Mathematics: CUET Mock Test - 3

Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:
P(E1 | A) = P(E1)P(A | E1)/( P(E1)P(A│E1)+P(E2)P(A | E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.

Bayes theorem formula is P(A | B) = 
The formula provides relationship between P(A | B) and P(B | A). It is mainly derived from conditional probability formula P(A | B) and P(B | A). Where,

In Bayesian statistics, we calculate new probability after information becomes available due to new events and this is known as Posterior Probability. There is no term like Independent probabilities and Dependent probabilities, there are only independent events and dependent events. Interior probabilities represent probabilities of the intersection between two events.

Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a red ball.
P(E1) = P(E2) = 1/2.
Also, P(A | E1) = P(drawing a red ball from Bag 1) = 3/8.
And P(A | E2) = P(drawing a red ball from Bag 2) = 4/10.
The probability of drawing a ball from bag 2, being given that it is red is P(E2 | A).
By using Bayes’ theorem,
P(E2 | A) = P(E2)P(A | E2)/( P(E1)P(A│E1)+P(E2)P(A | E2))
= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.

The angle between a line and a plane is the complement of the angle between the line L and a normal line to the plane π. If θ is the angle between line whose ratios are a₁, b₁, c₁ and the plane ax + by + cz + d = 0 then sin θ = .

Angle between a plane and a line sin θ = 
sin θ = – 0.49
θ = sin-1(- 0.49)
θ = – 29.34 

The angle between a line and a plane is the complement of the angle between the line L and a normal line to the plane π. If θ is the angle between line whose ratios are a₁, b₁, c₁ and the plane ax + by + cz + d = 0 then sin θ = 

The relation between the plane ax + by +cz + d = 0 and a₁, b₁, c₁ the direction ratios of a line, if the plane and line are parallel to each other is a₁a + b₁b + c₁c = 0.

The relation between the plane ax + by +cz + d = 0 and a₁, b₁, c₁ the direction ratios of a line, if the plane and line are parallel to each other is a₁a + b₁b + c₁c = 0.

Angle between a plane and a line sin θ = 
sinθ = 0.49
θ = sin-1(0.49)
θ = 29.34

The condition for a plane and a line are parallel to each other is a₁a + b₁b + c₁c = 0.
5(3) + 1(-1) + k(1) = 0
K(1) = -14
K = -14

The condition for a plane and a line are parallel to each other is a₁a + b₁b + c₁c = 0.
8(1) + 3(2) + 2(k) = 0
2(k) = -14
k = -7

A mathematical symbol θ is used to find the angle between line and a normal line to the plane π along with a trigonometric function called sine. Hence, the formula
sin θ = 

The trigonometric function is used to find the angle between a line and a plane is sine. If θ is the angle between line whose ratios are a₁, b₁, c₁ and the plane ax + by + cz + d = 0 then sin θ = .

A plane and A line which are perpendicular to each other or a plane and a line having an angle 90 degrees between them are called orthogonal. θ is equal to 90 degrees in sin θ = .

θ = 90 degrees
The relation between the plane ax + by + cz + d = 0 and a₁, b₁, c₁ the direction ratios of a line, if the plane and line are perpendicular to each other is a/a_, = b/b₁ = c/c₁.

Angle between a plane and a line sin θ = .
sinθ = 0.92
θ = 66.92

Angle between a plane and a line sin θ = .
sinθ = 0.06
θ = 3.43

The angle between the normals to two planes is called the angle between the planes. A trigonometric identity, cosine is used to find the angle called ‘θ’ between two planes.

The formula to find the angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is cos θ = .
θ is the angle between the normal of two planes.

If their normals are perpendicular to each other then a₁a₂ + b₁b₂ + c₁c₂ = 0.
2(3) + 2(1) + s(1) = 0
s(1) = - 8
k = - 8

θ = 90 degrees ⇒ cos θ
a₁a₂ + b₁b₂ – c₁c₂ = 0
Relation between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c2₁z + d₂ = 0, if their normal are perpendicular to each other is a₁a₂ + b₁b₂ + c₁c₂ = 0.

Relation between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c2₁z + d₂ = 0, if their normal are perpendicular to each other is a1a₂ + b₁b₂ + c₁c₂ = 0.
5(1) + 1(1) + 3(-k) = 0
-3k = -6
K = 2

Relation between the the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c2₁z + d₂ = 0, if their normal are perpendicular to each other is a₁a₂ + b₁b₂ + c₁c₂ = 0.
1(3) + 2(4) + k(-1) = 0
k(-1) = -11
k = 11

 , ax + by + cz = d, lx + my + nz = d are the various ways of representing a plane.
is the vector equation of the plane, where n^ is the unit vector normal to the plane.
ax + by + cz = d, lx + my + nz = d are the Cartesian equation of the plane in the normal form where, a, b, c are the direction ratios and l, m, n are the direction cosines of the normal to the plane respectively.

Let 

Hence, the required equation of the plane is 

Given that the equation of the plane is = 4.
We know that, 
∴ 

⇒ 2x + y - z = 4 is the Cartesian equation of the plane.

From the given equation, the direction ratios of the normal to the plane are 3, 4, -5; the direction cosines are
, i.e. 3√50,4√50,−5√50
Dividing the equation throughout by √50, we get

The above equation is in the form of lx + my + nz = d, where d is the distance of the plane from the origin. So, the distance of the plane from the origin is 7√50.

Let,
The vector equation of the plane passing through three points is given by
 = 0
= 0

The position vector of the point (3,2,-3) is  and the normal vector  perpendicular to the plane is 
Therefore, the vector equation of the plane is given by  = 0
Hence, 
 = 0
4(x - 3)-2(y-2) + 5(z + 3) = 0
4x - 2y + 5z + 7=0.

Given that the equation of the plane is 
We know that, 
∴  = 12
⇒ (λ + 2μ)x + (2λ - μ)y + (3λ - 2μ)z = 12 is the Cartesian equation of the plane.

The position vector of the point (3,2,-3) is  and the normal vector  perpendicular to the plane is 
Therefore, the vector equation of the plane is given by  = 0
Hence,  = 0
4(x - 3) - 2(y - 2) + 5(z + 3) = 0
4x - 2y + 5z + 7 = 0

To evaluate: 
= 
= 

Explanation: Given that,  and
Calculating the vector product, we get

= 
= 

Given that,  and
Calculating the vector product, we get

= 
= 

Given that: a)
b) θ = 60° and 
c)  = 14
∴  = 1/4
= | = 1/4
⇒  = 1/4 x 1/2
∴ 

Consider 
= 
= 
We know that,  and 
∴ 
= 
Hence, 

Direction of λ  and   is same if value λ is positive as it gives it a direction which is positive in nature. If the value of λ is negative then the direction of the result after multiplication becomes in opposite direction. Whereas the value of the product vector becomes zero if value of λ is 0.

|λ| times the magnitude of vector   is denoted as 
As we know that the magnitude of vector  is denoted by , if we multiply the magnitude of vector  with magnitude of λ we get .

As both the vectors are equal hence, we can equate their constants and get the value of x, y and z. Now we equate the coefficients of  of both the equations and get the values
x = 2, y = 2, z = 1.