Test: Distance Formula 3D Geometry - JEE MCQ

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.

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}

Let's break down the given problem step by step:

We are given two points:
A = (3, 4, 5)
B = (-1, -3, -7)

We need to find the equation of a set of points P = (x, y, z) such that the sum of the squared distances from P to A and from P to B is equal to a constant k squared. That is:

PA² + PB² = k²

Step 1: Use the distance formula

The square of the distance from point P to point A is:
PA² = (x - 3)² + (y - 4)² + (z - 5)²

The square of the distance from point P to point B is:
PB² = (x + 1)² + (y + 3)² + (z + 7)²

Step 2: Combine both expressions

Add both squared distances together:

PA² + PB² = [(x - 3)² + (y - 4)² + (z - 5)²] + [(x + 1)² + (y + 3)² + (z + 7)²]

Step 3: Expand and simplify

Now expand all the squared terms:

(x - 3)² = x² - 6x + 9
(y - 4)² = y² - 8y + 16
(z - 5)² = z² - 10z + 25

(x + 1)² = x² + 2x + 1
(y + 3)² = y² + 6y + 9
(z + 7)² = z² + 14z + 49

Now combine all the terms:

PA² + PB² =
x² - 6x + 9 + y² - 8y + 16 + z² - 10z + 25

• x² + 2x + 1 + y² + 6y + 9 + z² + 14z + 49

Now group like terms:

x² + x² = 2x²
y² + y² = 2y²
z² + z² = 2z²

-6x + 2x = -4x
-8y + 6y = -2y
-10z + 14z = 4z

Now add the constants:
9 + 16 + 25 + 1 + 9 + 49 = 109

So the simplified equation is:

2x² + 2y² + 2z² - 4x - 2y + 4z + 109 = k²

Step 4: Final Analysis

You can see that the entire left-hand side of the equation has no terms involving k, and the right-hand side is simply k squared.

That means the equation contains only k to the power 2, and no term with just k or any other powers.

Final Answer:

Option b) Have a term of k to the power 2 only

AB =  [(2−4)² +(3−2)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.

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)