Test: Distance Formula 3D Geometry


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


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QUESTION: 1

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

Solution:

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.

QUESTION: 2

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

Solution:

QUESTION: 3

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

Solution:

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}

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

Solution:
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

Solution:

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.

QUESTION: 6

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

Solution:

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)

QUESTION: 7

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

Solution:
QUESTION: 8

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

Solution:





 

QUESTION: 9

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

Solution:
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

Solution:

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