NCERT Solutions Class 11 Maths Chapter 11 – Introduction to Three Dimensional Geometry

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

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

 

If a point is in the XZ plane, then its y-co-ordinate is 0.

 

Here is the table which represents the octants:

(i) (1, 2, 3)

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

So, it lies in the I octant.

(ii) (4, -2, 3)

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

So, it lies in the IV octant.

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

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

So, it lies in the VIII octant.

(iv) (4, 2, -5)

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

So, it lies in the 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 the II octant.

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

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

So, it lies in the III octant.

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

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

So, it lies in the VIII octant.

 

(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.

 

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

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

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 2, y1 = 3, z1 = 5

x2 = 4, y2 = 3, z2 = 1

Distance PQ = √[(4 – 2)2 + (3 – 3)2 + (1 – 5)2]

= √[(2)2 + 02 + (-4)2]

= √[4 + 0 + 16]

= √20

= 2√5

∴ The required distance is 2√5 units.

(ii) (–3, 7, 2) and (2, 4, –1)

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

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 3, y1 = 7, z1 = 2

x2 = 2, y2 = 4, z2 = – 1

Distance PQ = √[(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]

= √[(5)2 + (-3)2 + (-3)2]

= √[25 + 9 + 9]

= √43

∴ The required distance is √43 units.

(iii) (–1, 3, –4) and (1, –3, 4)

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

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 1, y1 = 3, z1 = – 4

x2 = 1, y2 = – 3, z2 = 4

Distance PQ = √[(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]

= √[(2)2 + (-6)2 + (8)2]

= √[4 + 36 + 64]

= √104

= 2√26

∴ The required distance is 2√26 units.

(iv) (2, –1, 3) and (–2, 1, 3)

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

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = – 1, z1 = 3

x2 = – 2, y2 = 1, z2 = 3

Distance PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]

= √[(-4)2 + (2)2 + (0)2]

= √[16 + 4 + 0]

= √20

= 2√5

∴ The required distance is 2√5 units.

 

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

First, let us calculate the distance between the 3 points

i.e., PQ, QR and PR

Calculating PQ

P ≡ (–2, 3, 5) and Q ≡ (1, 2, 3)

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]

= √[(3)2 + (-1)2 + (-2)2]

= √[9 + 1 + 4]

= √14

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

By using the formula,

Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = – 1

Distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]

= √[(6)2 + (-2)2 + (-4)2]

= √[36 + 4 + 16]

= √56

= 2√14

Calculating PR

P ≡ (–2, 3, 5) and R ≡ (7, 0, – 1)

By using the formula,

Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = –2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = –1

Distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]

= √[(9)2 + (-3)2 + (-6)2]

= √[81 + 9 + 36]

= √126

= 3√14

Thus, PQ = √14, QR = 2√14 and PR = 3√14

So, PQ + QR = √14 + 2√14

= 3√14

= PR

∴ The points P, Q and R are collinear.

 

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

Let us consider the points,

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

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

So, first, let us calculate the distance of PQ, QR

Calculating PQ

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here,

x1 = 0, y1 = 7, z1 = – 10

x2 = 1, y2 = 6, z2 = – 6

Distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]

= √[(1)2 + (-1)2 + (4)2]

= √[1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= √[(3)2 + (3)2 + (-6+6)2]

= √[9 + 9 + 0]

= √18

Hence, PQ = QR

18 = 18

2 sides are equal

∴ PQR is an isosceles triangle.

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

Let the points be

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

First, let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

By using the formula,

Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = –1, y2 = 6, z2 = 6

Distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]

= √[(-1)2 + (-1)2 + (-4)2]

= √[1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, –6) and R ≡ (4, 9, –6)

By using the formula,

Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = 6, z1 = –6

x2 = 4, y2 = 9, z2 = –6

Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= √[(3)2 + (3)2 + (-6+6)2]

= √[9 + 9 + 0]

= √18

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

By using the formula,

Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 0, y1 = 7, z1 = 10

x2 = – 4, y2 = 9, z2 = 6

Distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]

= √[(-4)2 + (2)2 + (-4)2]

= √[16 + 4 + 16]

= √36

Now,

PQ2 + QR2 = 18 + 18

= 36

= PR2

By using the converse of Pythagoras theorem,

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

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

Let the points: 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 and BC = AD

First, let us calculate the distance

Calculating AB

A ≡ (–1, 2, 1) and B ≡ (1, –2, 5)

By using the formula,

Distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = – 1, y1 = 2, z1 = 1

x2 = 1, y2 = – 2, z2 = 5

Distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]

= √[(2)2 + (-4)2 + (4)2]

= √[4 + 16 + 16]

= √36

= 6

Calculating BC

B ≡ (1, –2, 5) and C ≡ (4, –7, 8)

