Distance between points MCQ Quiz in मल्याळम - Objective Question with Answer for Distance between points - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• The distance between the points A(x1, y1, z1) and B(x2, y2, z2) is given by: \(d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

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: \(d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

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

⇒ \(d = \sqrt {(2 + 3)^2 + (4 - 7)^2 + (-1 - 2)^2}\)

⇒ \(d = \sqrt {43}\)

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: \(\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

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: \(\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

Calculation

Let \( \dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{7-2}{12}=t \)

Any point on the line can be written in the parametric form as \( (3t+2, 4t-1 , 12t+2) \)

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

\( \Rightarrow 3t+2 -4t+1+12t+2 = 16 \)

\( \Rightarrow 11t = 11 \)

\( \Rightarrow t =1 \)

Hence, the point of intersection is \( (5,3,14) \)

The distance of \( (5,3,14) \) from \( (1,0,2) \) \( = \sqrt { 16+9+144 } = 13 \)

Hence option 4 is correct

Concept:

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

AB = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Explanation:

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

AB = \(\sqrt{(-1-1)^2+(-1-2)^2+(-1-3)^2}\) = \(\sqrt{4+9+16}\) = \(\sqrt{29}\)

BC = \(\sqrt{(3+1)^2+(5+1)^2+(7+1)^2}\) = \(\sqrt{16+36+64}\) = \(\sqrt{116}\) = 2\(\sqrt{29}\)

CA = \(\sqrt{(3-1)^2+(5-2)^2+(7-3)^2}\) = \(\sqrt{4+9+16}\) = \(\sqrt{29}\)

Here AB + CA = BC so

A, B, C are collinear 

(1) is correct

Answer : 3

Solution :

We have, equation of line.

\(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) 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

= \(\sqrt{(1-5)^{2}+(6-3)^{2}+(2-14)^{2}}\)

= \(\sqrt{16+9+144}=\sqrt{169}\) = 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

\( \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \)

Distance of P from A is 21

⇒ \( \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21\)

⇒ \(\left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right)\)

⇒ (22, 0, 7) = (a, b, c) 

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

\(\Rightarrow \sqrt{324+144+16}=\) 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) = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Explanation:

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

⇒ Distance (d) = \(\sqrt{(4 - 2)^2 + (1 - 3)^2}\)

⇒ Distance (d) = \(\sqrt{(2)^2 + (-2)^2}\)

⇒ 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,

\(\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2 +(z_2-z_1)^2 }\)

Calculation:

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

\(\sqrt{(0-0)^2 +(5-0)^2 +(5-0)^2 } = \sqrt{50}\)

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

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

\(\sqrt{(5-0)^2 +(8-0)^2 +(6-0)^2 } = \sqrt{125}\)

⇒ 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

 \(\rm\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\).

Calculation:

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

Given distance between them is \(\sqrt{27}\).

⇒ \(\sqrt{27}\) = \(\rm\sqrt{(a-0)^2+(0-1)^2+(1-2)^2}\)

⇒ \(\sqrt{27}\) = \(\rm\sqrt{a^2+2}\) .

Squaring both sides, we get:

a² + 2 = 27

⇒ a² = 25

⇒ a = ± 5.

∴ Required value of a is ± 5.