(5μ - 4)

is parallel to x - 4y + 3z = 1

1(- 2 - 3μ) + 4(μ - 3) + 3(5μ - 4) = 0

μ = 1/4

QUESTION: 26

The equation of the plane through the point (-1, 2, 0) and parallel to the line and

Solution:

QUESTION: 27

Let A (1,1,1) , B (2, 3, 5) and C (-1, 0, 2) be three points, then equation of a plane parallel to the plane ABC which is at a distance 2 from origin is

Solution:

A (1,1,1), B (2,3,5), C(-1,0,2) direction ratios of AB are < 1,2,4 >

Therefore, direction ratios of normal to plane ABC are < 2, -3,1 >

As a result, equation of the required plane is 2x – 3y +z = k then

Hence, equation of the required plane is

QUESTION: 28

The distance between the line and the plane

Solution:

Therefore, the line and the plane are parallel. A point on the line is (2, -2, 3). Required distance = distance of (2, -2, 3) from the given plane x + 5y + z - 5 = 0

QUESTION: 29

If the distance between the plane x - 2y + z = d and the plane containing the line and then |d| is

Solution:

Equation of plane containing the given lines is

⇒ (x - 1) (-1) - (y - 2) (-2) + (z - 3) (-1) = 0

⇒ -x + 1 + 2y - 4 - z + 3

⇒ -x + 2y - z = 0

Given plane is

x - 2y + z = d .....(ii)

Eqs. (i) and (ii) are parallel

Now, distance between planes

⇒ |d| = 6

QUESTION: 30

A plane π passes through the point (1, 1, 1). If b, c, a are the direction ratios of a normal to the plane, where a, b, c (a < b < c) are the prime factors of 2001, then the equation of the plane π is

Solution:

2001 = 3 x 23 x 29 and (3+23+29)

= 55 ⇒ a = 3, b = 23, c = 29

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