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Test: Distance between two points in 3D Space - Class 10 MCQ


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10 Questions MCQ Test The Complete SAT Course - Test: Distance between two points in 3D Space

Test: Distance between two points in 3D Space for Class 10 2024 is part of The Complete SAT Course preparation. The Test: Distance between two points in 3D Space questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Distance between two points in 3D Space MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Distance between two points in 3D Space below.
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Test: Distance between two points in 3D Space - Question 1

Find the distance between two points (5, 6, 7) and (2, 6, 3).

Detailed Solution for Test: Distance between two points in 3D Space - Question 1

We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is 
So, distance between two points (5, 6, 7) and (2, 6, 3) will be 

Test: Distance between two points in 3D Space - Question 2

The three points A (1, 2, 3), B (3, 1, 2), C (2, 3, 1) form _________.

Detailed Solution for Test: Distance between two points in 3D Space - Question 2

We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is 

Since AB = BC = CA so, it forms equilateral triangle.

Test: Distance between two points in 3D Space - Question 3

The three points A (7, 0, 10), B (6, -1, 6), C (9, -4, 6) form ________.

Detailed Solution for Test: Distance between two points in 3D Space - Question 3

We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is 

Since AB = BC and AB2+BC2 = AC2 so, it forms right angled isosceles triangle.

Test: Distance between two points in 3D Space - Question 4

The three points A (3, 0, 3), B (5, 3, 2), C (6, 5, 5) form ________.

Detailed Solution for Test: Distance between two points in 3D Space - Question 4

We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is 

Since AB = BC so, it forms isosceles triangle.

Test: Distance between two points in 3D Space - Question 5

The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form __________.

Detailed Solution for Test: Distance between two points in 3D Space - Question 5

We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is

Since AB = CD and BC = AD i.e. opposite two sides are equal so, it is a parallelogram.

Test: Distance between two points in 3D Space - Question 6

Find the cartesian equation of the plane which passes through the point (5, 2, -4) and perpendicular to the line with direction ratios 2, 3, -1?

Detailed Solution for Test: Distance between two points in 3D Space - Question 6

Given:
The plane which passes through the point (5, 2, -4)
Plane is perpendicular to the line with direction ratios 2, 3, -1

Concept:
Cartesian equation of the plane passing through (x1 , y1 , z1) and perpendicular to line having drs a, b, c is given by :
a(x - x1) + b(y - y1) + c(z - z1) = 0

Solution:
Equation of plane :
2(x - 5) + 3(y - 2) -(z - (-4)) = 0
⇒ 2x + 3y - z = 20

Test: Distance between two points in 3D Space - Question 7

If the points (k, 4, 2), (6, 2, - 1) and (8, - 2, - 7) are collinear, then find the value of k.

Detailed Solution for Test: Distance between two points in 3D Space - Question 7

CONCEPT:
If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then 

CALCULATION:
 Given: The points (k, 4, 2), (6, 2, - 1) and (8, - 2, - 7) are collinear
As we know that, if the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then 

Here, x1 = k, y1 = 4, z1 = 2, x2 = 6, y2 = 2, z2 = - 1, x3 = 8, y3 = - 2 and z3 = - 7

⇒ k × (- 14 - 2) - 4 × (- 42 + 8) + 2 × (- 12 - 16) = 0
⇒ - 16k + 136 - 56 = 0
⇒ 16k = 80
⇒ k = 5
Hence, option D is the correct answer.

Test: Distance between two points in 3D Space - Question 8

What is the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0?

Detailed Solution for Test: Distance between two points in 3D Space - Question 8

Concept:

Perpendicular Distance of a Point from a Plane 

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d

And a point whose coordinate is, (x1, y1, z1)

Now, distance = 
Calculation:

We have to find the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0

Distance =  

Test: Distance between two points in 3D Space - Question 9

Find the values of k so the line are parallel.

Detailed Solution for Test: Distance between two points in 3D Space - Question 9

Concept:
Let two lines having direction ratios a1, b1, c1, and a2, b2, c2 respectively.
Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0
Condition for parallel lines: 
 

Calculation:
Given lines are 
 
The direction ratio of the first line is (2k, 3, -1) and the direction ratio of second line is (8, 6, -2)
Lines are parallel;

Test: Distance between two points in 3D Space - Question 10

The sum of the direction cosine of z-axis is

Detailed Solution for Test: Distance between two points in 3D Space - Question 10

Concept:
The direction cosines of the vector are the cosines of angles that the vector forms with the coordinate axes.
Calculation:
Z-axis makes an angle 90° with X-axis, 90° with Y-axis, and 0° with Z-axis
∴ Direction cosines of Z-axis: cos 90, cos 90, cos 0
i.e., 0, 0, 1
Now sum of the direction cosine of z-axis = 0 +  0 + 1 = 1
Hence, option (3) is correct. 

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