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Introduction To Three Dimensional Geometry

Class 11th Mathematics CBSE Solution
Exercise 12.1
  1. A point is on the x-axis. What are its y coordinate and z-coordinates?…
  2. A point is in the XZ-plane. What can you say about its y-coordinate?…
  3. Name the octants in which the following points lie: (1, 2, 3), (4, -2, 3), (4,…
  4. Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane…
Exercise 12.2
  1. (2, 3, 5) and (4, 3, 1) Find the distance between the following pairs of…
  2. (-3, 7, 2) and (2, 4, -1) Find the distance between the following pairs of…
  3. (-1, 3, - 4) and (1, -3, 4) Find the distance between the following pairs of…
  4. (2, -1, 3) and (-2, 1, 3). Find the distance between the following pairs of…
  5. Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.…
  6. (0, 7, -10), (1, 6, - 6) and (4, 9, - 6) are the vertices of an isosceles…
  7. (0, 7, 10), (-1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled…
  8. (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a…
  9. Find the equation of the set of points which are equidistant from the points…
  10. Find the equation of the set of points P, the sum of whose distances from A (4,…
Exercise 12.3
  1. Find the coordinates of the point which divides the line segment joining the…
  2. Given that P (3, 2, - 4), Q (5, 4, - 6) and R (9, 8, -10) are collinear. Find…
  3. Find the ratio in which the YZ-plane divides the line segment formed by joining…
  4. Using section formula, show that the points A (2, -3, 4), B (-1, 2, 1) and c (0…
  5. Find the coordinates of the points which trisect the line segment joining the…
Miscellaneous Exercise
  1. Three vertices of a parallelogram ABCD are A(3, - 1, 2), B (1, 2, - 4) and C (-…
  2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B…
  3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q…
  4. Find the coordinates of a point on y-axis which are at a distance of 5√2 from…
  5. A point R with x-coordinate 4 lies on the line segment joining the points P(2,…
  6. If A and B be the points (3, 4, 5) and (-1, 3, -7), respectively, find the…

Exercise 12.1
Question 1.

A point is on the x-axis. What are its y coordinate and z-coordinates?


Answer:

If a point is on the x-axis, then the coordinates of y and z are 0.


So the point is (x, 0, 0).



Question 2.

A point is in the XZ-plane. What can you say about its y-coordinate?


Answer:

If a point is in XZ plane, then its y-coordinate is 0.



Question 3.

Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, – 4, –7).


Answer:


(i) (1, 2, 3)


Here x is positive, y is positive and z is positive.


So it lies in I octant.


(ii) (4, -2, 3)


Here x is positive, y is negative and z is positive.


So it lies in IV octant.


(iii) (4, -2, -5)


Here x is positive, y is negative and z is negative.


So it lies in VIII octant.


(iv) (4, 2, -5)


Here x is positive, y is positive and z is negative.


So it lies in V octant.


(v) (-4, 2, -5)


Here x is negative, y is positive and z is negative.


So it lies in VI octant.


(vi) (-4, 2, 5)


Here x is negative, y is positive and z is positive.


So it lies in II octant.


(vii) (-3, -1, 6)


Here x is negative, y is negative and z is positive.


So it lies in III octant.


(viii) (2, -4, -7)


Here x is positive, y is negative and z is negative.


So it lies in VIII octant.



Question 4.

Fill in the blanks:

(i) The x-axis and y-axis taken together determine a plane known as _______.

(ii) The coordinates of points in the XY-plane are of the form _______.

(iii) Coordinate planes divide the space into ______ octants.


Answer:

(i) The x-axis and y-axis taken together determine a plane known as XY Plane.


(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).


(iii) Coordinate planes divide the space into eight octants.




Exercise 12.2
Question 1.

Find the distance between the following pairs of points:

(2, 3, 5) and (4, 3, 1)


Answer:

Let P be (2, 3, 5) and Q be (4, 3, 1)


Distance PQ


Here,


x1 = 2, y1 = 3, z1 = 5


x2 = 4, y2 = 3, z2 = 1


Distance PQ






Thus, the required distance is units.



Question 2.

Find the distance between the following pairs of points:

(–3, 7, 2) and (2, 4, –1)


Answer:

Let P be (– 3, 7, 2) and Q be (2, 4, – 1)


Distance PQ


Here,


x1 = – 3, y1 = 7, z1 = 2


x2 = 2, y2 = 4, z2 = – 1


Distance PQ





Thus, the required distance is units.



Question 3.

