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Test: Distance Formula 3D Geometry - Commerce MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Distance Formula 3D Geometry

Test: Distance Formula 3D Geometry for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Distance Formula 3D Geometry questions and answers have been prepared according to the Commerce exam syllabus.The Test: Distance Formula 3D Geometry MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Distance Formula 3D Geometry below.
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Test: Distance Formula 3D Geometry - Question 1

The equation representing the set of points which are equidistant from the points (1, 2 , 3) and (3 , 2 , -1) is

Detailed Solution for Test: Distance Formula 3D Geometry - Question 1

Let the given points be A (1, 2, 3) and B (3, 2, -1) Let P(x, y, z) be any point which is equidistant from the points A and B.
Then PA = PB 


⇒ (x - 1)2 + (y - 2)2 + (z - 3)2
= (x - 3)2 + (y - 2)2 + (z +1)2
⇒ x2 +1 - 2x + y2 + 4 - 4y + z2 + 9 - 6z
= x2 + 9 - 6x + y2 + 4 - 4y + z2 +1+2z
⇒ 6x - 2x - 4y + 4y -  6z - 2z = 0
⇒ 4x - 8z = 0
⇒ x - 2z = 0
This is the required equation of the set of points in reference.

Test: Distance Formula 3D Geometry - Question 2

The locus of the point which is equidistant from the points A(0, 2, 3) and B(2, -2, 1) is:

Detailed Solution for Test: Distance Formula 3D Geometry - Question 2

Test: Distance Formula 3D Geometry - Question 3

The distances of the point (1, 2, 3) from the coordinate axes are A, B and C respectively. Which option is correct?

Detailed Solution for Test: Distance Formula 3D Geometry - Question 3

Given points (1,2,3)
Pt A (1,0,0) on x axis
Pt B (0,2,0) on y axis
Pt C (0,0,3) on z axis
A2 = [0 + 2 + 3]
A2 = 13
B2 = [1 + 0 + 3]
B2 = 10
C2 = [1 + 2 + 0]
C2 = 5
 
a)2A2C2 = 13B2
⇒ 2(13)(5) = 13(10)
⇒ 130 = 130 {true}
 
b) A2 = 2C2
⇒ (13) = 2(5)
⇒ 13 = 10 {false}
 
c)B2 = 3C2
⇒ 10 = 3(5)
⇒ 10 = 15 {false}
 
d)A2 = B2 + C2
⇒ (13) = 10 + 5
⇒ 13 = 15 {false}

Test: Distance Formula 3D Geometry - Question 4

A and B be the points (3, 4, 5) and (-1, -3, -7), respectively, the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant will

Test: Distance Formula 3D Geometry - Question 5

A(4,7,8) B(2,3,4) , C (-1,-2,1) and D(1,2,5) are vertices of a quadrilateral. The quadrilateral is a

Detailed Solution for Test: Distance Formula 3D Geometry - Question 5

AB =  [(2−4)2 +(3−2)+(4−8)2]1/2
 AB=  [(−2)2 + (1)2 + (−4)2]^1/2
 AB =  (21)1/2
Similarly you find that BC=  (43)1/2
CD= (33)1/2  and DA= (43)1/2 
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC=  [(-1-4)2 + (−2-7)2 + (1−8)2]1/2
AC=  (155)1/2
similarly BD=  (3)1/2
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through  CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
Hence it is parallelogram.

Test: Distance Formula 3D Geometry - Question 6

Find the points on z-axis which are at a distance  from the point (1, 2, 3).

Detailed Solution for Test: Distance Formula 3D Geometry - Question 6

Let the point on Z axis be given as (0,0,z).  The distance between (1,2,3) and (0,0,z) is given as [(1)2 + (2)2 + (3-z)2]½ = (21)1/2
5+(3−z)2=21
z2−6z−7=0
z=7,z = −1
Hence points are (0,0,7),(0,0,−1)

Test: Distance Formula 3D Geometry - Question 7

The distance of the point (3, 4, 5) from X-axis is:

Test: Distance Formula 3D Geometry - Question 8

The distance between the points P(x1, y1, z1) and Q (x2, y2, z2) is given by:

Detailed Solution for Test: Distance Formula 3D Geometry - Question 8





 

Test: Distance Formula 3D Geometry - Question 9

The points (1, -1, 3), (2, -4, 5) and (5, -13, 11) are:

Test: Distance Formula 3D Geometry - Question 10

If P,Q,R be the mid points of sides AB, AC and BC of triangle ABC. Where , A (0, 0, 6), B (0,4, 0)and (6, 0, 0).Then

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