Distance between points MCQ Quiz in বাংলা - Objective Question with Answer for Distance between points - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

• The distance between the points A(x1, y1, z1) and B(x2, y2, z2) is given by: 

CALCULATION:

Given: A(-3, 7, 2) and B(2, 4, -1) are two points in a 3D plane.

Let d denote the distance between the given points.

As we know that, the distance between the points A(x1, y1, z1) and B(x2, y2, z2) is given by: 

Here, x1 = -3, y1 = 7, z1 = 2, x2 = 2, y2 = 4 and z2 = -1.

⇒ 

⇒ 

Hence, correct option is 4.

CONCEPT:

If A(x1, y1, z1) and B(x2, y2, z2) then the distance between the points A and B is given by: 

CALCULATION:

Given: A (0, 0, 0) and B (1, 2, 3) are two points in a 3D space.

Here, we have to find the distance between the the given points.

As we know that, if A(x1, y1, z1) and B(x2, y2, z2) then the distance between the points A and B is given by: 

Calculation

Let

Any point on the line can be written in the parametric form as

To find the point of intersection, let us substitute the point in the equation of the plane.

Hence, the point of intersection is

The distance of from

Hence option 4 is correct

Concept:

Distance between the points A(x₁, y₁, z₁), B(x2, y2, z2) is

AB = 

Explanation:

The points are A(1, 2, 3), B(-1, -1, -1) C(3, 5, 7) 

AB =  =  = 

BC =  =  =  = 2

CA =  =  = 

Here AB + CA = BC so

A, B, C are collinear 

(1) is correct

Answer : 3

Solution :

We have, equation of line.

 and plane : x - y + z = 16

any point on line is (3t + 2, 4t - 1, 12t + 2) and this point will satisfy the plane.

∴ 3t + 2 - 4t + 1 + 12t + 2 = 16 ⇒ 11t = 11 ⇒ t = 1

So, point will be (5, 3, 14)

Hence, distance between (5, 3, 14) and (1, 6, 2) is

= 

=  = 13 units

Concept:

• Line and Plane Intersection:
  • A line in 3D can be written in symmetric form: (x − x₁)/a = (y − y₁)/b = (z − z₁)/c
  • A plane in 3D has the general form: Ax + By + Cz + D = 0
  • To find the intersection point, substitute parametric equations of the line into the plane's equation.

Calculation:

Given, line: (x + 1) = (y + 3)/3 = (−z + 2)/2

Let the common value = t

⇒ x = t − 1

⇒ y = 3t − 3

⇒ z = 2 − 2t

Given plane: 3x + 4y + 5z = 10

Substitute values of x, y, z into the plane:

⇒ 3(t − 1) + 4(3t − 3) + 5(2 − 2t) = 10

⇒ 3t − 3 + 12t − 12 + 10 − 10t = 10

⇒ (3t + 12t − 10t) + (−3 −12 + 10) = 10

⇒ 5t − 5 = 10

⇒ 5t = 15

⇒ t = 3

Now, find coordinates:

⇒ x = 3 − 1 = 2

⇒ y = 3×3 − 3 = 6

⇒ z = 2 − 2×3 = −4

∴ The point of intersection is (2, 6, −4)

Calculation

Equation of line AB

Distance of P from A is 21

⇒ 

⇒ 

⇒ (22, 0, 7) = (a, b, c) 

The distance between the points P(22, 0, 7) and Q(4, –12, 3) 

 22

Formula Used:

To find the distance between two points (x₁, y₁) and (x₂, y₂) in a Cartesian coordinate system, you can use the distance formula:

Distance (d) = 

Explanation:

In this case, the points are (2, 3) and (4, 1), so you can use these coordinates in the distance formula:

⇒ Distance (d) = 

⇒ Distance (d) = 

⇒ Distance (d) = √(4 + 4)

⇒ Distance (d) = √8

⇒ Distance (d) = 2√2

So, the distance between the points (2, 3) and (4, 1) is 2√2 units. 

Concept:

Distance formula- The distance between (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by,

Calculation:

Distance from the origin to (0,5,5) is given by

⇒ Distance from the origin to (0,5,5) = 5√2

And Distance from origin to (5,8,6) is given by,

⇒ Distance from the origin to (5,8,6) = 5√5

⇒ The sum of distances from origin to (0, 5, 5) and (5, 8, 6) = 5√2 + 5√5 = 5(√2 + √5)

∴ The correct option is (4).

Concept:

Distance between point (x1, y1, z1) and (x2, y2, z2) is

 .

Calculation:

Given points are (a, 0, 1) and (0, 1, 2).

Given distance between them is .

⇒  = 

⇒  =  .

Squaring both sides, we get:

a² + 2 = 27

⇒ a² = 25

⇒ a = ± 5.

∴ Required value of a is ± 5.