Find the distance between the following pairs of points:

(–1, 3, – 4) and (1, –3, 4)


Answer:

Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)


Distance PQ


Here,


x1 = – 1, y1 = 3, z1 = – 4


x2 = 1, y2 = – 3, z2 = 4


Distance PQ






Thus, the required distance is units.



Question 4.

Find the distance between the following pairs of points:

(2, –1, 3) and (–2, 1, 3).


Answer:

Let P be (2, – 1, 3) and Q be (– 2, 1, 3)


Distance PQ


Here,


x1 = 2, y1 = – 1, z1 = 3


x2 = – 2, y2 = 1, z2 = 3


Distance PQ






Thus, the required distance is units.



Question 5.

Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.


Answer:

If three points are collinear, then they lie on a line.


Let first calculate distance between the 3 points


i.e. PQ, QR and PR


Calculating PQ


P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)


Distance PQ


Here,


x1 = – 2, y1 = 3, z1 = 5


x2 = 1, y2 = 2, z2 = 3


Distance PQ





Calculating QR


Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)


Distance QR


Here,


x1 = 1, y1 = 2, z1 = 3


x2 = 7, y2 = 0, z2 = – 1


Distance QR






Calculating PR


P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)


Distance PR


Here,


x1 = – 2, y1 = 3, z1 = 5


x2 = 7, y2 = 0, z2 = – 1


Distance PR






Thus, PQ = , QR = & PR =


So, PQ + QR = + = = PR


Thus, Points P, Q and R are collinear.



Question 6.

Verify the following:

(0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.


Answer:

Let points be


P(0, 7, –10), Q(1, 6, – 6) & R(4, 9, – 6)


If any 2 sides are equal, it will be an isosceles triangle


Calculating PQ


P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)


Distance PQ


Here,


x1 = 0, y1 = 7, z1 = – 10


x2 = 1, y2 = 6, z2 = – 6


Distance PQ





Calculating QR


Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)


Distance QR


Here,


x1 = 1, y1 = 6, z1 = – 6


x2 = 4, y2 = 9, z2 = – 6


Distance QR





Since PQ = QR


∴ 2 sides are equal


Hence PQR is an isosceles triangle.



Question 7.

Verify the following:

(0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.


Answer:

Let points be


P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)


Calculating PQ


P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)


Distance PQ


Here,


x1 = 0, y1 = 7, z1 = 10


x2 = – 1, y2 = 6, z2 = 6


Distance PQ





Calculating QR


Q ≡ (– 1, 6, 6) and R ≡ (– 4, 9, 6)


Distance QR


Here,


x1 = – 1, y1 = 6, z1 = 6


x2 = – 4, y2 = 9, z2 = 6


Distance QR





Calculating PR


P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)


Distance PR


Here,


x1 = 0, y1 = 7, z1 = 10


x2 = – 4, y2 = 9, z2 = 6


Distance PR





Now,


PQ2 + QR2 = 18 + 18 = 36 = PR2


Hence, According to converse of pythagoras theorem,


the given vertices P, Q & R are the vertices of a right – angled triangle at Q.



Question 8.

Verify the following:

(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.


Answer:

Let A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)


ABCD can be vertices of parallelogram only if opposite sides are equal.


i.e. AB = CD & BC = AD


Calculating AB


A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)


Distance AB


Here,


x1 = – 1, y1 = 2, z1 = 1


x2 = 1, y2 = – 2, z2 = 5


Distance AB






Calculating BC


B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)


Distance BC


Here,


x1 = 1, y1 = – 2, z1 = 5


x2 = 4, y2 = – 7, z2 = 8


Distance BC





Calculating CD


C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)


Distance CD


Here,


x1 = 4, y1 = – 7, z1 = 8


x2 = 2, y2 = – 3, z2 = 4


Distance CD






Calculating DA


D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)


Distance DA


Here,


x1 = 2, y1 = – 3, z1 = 4


x2 = – 1, y2 = 2, z2 = 1


Distance DA





Since AB = CD & BC = DA


So, In ABCD both pairs of opposite sides are equal.


Thus, ABCD is a parallelogram.



Question 9.

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).


