NCERT Solutions for Class 11 Maths Chapter 11 - Introduction to Three Dimensional Geometry

Exercise 12.1

1. A point is one the X-axis. What are its y-coordinate and z-coordinate?

Sol. Coordinate of any point on the X-axis is (x, 0, 0) so its y and z-coordinates are 0.

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

Sol. When point on theXZ-plane will have the coordinate (x, 0, z), then its y-coordinate is 0.

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).

Sol.

Point	Octant	Name
(1, 2, 3)	I (all the three coordinates are positive)	XOYZ
(4, – 2, 3)	IV (x and z are positve, y-coordinate in negative)	XOY′Z
(4, – 2, – 5)	VIII (y and z-coordinate are negative, x is positive)	XOY′Z′
(4, 2, – 5)	V (x and y are positvie, z-coordinate is negative)	XOYZ′
(– 4, 2, – 5)	VI (x and z-coordinates are negative, y is positive)	X′OYZ′
(– 4, 2, 5)	II (x-coordinate is negative, y and z are positive)	X′OYZ
(– 3, – 1, 6)	III (x and y are negative, z is positive)	X′OY′Z
(2, – 4, – 7)	VIII (y and z-coordinates are negative, x is positive)	XOY′Z′

4. Fill in the blanks:

(i) The X-axis and Y-axis taken together determine a plane known as ............... .

(ii) The coordinates of a point in the XY-plane are of the form ............... .

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

Sol. (i) XY-plane

(ii) (x, y, 0)

(iii) eight

Exercise 12.2

1. Find the distance between the following pairs of points:

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

(ii) (– 3, 7, 2) and (2, 4, – 1)

(iii) (– 1, 3, – 4) and (1, – 3, 4)

(iv) (2, – 1, 3) and (– 2, 1, 3)

Sol. (i) Let the given points are P(2, 3, 5) and Q(4, 3, 1).

∴ x₁ = 2, y₁ = 3, z₁ = 5

x₂ = 4, y₂ = 3, z₂ = 1

∴ Required distance

$$\text{PQ}=\sqrt{(x_2-x_1)^2+(y_2=y_1)^2+(z_2-z_1)^2}\\\Rarr\space\text{PQ}=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}\\=\sqrt{2^2+0^2+(-4)^2}\\=\sqrt{4+0+16}\\=\sqrt{20}\\=2\sqrt{5}.$$

(ii) Let the given points are A(– 3, 7, 2) and
B(2, 4, – 1).

Here, x₁ = – 3, y₁ = 7, z₁ = 2

x₂ = 2, y₂ = 4, z₂ = – 1

∴ Required distance

$$\text{AB}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\\=\sqrt{[2-(-3)^2]+(4-7)^2+(-1-2)^2}\\=\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}\\=\sqrt{5^2+(-3)^2+(-3)^2}\\=\sqrt{25+9+9}\\=\sqrt{43}.$$

(iii) Let the given points are P(– 1, 3, – 4) and Q(1, – 3, 4).

∴ Required distance

$$=\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}\\=\sqrt{22+(-6)^2+8^2}\\=\sqrt{4+36+64}\\=\sqrt{104}=2\sqrt{26}.$$

(iv) Let the given points are A(2, – 1, 3) and B(– 2, 1, 3).

∴ Required distance

$$\text{AB}=\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}\\=\sqrt{(-4)^2+2^2+0}\\=\sqrt{16+4}\\=\sqrt{20}\\=2\sqrt{5}.$$

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

Sol. Let the given points are

P(– 2, 3, 5); Q(1, 2, 3); R(7, 0, – 1)

Distance between P and Q

$$\text{PQ}=\sqrt{(-2-1)^2+(3-2)^2+(5-3)^2}\\=\sqrt{(-3)^2+(1)^2+(2)^2}\\=\sqrt{9+1+4}=\sqrt{14}$$

Distance between Q and R

$$\text{QR}=\sqrt{(1-7)^2+(2-0)^2+(3+1)^2}\\=\sqrt{(-6)^2+(2)^2+(4)^2}\\=\sqrt{36+4+16}=\sqrt{56}\\=2\sqrt{14}$$

and distance between P and R

$$\text{PR}=\sqrt{(-2-7)^2+(3-0)^2+(5+1)^2}\\=\sqrt{(-9)^2+(3)^2+(6)^2}\\=\sqrt{81+9+36}\\=\sqrt{126}=3\sqrt{14}$$

Clearly,

PQ + QR = PR

Hence, the given points are collinear.

