**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)**

Solution:

The distance between points P(x_{1}, y_{1}, z_{1}) and P(x_{2}, y_{2}, z_{2}) is given by

PQ = √[(x_{2} – x_{1})^{2}+(y_{2} – y_{1})^{2}+(z_{2} – z_{1})^{2}]

(i) Distance between points (2, 3, 5 ) and (4, 3, 1)

= √[(4 – 2)^{2}+(3 – 3)^{2}+(1 – 5)^{2}]

= √[(2)^{2}+(0)^{2}+(4)^{2}]

= √[4+16]

= 2√5

(ii) Distance between points (-3, 7, 2 ) and (2, 4, -1)

= √[(2+3)^{2}+(4 – 7)^{2}+(-1 – 2)^{2}]

= √[(5)^{2}+(-3)^{2}+(-3)^{2}]

= √[25+9+9]

= √43

(iii) Distance between points (2-1, 3, -4) and (1, -3, 4)

= √[(1+1)^{2}+(-3 – 3)^{2}+(4+4)^{2}]

= √[(2)^{2}+(-6)^{2}+(8)^{2}]

= √[4+36+64]

= √104

= 2√26

(iv) Distance between points (2, -1,3) and (-2, 1, 3)

= √[(-2 – 2)^{2}+(1+1)^{2}+(3 – 3)^{2}]

= √[(-4)^{2}+(2)^{2}+(0)^{2}]

= √[16+4]

= √20

= 2√5

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

Solution:

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively. Points P, Q, and R are collinear if they lie on a line.

PQ = √[(1+2)^{2}+(2 – 3)^{2}+(3 – 5)^{2}]

= √[(3)^{2}+(-2)^{2}+(-2)^{2}]

= √(9+1+4)

= √14

QR = √[(7 – 1)^{2}+(0 – 3)^{2}+(-1 – 5)^{2}]

= √[(6)^{2}+(-2)^{2}+(-4)^{2}]

= √(36+4+16)

= √56

= 2√14

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

= √[(9)^{2}+(-3)^{2}+(-6)^{2}]

= √(81+9+36)

= √126

= 3√14

Here, PQ + QR = PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

**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.**

Solution:

(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.

AB = √[(7 – 1)^{2}+(0 – 3)^{2}+(- 6 + 10)^{2}]

= √[(1)^{2}+(-1)^{2}+(4)^{2}]

= √(1+1+16)

= √18

= 3√2

BC = √[(4 – 1)^{2}+(9 – 6)^{2}+(- 6 + 6)^{2}]

= √[(3)^{2}+(3)^{2}]

= √(9+9)

= √18

= 3√2

CA = √[(0 – 4)^{2}+(7 – 9)^{2}+(- 10 + 6^{2}]

= √[(-4)^{2}+(-2)^{2}+(-4)^{2}]

= √(16+4+16)

= √36

= 6

Here AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

(i) Let points (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) be denoted by A, B, and C respectively.

AB = √[(-1 – 0)^{2}+(6 – 7)^{2}+(6 + 10)^{2}]

= √[(-1)^{2}+(-1)^{2}+(-4)^{2}]

= √(1+1+16)

= √18

= 3√2

BC = √[(-4 + 1)^{2}+(9 – 6)^{2}+(6 – 6)^{2}]

= √[(-3)^{2}+(3)^{2}+(0)^{2}]

= √(9+9)

= √18

= 3√2

CA = √[(0 + 4)^{2}+(7 – 9)^{2}+(10 – 6^{2}]

= √[(-4)^{2}+(-2)^{2}+(4)^{2}]

= √(16+4+16)

= √36

= 6

Here AB^{2} + BC^{2} = (3√2)^{2}+(3√2)^{2} = 18 + 18 = 36 = AC^{2}

Thus, the given points are the vertices of an isosceles triangle.

Hence, the given points are the vertices of a right-angled triangle.

(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.

AB = √[(1+1)^{2}+(-2-2)^{2}+(5-1)^{2}]

= √[(2)^{2}+(-4)^{2}+(4)^{2}]

= √(4+16+16)

= √36

= 6

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

= √(9+25+9)

= √43

CD = √[(2 – 4)^{2}+(-3 + 7)^{2}+(4 – 8)^{2}]

= √(9+25+9)

= √43

Here AB = CD = 6; BC= AD = √43

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Thus, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

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

Solution:

Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).

Therefore, PA = PB

PA^{2} = PB^{2}

( x – 1)^{2} + ( y – 2)^{2} + ( z – 3)^{2} = ( x – 3)^{2} + ( y – 2)^{2} + ( z + 1)^{2}

x^{2} – 2x + 1 + y^{2 }– 4y + 4 + z^{2} – 6z + 9 = x^{2} – 6x + 9 + y^{2 }– 4y + 4 + z^{2} + 2z + 1

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

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

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

**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.**

Solution:

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

The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.

It is given that PA + PB = 10.

√[(x – 4)^{2} + y^{2} + z^{2}] +√[(x + 4)^{2} + y^{2} + z^{2}] = 10

√[(x – 4)^{2} + y^{2} + z^{2}] = 10 – √[(x + 4)^{2} + y^{2} + z^{2}]

Squaring on both the sides, we obtain

{√[(x – 4)^{2} + y^{2} + z^{2}]}^{2}= {10 – √[(x + 4)^{2} + y^{2} + z^{2}] }^{2}

(x – 4)^{2} + y^{2} + z^{2} = 10^{2} – 2x10x(√[(x + 4)^{2} + y^{2} + z^{2})+ (√[(x + 4)^{2} + y^{2} + z^{2})^{2}

(x – 4)^{2} + y^{2} + z^{2} = 100 – 20(√[x^{2} + 8x + 16 + y^{2} + z^{2}]+ [(x + 4)^{2} + y^{2} + z^{2}]

20(√[x^{2} + 8x + 16 + y^{2} + z^{2}] = 100 + x^{2} + 8x + 16 + y^{2} + z^{2} – x^{2} + 8x – 16 – y^{2} – z^{2}

20(√[x^{2} + 8x + 16 + y^{2} + z^{2}]) = 100 + 16x

5(√[x^{2} + 8x + 16 + y^{2} + z^{2}]) = (25 + 4x)

Again squaring on both the sides

{5(√[x^{2} + 8x + 16 + y^{2} + z^{2}])}^{2} = {(25 + 4x)}^{2}

25(x^{2} + 8x + 16 + y^{2} + z^{2}) = 625 + 200x + 16x^{2}

25x^{2} + 200x + 400 + 25y^{2} + 25z^{2} = 625 + 200x + 16x^{2}

9x^{2} + 25y^{2} + 25y^{2} – 225 = 0

Thus the required equation is 9x^{2} + 25y^{2} + 25y^{2} – 225 = 0