Answer:

Let A (1, 2, 3) & B (3, 2, – 1)


Let point P be (x, y, z)


Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1)


i.e. PA = PB


Calculating PA


P ≡ (x, y, z) and A ≡ (1, 2, 3)


Distance PA


Here,


x1 = x, y1 = y, z1 = z


x2 = 1, y2 = 2, z2 = 3


Distance PA


Calculating PB


P ≡ (x, y, z) and B ≡ (3, 2, – 1)


Distance PB


Here,


x1 = x, y1 = y, z1 = z


x2 = 3, y2 = 2, z2 = – 1



Since PA = PB


Squaring both sides, we get –


PA2 = PB2


⇒ (1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2


⇒ (1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)


= (9 + x2 – 6x) + (4 + y2 - 4y) + (1 + z2 + 2z)


⇒ – 2x – 4y – 6z + 14 = – 6x - 4y + 2z + 14


⇒ 4x – 8z = 0


⇒ x – 2z = 0


This is the required equation.


Question 10.

Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.


Answer:

Let A (4, 0, 0) & B (– 4, 0, 0)


Let the coordinates of point P be (x, y, z)


Calculating PA


P ≡ (x, y, z) and A ≡ (4, 0, 0)


Distance PA


Here,


x1 = x, y1 = y, z1 = z


x2 = 4, y2 = 0, z2 = 0


Distance PA


Calculating PB


P ≡ (x, y, z) and B ≡ (– 4, 0, 0)


Distance PB


Here,


x1 = x, y1 = y, z1 = z


x2 = – 4, y2 = 0, z2 = 0


Distance PB


Given that –


PA + PB = 10


⇒ PA = 10 – PB


Squaring both sides, we get –


PA2 = (10 – PB)2


⇒ PA2 = 100 + PB2 – 20 PB


⇒ (4 – x)2 + (0 – y)2 + (0 – z)2


= 100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB


⇒ (16 + x2 – 8x) + (y2) + (z2)


= 100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB


⇒ 20 PB = 16x + 100


⇒ 5 PB = (4x + 25)


Squaring both sides again, we get –


⇒ 25 PB2 = 16x2 + 200x + 625


⇒ 25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625


⇒ 25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625


⇒ 25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625


⇒ 9x2 + 25y2 + 25z2 – 225 = 0


This is the required equation.



Exercise 12.3
Question 1.

Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.


Answer:

Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.


(i) By Section Formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



Comparing this information with the details given in the question, we have


x1 = -2, y1 = 3, z1 = 5; x2 = 1, y2 = -4, z2 = 6 and m = 2, n = 3


So,


The coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally is given by:



Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is .


(ii) By Section Formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m : n is given by:



Comparing this information with the details given in the question, we have


x1 = -2, y1 = 3, z1 = 5; x2 = 1, y2 = -4, z2 = 6 and m = 2, n = 3


So,


The coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 externally is given by:



Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).



Question 2.

Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.


Answer:

Let Q divides PR in the ratio k : 1.

By Section Formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



Comparing this information with the details given in the question, we have


x1 = 3, y1 = 2, z1 = -4; x2 = 9, y2 = 8, z2 = -10 and m = k, n = 1


So, we have




⇒ 9k + 3 = 5 (k+1) ⇒ 9k + 3 = 5k + 5 ⇒ 4k = 2



Hence, the ratio in which Q divides PR is 1 : 2.



Question 3.

Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).


Answer:

Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.

We know that any point on the YZ-plane is of the form (0, y, z).


Now, let R (0, y, z) divides the line segment PQ in the ratio k : 1.


Then,


Comparing this information with the details given in the question, we have


x1 = -2, y1 = 4, z1 = 7; x2 = 3, y2 = -5, z2 = 8 and m = k, n = 1


By section formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



So, we have




⇒ 3k – 2 = 0 ⇒ 3k = 2



Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2 :3.



Question 4.

Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and are collinear.


Answer:

Let the point P divides AB in the ratio k : 1.

Then,


Comparing this information with the details given in the question, we have


x1 = 2, y1 = -3, z1 = 4; x2 = -1, y2 = 2, z2 = 1 and m = k, n = 1


By section formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



So, we have,


The coordinates of P =


Now, we check if for some value of k, the point coincides with the point C.


Put


⇒ -k + 2 = 0 ⇒ k = 2


When k =2, then


And


Therefore, C is a point which divides AB in the ratio 2 : 1 and is same as P.


Hence, A, B, C are collinear.



Question 5.

Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).


Answer:

Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).

⇒ A divides the line segment PQ in the ratio 1 : 2.


Then,


Comparing this information with the details given in the question, we have


x1 = 4, y1 = 2, z1 = -6; x2 = 10, y2 = -16, z2 = 6 and m = 1, n = 2


By section formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



So, we have


The coordinates of A =


Similarly, We know that B divides the line segment PQ in the ratio 2 : 1.