3. Verify the following:

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

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

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

Sol. (i) Let A(0, 7, – 10), B(1, 6, – 6) and C(4, 9, – 6) are the vertices of a triangle. Then

Side AB = distance between points A and B

$$=\sqrt{(0-1)^2+(7-6)^2+(-10+6)^2}\\=\sqrt{(-1)^2+(1)^2+(-4)^2}\\=\sqrt{1+1+16}\\=\sqrt{18}=3\sqrt{2}$$

and side BC = distance between points B and C

$$=\sqrt{(1-4)^2+(6-9)^2+(-6+6)^2}\\=\sqrt{(-3)^2+(-3)^2+0^2}\\=\sqrt{9+9+0}\\=\sqrt{18}=3\sqrt{2}$$

Clearly, AB = BC. Hence, triangle is an isosceles triangle.

(ii) Let A(0, 7, 10), B(– 1, 6, 6) and C(– 4, 9, 6) and C(– 4, 9, 6) are the vertices of a triangle.

Then,

$$\text{Side AB}=\\\sqrt{(0+1)^2+(7-6)^2+(10-6)^2}\\=\sqrt{1^2+1^2+4^2}\\\Rarr\space\text{AB}=\sqrt{1+1+16}\\=\sqrt{18}=3\sqrt{2}\\\text{Side BC}=\\ \sqrt{(-1+4)^2+(6-9)^2+(6-6)^2}\\=\sqrt{3^2+(-3)^2+0}\\\Rarr\space\text{BC}=\sqrt{9+9+0}=\sqrt{18}=3\sqrt{2}\\\text{and side CA}\\=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}\\=\sqrt{(-4)^2+(2)^2+(-4)^2}\\\Rarr\space\text{CA}=\sqrt{16+4+16}$$

$$=\sqrt{36}=6$$

Now, AB² + BC²

$$=(3\sqrt{2})^2+(3\sqrt{2})^2=6^2=\text{CA}^2$$

∴ ΔABC is right angled triangle at B.

(iii) Let P(– 1, 2, 1), Q(1, – 2, 5), R(4, – 7, 8) and S(2, – 3, 4) are the vertices of a quadrilatearl PQRS.

Introduction to Three Dimensional Geometry_12.2_3.1

Then, mid-point of PR

$$=\bigg(\frac{-1+4}{2},\frac{2-7}{2},\frac{1+8}{2}\bigg)\\=\bigg(\frac{3}{2},\frac{-5}{2},\frac{9}{2}\bigg)\\\text{Mid-point of QS}\\=\bigg(\frac{1+2}{2},\frac{-2-3}{2},\frac{5+4}{2}\bigg)\\=\bigg(\frac{3}{2},\frac{-5}{2},\frac{9}{2}\bigg)$$

Mid-points of both the diagonals are same (i.e., they bisect each other).

Hence, PQRS is a parallelogram.

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

Sol. Let the given points are A and B. Let P(x, y, z) be any point equidistant from A and B.

∴ PA = PB,

Introduction to Three Dimensional Geometry_12.3_ans 4

$$\Rarr\space\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}\\=\sqrt{(x-3)^2+(y-2)^2+(z+1)^2}$$

⇒ (x – 1)² + (y – 2)² + (z – 3)² [∵ PA² = PB²]

= (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

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

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8x = 0

⇒ x – 2z = 0

Thus, the required equation is x – 2z = 0.

5. 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.

Sol. Let the coordinates be P(x, y, z) then it is given

PA + PB = 10

$$\Rarr\space\sqrt{(x-4)^2+(y-0)^2+(z-0)^2}\\+\sqrt{(x+4)^2+(y-0)^2+(z-0)^2}=10\\\Rarr\space\sqrt{(x-4)^2+y^2+z^2}\\=10-\sqrt{(x+4)^2+y^2+z^2}$$

