Distance between points MCQ Quiz - Objective Question with Answer for Distance between points - Download Free PDF

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

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:

• 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:

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 (2, 0, 3) and B (- 4, a, - 1) are two points in a 3D space such that distance between them is 8 units.

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

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: P (6, 4, - 3) and Q (2, - 8, 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: \(\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

⇒ \(d = \sqrt {(-4)^2+(-12)^2+(6)^2}\)

\(d = \sqrt {{{\left( {{2} - {6}} \right)}^2} + {{\left( {{-8} - {4}} \right)}^2} + {{\left( {{3} + {3}} \right)}^2}} = 14 \ units\)

Concept:

Distance formula between the points (x1, y1, z1) and (x2, y2, z2) is = 

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

Explanation:

We have to find the distance between the point P (2m, 3m, 4m) and the x-axis.

Now,

The General point in the x-axis can be represented as (x, 0, 0) (As the point is in x-axis, its y and z

coordinates will be 0)

⇒ We have to find the distance between the point P (2m, 3m, 4m) and (2m, 0, 0).

Thus,

Distance between the point P (2m, 3m, 4m) and the x-axis =

Distance between the point P (2m, 3m, 4m) and the (2m, 0, 0) =

\(\Rightarrow \sqrt{(2m-2m)^2+(3m-0)^2+(4m-0)^2} \)

\(⇒ \sqrt{9m^2+16m^2} \)

\(⇒ \sqrt{25m^2}\)

⇒ 5m

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:

Let P(x, y, z) be the point which is equidistant from the points A(3, 4, -5) and B(-2, 1, 4)

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

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

First let's find out the distance between the the points P and A

⇒ \(PA = \sqrt {(3- x)^2 + (4 - y)^2 + (-5 - z)^2}\)

Similarly, let's find out the distance between the the points P and B

⇒ \(PB = \sqrt {(-2- x)^2 + (1 - y)^2 + (4 - z)^2}\)

∵ PA = PB

⇒ \(\sqrt {(3- x)^2 + (4 - y)^2 + (-5 - z)^2} = \sqrt {(-2- x)^2 + (1 - y)^2 + (4 - z)^2}\)

By squaring both the sides we get,

⇒ (3 - x)2 + (4 - y)2 + (-5 - z)2 = (-2 - x)2 + (1 - y)2 + (4 - z)2

⇒ x2 + y2 + z2 - 6x - 8y + 10z + 50 = x2 + y2 + z2 + 4x - 2y - 8z + 21

⇒ 10x + 6y - 18z - 29 = 0

So, the set of the points which are equidistant from the points A(3, 4, -5) and B(-2, 1, 4) is given by: 10x + 6y - 18z - 29 = 0.

Hence, correct option is 2.

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(2, - 1, 3) and B(-2, 1, 3) 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 = 2, y1 = - 1, z1 = 3, x2 = - 2, y2 = 1 and z2 = 3.

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

⇒ \(d = 2\sqrt 5\)

Hence, correct option is 1.

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: The distance between the points A(- 1, 3, - 4) and B(1, - 3, a) is \(2\sqrt{26}\)

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 = -1, y1 = 3, z1 = -4, x2 = 1, y2 = - 3, z2 = a and \(d = 2\sqrt{26}\)

⇒ \(2\sqrt {26} = \sqrt {(1+1)^2 + (-3 - 3)^2 + (a + 4)^2}\)

⇒ \(2\sqrt {26} = \sqrt {4 + 36 + (a + 4)^2}\)

By squaring both the sides we get,

⇒ 104 = 40 + a² + 8a + 16

⇒ a² + 8a - 48 = 0

⇒ a = 4, - 12

Hence, correct option is 2.

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: P (2, - 5, 7) and Q (3, 4, 5) 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: \(\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

Given:

Point 1 = (2, -1, 3)

Point 2 = (-5, 2, 1)

Formula:

Distance between two points having coordinates (x₁, y₁, z₁) and (x₂, y₂ , z₂) is given by :

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

Solution:

d = √[(2 - (-5))² + ((-1) - 2)² + (3 - 1)²]

= √(49 + 9 + 4)

= √62 units

Note: The options given in the Official Paper were wrong. We have corrected the option.

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 (1, 2, 5) and B (3, - 5, 0) 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: \(\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

Given:

Two points are (0, 5, 5) and (5, 8, 6).

Concept:

Distance between two points \(\rm (x_1,y_2,z_3) and(x_2,y_2,z_3)\) is

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

Calculation:

Distance form origin (0,0,0) to point (0,5,5) is

\(\rm d_1=\sqrt{(0)^2+(5)^2+(5)^2}\)

\(\rm d_1=5\sqrt{2}\)

Distance form origin (0,0,0) to point (5,8,6) is

\(\rm d_2=\sqrt{(5)^2+(8)^2+(6)^2}\)

\(\rm d_2=5\sqrt5\)

Now the sum of distances from origin to (0, 5, 5) and (5, 8, 6) is

\(\rm d=d_1+d_2\)

\(\rm d=5\sqrt{2}+5\sqrt{5}\)

\(\rm d=5(\sqrt{2}+\sqrt{5})\)

Hence the option (4) is correct.