Then,


Comparing this information with the details given in the question, we have


x1 = 4, y1 = 2, z1 = -6; x2 = 10, y2 = -16, z2 = 6 and m = 2, n = 1


By section formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



So, we have


The coordinates of B =


Hence, the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).




Miscellaneous Exercise
Question 1.

Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.


Answer:

Given: ABCD is a parallelogram, with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2).

⇒ x1 = 3, y1 = -1, z1 = 2; x2 = 1, y2 = 2, z2 = -4; x3 = -1, y3 = 1, z3 = 2



Let the coordinates of the fourth vertex be D (x, y, z).


We know that the diagonals of a parallelogram bisect each other, so the mid points of AC and BD are equal, i.e. Midpoint of AC = Midpoint of BD ……….(i)


Now, by Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are .


So, we have


Coordinates of the midpoint of AC


Coordinates of the midpoint of BD


So, using (i), we have




⇒ 1 + x = 2, 2 + y = 0, -4 + z = 4


⇒ x = 1, y = -2, z = 8


Hence, the coordinates of the fourth vertex is D (1, -2, 8).



Question 2.

Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).


Answer:

Given: The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

⇒ x1 = 0, y1 = 0, z1 = 6; x2 = 0, y2 = 4, z2 = 0; x3 = 6, y3 = 0, z3 = 0



We know that the median is a line segment through a vertex of a triangle to the midpoint of the side opposite to the vertex.


So, let the medians of this triangle be AD, BE and CF corresponding to the vertices A, B and C respectively.


⇒ D, E and F are the midpoints of the sides BC, AC and AB respectively.


By Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are .


So, we have


The coordinates of D = (3, 2, 0)


The coordinates of E = (3, 0, 3)


And the coordinates of F = (0, 2, 3)


By Distance Formula, we know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by .


The lengths of the medians are:





So, the lengths of the medians of the given triangle are 7, and 7.



Question 3.

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.


Answer:

Given: The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

⇒ x1 = 2a, y1 = 2, z1 = 6; x2 = -4, y2 = 3b, z2 = -10; x3 = 8, y3 = 14, z3 = 2c


We know that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), are .


So, the coordinates of the centroid of the triangle PQR are



Now, it is given that the origin (0, 0, 0) is the centroid.


So, we have



⇒ 2a +4 = 0, 3b + 16 = 0, 2c – 4 = 0




Question 4.

Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3, –2, 5).


Answer:

Let the point on y-axis be A (0, y, 0).

Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is .


Now, by Distance Formula, we know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by .


So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by





Squaring both sides, we get


(-2 -y)2 + 34 = 25 × 2


⇒ (-2 -y)2 = 50 – 34


⇒ 4 + y2 + (2 × -2 × -y) = 16


⇒ y2 + 4y + 4 -16 = 0


⇒ y2 + 4y – 12 = 0


⇒ y2 + 6y – 2y – 12 = 0


⇒ y (y + 6) – 2 (y + 6) = 0


⇒ (y + 6) (y - 2) = 0


⇒ y = -6, y = 2


So, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.



Question 5.

A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

[Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by



Answer:

Given: The coordinates of the points P (2, -3, 4) and Q (8, 0, 10).

x1 = 2, y1 = -3, z1 = 4; x2 = 8, y2 = 0, z2 = 10


Let the coordinates of the required point be (4, y, z).


Now, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k : 1.


Also, by Section Formula,


We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n is given by:



So, the coordinates of the point R are given by


So, we have



⇒ 8k + 2 = 4 (k + 1)


⇒ 8k + 2 = 4k + 4


⇒ 8k – 4k = 4 – 2


⇒ 4k = 2




Hence, the coordinates of the required point are (4, -2, 6).


Question 6.

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


Answer:

Given: The points A (3, 4, 5) and B (-1, 3, -7)

⇒ x1 = 3, y1 = 4, z1 = 5; x2 = -1, y2 = 3, z2 = -7;


PA2 + PB2 = k2 ……….(i)


Let the point be P (x, y, z).


Now, by Distance Formula, we know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by .


So,


And


Now, substituting these values in (i), we have


[(3 - x)2 + (4 - y)2 + (5 - z)2] + [(-1 - x)2 + (3 - y)2 + (-7 - z)2] = k2


⇒ [(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] + [(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2


⇒ 9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2


⇒ 2x2 + 2y2 + 2z2 - 4x - 14y + 4z + 109 = k2


⇒ 2x2 + 2y2 + 2z2 - 4x - 14y + 4z = k2 – 109


⇒ 2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109



Hence, the required equation is .