Squaring on both sides, we get

$$(x-4)^2+y^2+z^2=\\100+(x+4)^2+y^2+z^2\\-20\sqrt{(x+4)^2+y^2+z^2}$$

$$\Rarr\space x^2+16 -8x\\= 100+x^2+16+8x\\- 20\sqrt{(x+4)^2+y^2+z^2}\\\Rarr\space-8x-8x-100=\\-20\sqrt{(x+4)^2+y^2+z^2}\\\Rarr\space -8x-8x-100\\-20\sqrt{(x+4)^2+y^2+z^2}\\\Rarr\space-16x-100=-20\sqrt{(x+4)^2+y^2+z^2}$$

$$16x + 100=20\sqrt{x^2+8x+16+y^2+z^2}\\\Rarr\space 4x+25=5\sqrt{(x^2+8x+16)+y^2+z^2}$$

Again squaring both sides, we get

(4x + 25)² = 25[(x² + 8x + 16) + y² + z²]

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

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

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

⇒ 9x² + 25y² + 25z² – 225 = 0

which is the required equation.

Exercise 12.3

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

Sol. The given points are A(– 2, 3, 5) and B(1, – 4, 6).

(i) The given points are A(– 2, 3, 5) and B(1, – 4, 6).

Introduction to Three Dimensional Geometry_12.3_ans1

Here, the ratio is 2 : 3.

∴ m = 2, n = 3

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

$$=\bigg[\bigg(\frac{m_1x_2+nx_1}{m+n}\bigg),\bigg(\frac{m_1y_2+ny_1}{m+n}\bigg),\\\bigg(\frac{m_1z_2+nz_1}{m+n}\bigg)\bigg]\\\Rarr\space\text{C}=\bigg(\frac{2×(+1)+3×(-2)}{(2+3)},\\\frac{2×(-4)+3×3}{(2+3)},\\\frac{2×6+3×5}{(2+3)}\bigg)\\=\bigg(\frac{2-6}{5},\frac{-8+9}{5},\frac{12+15}{5}\bigg)\\=\space\bigg(\frac{-4}{5},\frac{1}{5},\frac{27}{5}\bigg)$$

$$\Rarr\space x=\frac{-4}{5},y=\frac{1}{5},z=\frac{27}{5}$$

(ii) Let the point C divides the line externally in the ratio 2 : 3.

Introduction to Three Dimensional Geometry_12.3_ii

Hence the ratio is 2 : 3.

∴ m = 2, n = 3

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

$$\bigg[\bigg(\frac{mx_2-nx_1}{m-n}\bigg),\bigg(\frac{my_2-ny_1}{m-n},\bigg)\\\bigg(\frac{mz_2-nz_1}{m-n}\bigg)\bigg]\\\Rarr\space\text{C}=\bigg(\frac{2×(1)-3×(-2)}{(2-3)},\frac{2×(-4)-3×3}{(2-3)},\\\frac{2×6-3×5}{(2-3)}\bigg)\\=\bigg(\frac{2+6}{(-1)},\frac{-8-9}{(-1)},\frac{12-15}{(=1)}\bigg)\\=(-8,17,3)$$

⇒ x = – 8, y = 17, z = 3.

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.

Sol. Let point Q divides PR is the ratio K : 1.

Introduction to Three Dimensional Geometry_ans2_12.3

Here, the point Q divides the line PR internally, so its coordinates are

$$\bigg[\bigg(\frac{mx_2+nx_1}{m+n}\bigg),\bigg(\frac{my_2+ny_1}{m+n}\bigg),\\\bigg(\frac{mz_2+nz_1}{m+n}\bigg)\bigg]\\\Rarr\space\text{Q}=\bigg(\frac{k×9+1×3}{K+1},\frac{K×8+1×2}{K+1},\\\frac{K(-10)+1×(-4)}{K+1}\bigg)\\=\bigg(\frac{9K+3}{K-1},\frac{8K+2}{K+1},\frac{-10K-4}{K+1}\bigg)$$

But given Q = (5, 4, – 6).

On comparing the corresponding coordinates

$$\therefore\space\frac{9K+3}{K+1}=5,\frac{8K+2}{K+1}=4,\\\frac{-10K-4}{\text{K+1}}=-6$$

⇒ 9K + 3 = 5K + 5, 8K + 2

= 4K + 4, – 10K + 4 = – 6K – 6

⇒ 4K = 2

$$\Rarr\space K=\frac{1}{2}$$

Hence, point Q divides PR internally in the ratio 1 : 2.

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).

Sol. The given points are A(– 2, 4, 7) and B(3, – 5, 8).