By using the formula,

Distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 1, y1 = – 2, z1 = 5

x2 = 4, y2 = – 7, z2 = 8

Distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]

= √[(3)2 + (-5)2 + (3)2]

= √[9 + 25 + 9]

= √43

Calculating CD

C ≡ (4, –7, 8) and D ≡ (2, –3, 4)

By using the formula,

Distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 4, y1 = –7, z1 = 8

x2 = 2, y2 = –3, z2 = 4

Distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]

= √[(-2)2 + (4)2 + (-4)2]

= √[4 + 16 + 16]

= √36

= 6

Calculating DA

D ≡ (2, –3, 4) and A ≡ (–1, 2, 1)

By using the formula,

Distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = 2, y1 = – 3, z1 = 4

x2 = –1, y2 = 2, z2 = 1

Distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]

= √[(-3)2 + (5)2 + (-3)2]

= √[9 + 25 + 9]

= √43

Since AB = CD and BC = DA (given),

In ABCD, both pairs of opposite sides are equal.

∴ ABCD is a parallelogram.

 

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

First, let us calculate

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

By using the formula,

Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Distance PA = √[(1 – x)2 + (2 – y)2 + (3 – z)2]

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, –1)

By using the formula,

Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = –1

Distance PB = √[(3 – x)2 + (2 – y)2 + (-1 – z)2]

Since PA = PB

Square on 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

∴ The required equation is x – 2z = 0.

 

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)

By using the formula,

Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So, here

x1 = x, y1 = y, z1 = z

x2 = 4, y2 = 0, z2 = 0

Distance PA = √[(4– x)2 + (0 – y)2 + (0 – z)2]

Calculating PB,

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

By using the formula,

Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = – 4, y2 = 0, z2 = 0

Distance PB = √[(-4– x)2 + (0 – y)2 + (0 – z)2]

It is given that,

PA + PB = 10

PA = 10 – PB

Square on 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)

Square on 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

∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0.

 

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

(i) 2:3 internally

By using the 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:

Upon comparing, 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 divide 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 (-4/5, 1/5, 27/5).

(ii) 2:3 externally

By using the 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:

Upon comparing, 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 divide the line segment joining the points P (–2, 3, 5) and Q (1, –4, 6) in the ratio 2:3 externally is given by:

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

 

Let us consider Q divides PR in the ratio k:1.

By using the 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:

Upon comparing, 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

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

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

 

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).

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

Then,

Upon comparing, we have,

x1 = -2, y1 = 4, z1 = 7;

x2 = 3, y2 = -5, z2 = 8 and

m = k, n = 1

By using the 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

k = 2/3

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.

 

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

Upon comparing, we have,

x1 = 2, y1 = -3, z1 = 4;

x2 = -1, y2 = 2, z2 = 1 and

m = k, n = 1

By using the 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,

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

Put (-k+2)/(k+1) = 0

-k + 2 = 0

k = 2

When k = 2, then (2k-3)/(k+1) = (2(2)-3)/(2+1)

= (4-3)/3

= 1/3

And, (k+4)/(k+1) = (2+4)/(2+1)

= 6/3

= 2

∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2:1 and is the same as P.

Hence, A, B, and C are collinear.

 

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.

Upon comparing, we have,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 1, n = 2

By using the 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

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

Upon comparing, we have,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 2, n = 1

By using the 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 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).

 

Given:

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

Where, 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 midpoints of AC and BD are equal, i.e., Midpoint of AC = Midpoint of BD ……….(1)

Now, by the 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 [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So we have,

= (2/2, 0/2, 4/2)

= (1, 0, 2)

1 + x = 2, 2 + y = 0, -4 + z = 4

x = 1, y = -2, z = 8

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

 

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

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 the midpoint formula, the coordinates of the midpoint of the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) are [(x1+x2)/2, (y1+y2)/2, (z1+z2)/2]

So, we have

By the distance formula, the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

∴ The lengths of the medians of the given triangle are 7, √34 and 7.

 

Given:

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

Where,

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 [(x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3]

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

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0

a = -2, b = -16/3, c = 2

∴ The values of a, b and c are a = -2, b = -16/3, and c = 2.

 

Let the point on the 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 5√2.

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

Distance of PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

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

Distance of AP = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

= √[(3-0)2 + (-2-y)2 + (5-0)2]

= √[32 + (-2-y)2 + 52]

= √[(-2-y)2 + 9 + 25]

5√2 = √[(-2-y)2 + 34]

Squaring on 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

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

 

Given:

The coordinates of the points are 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).

So, 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.

By using the 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

k = 2/4

= 1/2

Now, let us substitute the values, and we get

= 6

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

 

Now, substituting these values in (1), 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

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

Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z) = (k2 – 109)/2.Given:

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 ……….(1)

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

Now, by using the distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

So,