Let the point P(0, y, z) in YZ-plane divides AB in the ratio K : 1, then

$$\text{X-coordinate of point P =}\frac{mx_2+nx_1}{m+n}\\\frac{K×3+1×(-2)}{K+1}=0$$

(∵ x-coordinate of P is zero)

⇒ 3K – 2 = 0

$$\Rarr\space K=\frac{2}{3}$$

⇒ K : 1 = 2 : 3

Hence, YZ-plane divides the line segment AB internally in the ratio 2 : 3.

4. Using section formula, show that the points

$$\textbf{A(2, – 3, 4), B(– 1, 2, 1) and}\\\bigg(\textbf{0},\frac{\textbf{1}}{\textbf{3}}\textbf{, 2}\bigg)\space\textbf{are collinear.}$$

Sol. Let P divides AB in the ratio K : 1.

Introduction to Three Dimensional Geometry_ans4_12.3

Now, x-coordinate of P = 0

$$\text{i.e.}\space\frac{K×(-1)+1×(2)}{K+1}=0\\\Rarr\space\frac{-\text{K+2}}{\text{K+1}}=0$$

⇒ – K + 2 = 0

⇒ K = 2

⇒ K : 1 = 2 : 1

$$\text{and y-coordinate of P =}\frac{1}{3}\\\Rarr\space\frac{K×2+1×(-3)}{\text{K}+1}=\frac{1}{3}\\\Rarr\space\frac{2K-3}{\text{K+1}}=\frac{1}{3}$$

⇒ 6K : 9 = K + 1

⇒ 5K = 10

⇒ K = 2

⇒ K : 1 = 2 : 1

and z-coordinate of P = 2

$$\Rarr\space\frac{K×1+1×4}{K+1}=2\\\frac{\text{K+4}}{\text{K+1}}=2$$

⇒ K + 4 = 2K + 2

⇒ K = 2

⇒ K : 1 = 2 : 1

Thus, P divides AB internally in the ratio 2 : 1.

Hence, points A, B and P are collinear.

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

Sol. Let the point R1 trisects the line PQ i.e., it divides the line in the ratio 1 : 2.

$$\Rarr\space \text{R}_1=\bigg(\frac{1×10+2×4}{1+2},\frac{1×(-16)+2×2}{1+2},\\\frac{1×6+2×(-6)}{1+2}\bigg)\\=\bigg(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\bigg)$$

$$=\bigg(\frac{18}{3},\frac{-12}{3},\frac{-6}{3}\bigg)=(6,-4,-2)$$

Again, let the point R₂ divides PQ internally in the ratio 2 : 1. Then

Introduction to Three Dimensional Geometry_12.3_ans5_ii

$$\text{R}_2=\bigg(\frac{2×10+1×4}{2+1},\frac{2×(-16)+1×2}{2+1},\\\frac{2×6+1×(-6)}{1+2}\bigg)\\=\bigg(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\bigg)\\=\bigg(\frac{24}{3},\frac{-30}{3},\frac{6}{3}\bigg)$$

= (8, – 10, 2)

Hence, required points are (6, – 4, – 2) and B(8, –10, 2).

Miscellaneous Exercise

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.

Sol. Let ABCD be a parallelogram and coordinates of point D(x, y, z) and the

In a parallelogram diagonals bisect each other.

∴ mid-point of BD = mid-point of AC

Introduction to Three Dimensional Geometry_ans1_Miscellaneous Exercise

$$\Rarr\space\bigg(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2}\bigg)\\=\bigg(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\bigg)\\\Rarr\space\bigg(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2}\bigg)\\=\bigg(\frac{2}{2},0,\frac{4}{2}\bigg)\\\Rarr\space\bigg(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2}\bigg)=(1,0,2)$$

On comparing the corresponding coordinates, we get

$$\frac{x+1}{2}=1,\frac{y+2}{2}=0,\frac{z-4}{2}=2$$

⇒ x + 1 = 2, y + 2 = 0, z – 4 = 4

⇒ x = 1, y = – 2, z = 8

∴ Coordinates of point D are (1, – 2, 8).

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

Sol. ABC is a triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).

Let points D, E and F are the mid-points of BC, AC and AB, respectively. So AD, Be and CF will be the medians of the triangle.

Introduction to Three Dimensional Geometry_ch-12_2_Miscellaneous Exercise

⇒ Coordinates of points D

$$=\bigg(\frac{0+6}{2},\frac{0+6}{2},\frac{0-0}{2}\bigg)=(3,2,0)\\\text{Coordinates of point E =}\\\bigg(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2}\bigg)=(3,0,3)\\\text{and coordinates of point F =}\\\bigg(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\bigg)=(0,2,3)$$

Now, Length of median AD = Distance between A and D

$$\text{AD}=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}\\=\sqrt{9+4+36}=\sqrt{49}=7\\\text{Similarly, BE}=\\\sqrt{(0-3)^2+(4-0)^2+(0-3)^2}\\=\sqrt{9+16+9}=\sqrt{34}\\\text{and CF=}\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}\\=\sqrt{36+4+9}=\sqrt{49}=7$$

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 the values of a, b and c.

Sol. Centroid of the ΔPQR is

$$\bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\frac{z_1+z_2+z_3}{3}\bigg)\\\text{i.e,\space}\bigg(\frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3}\bigg)$$

Origin is the centroid, i.e., the coordinates of the centroid are (0, 0, 0), then

$$\frac{2a-4+8}{3}=0$$

⇒ 2a + 4 = 0

⇒ a = – 2

$$\text{Also}\space\frac{2+3b+14}{3}=0$$

⇒ 3b + 16 = 0

$$\Rarr\space b=-\frac{16}{3}\\\text{And}\space\frac{2c-4}{3}=0$$

⇒ 2c – 4 = 0

⇒ c = 2

$$\therefore\space a=-2,b=\frac{-16}{3},c=2.$$

4. Find the coordinates of a point on Y-axis which are at a distance

$$\textbf{5}\sqrt{\textbf{2}}\space\textbf{of from the point P(3, – 2, 5).}$$

Sol. Let any point on the Y-axis is A(0, y, 0).

Given, distance between P and A,

$$\text{PA}=5\sqrt{2}\\\Rarr\space\sqrt{(3-0)^2+(-2-y)^2+(5-0)^2}\\=5\sqrt{2}$$

Squaring both sides, we get

⇒ (3 – 0)² + (– 2 – y)² + (5 –0)² = 50 [∵ PA² = 50]

⇒ 9 + 4 + y² + 4y + 25 = 50

⇒ y²+ 4y + 38 – 50 = 0

⇒ y² + 4y – 12 = 0

Factorizing it by splitting i.e., middle term,

⇒ y2 + 6y – 2y – 12 = 0

⇒ y(y + 6) – 2(y + 6) = 0

⇒ (y + 6) (y – 2) = 0

⇒ y = – 6, 2

Hence, coordinates on Y-axis are (0, – 6, 0) or (0, 2, 0).

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

Sol. Let the point R(4, y, z) divides PQ in the ratio K : 1.

Now, x-coordinate of R.

$$\Rarr\space\frac{\text{K×8}+1×2}{\text{K+1}}=4\space\text{(Given)}\\\Rarr\space\frac{\text{8K+2}}{\text{K+1}}=4$$

⇒ 8K + 2 = 4K + 4

⇒ 8K – 4K = 4 – 2

⇒ 4K = 2

⇒ K : 1 = 1 : 2

Hence, the point R divides PQ internally in the ratio 1 : 2.

$$\text{Therefore, y-coordinates of}\space\\\text{R}=\bigg(\frac{1×0+2×(-3)}{1+2}\bigg)\\=\bigg(\frac{-6}{3}\bigg)=-2\\\text{And z-coordinate of}\\\space\text{R}=\bigg(\frac{1×10+2×4}{1+2}\bigg)\\=\bigg(\frac{10+8}{3}\bigg)=\frac{18}{3}=6$$

Hence, coordinates of point R are (4, – 2, 6).

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 (PA)² + (PB)² = K², where K is a constant.

Sol. Let the point P is (x, y, z).

Given, (PA)² + (PB)² = K²

⇒ (x – 3)² + (y – 4)² + (z – 5)² + (x + 1)² + (y – 3)² + (z + 7)² = K²

⇒ x² + 9 – 6x + y² + 16 – 8y + z² + 25 – 10z + x² + 1 + 2x + y² + 9 – 6y + z2 + 49 + 14z = K²

⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z + 109 – K² = 0

⇒ 2(x² + y² + z²) – 4x – 14y + 4z + 109 – K² = 0

⇒ 2(x² + y² + z² – 2x – 7y + 2z) + 109 – k² = 0

which is the required equation.