`star` Distance between Two Points

`\color{red} ✍️` We have studied about the distance between two points in two-dimensional coordinate system. Let us now extend this study to three-dimensional system.

`\color{red} ✍️` Let `P(x_1, y_1, z_1)` and `Q ( x_2, y_2, z_2)` be two points referred to a system of rectangular axes `OX, OY` and `OZ.`

`\color{red} ✍️` Through the points `P` and `Q` draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig 12.4).

Now, since `∠PAQ` is a right angle, it follows that, in triangle `PAQ,`

`color(blue)(PQ^2 = PA^2 + AQ^2)` ................................ (1)

Also, triangle `ANQ` is right angle triangle with `∠ANQ` a right angle.

Therefore `color(blue)(AQ^2 = AN^2 + NQ^2)` ................................. (2)

From `(1)` and `(2),` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(green)(PQ^2 = PA^2 + AN^2 + NQ^2)`

Now `PA = y_2 – y_1,` ` \ \ AN = x_2 – x_1` and `NQ = z_2 – z_1`

Hence `PQ^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2`

Therefore `color{red}(PQ = sqrt ((x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2))`

This gives us the distance between two points `(x_1, y_1, z_1)` and `(x_2, y_2, z_2).`

`\color{red} ✍️` In particular, if `color(blue)(x_1 = y_1 = z_1 = 0)`, i.e., point `P` is origin `O,`

then `color{orange}(OQ = sqrt( x_2^2 +y_2^2 +z_2^2) ,)` which gives the distance between the origin `O` and any point `Q (x_2, y_2, z_2).`

`\color{red} ✍️` Let `P(x_1, y_1, z_1)` and `Q ( x_2, y_2, z_2)` be two points referred to a system of rectangular axes `OX, OY` and `OZ.`

`\color{red} ✍️` Through the points `P` and `Q` draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig 12.4).

Now, since `∠PAQ` is a right angle, it follows that, in triangle `PAQ,`

`color(blue)(PQ^2 = PA^2 + AQ^2)` ................................ (1)

Also, triangle `ANQ` is right angle triangle with `∠ANQ` a right angle.

Therefore `color(blue)(AQ^2 = AN^2 + NQ^2)` ................................. (2)

From `(1)` and `(2),` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(green)(PQ^2 = PA^2 + AN^2 + NQ^2)`

Now `PA = y_2 – y_1,` ` \ \ AN = x_2 – x_1` and `NQ = z_2 – z_1`

Hence `PQ^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2`

Therefore `color{red}(PQ = sqrt ((x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2))`

This gives us the distance between two points `(x_1, y_1, z_1)` and `(x_2, y_2, z_2).`

`\color{red} ✍️` In particular, if `color(blue)(x_1 = y_1 = z_1 = 0)`, i.e., point `P` is origin `O,`

then `color{orange}(OQ = sqrt( x_2^2 +y_2^2 +z_2^2) ,)` which gives the distance between the origin `O` and any point `Q (x_2, y_2, z_2).`

Q 3059634514

Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2).

The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is

`PQ = sqrt((-4-1)^2+(1+3)^2+(2-4)^2)`

`= sqrt(25+16+4)`

`= sqrt(45) =3 sqrt3` units

Q 3089834717

Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices of a right angled triangle?

By the distance formula, we have

`AB^2 = (10 – 3)^2 + (20 – 6)^2 + (30 – 9)^2`

`= 49 + 196 + 441 = 686`

`BC^2 = (25 – 10)^2 + (– 41 – 20)^2 + (5 – 30)^2`

`= 225 + 3721 + 625 = 4571`

`CA^2 = (3 – 25)^2 + (6 + 41)^2 + (9 – 5)^2`

`= 484 + 2209 + 16 = 2709`

We find that `CA^2 + AB^2 ≠ BC^2.`

Hence, the triangle ABC is not a right angled triangle.

Q 3019034810

Find the equation of set of points P such that `PA^2 + PB^2 = 2k^2`, where A and B are the points `(3, 4, 5)` and `(–1, 3, –7)`, respectively.

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

Here `PA^2 = (x – 3)^2 + (y – 4)^2 + ( z – 5)^2`

`PB^2 = (x + 1)^2 + (y – 3)^2 + (z + 7)^2`

By the given condition `PA^2 + PB^2 = 2k^2`, we have

`(x – 3)^2 + (y – 4)^2 + (z – 5)^2 + (x + 1)^2 + (y – 3)^2 + (z + 7)^2 = 2k^2`

i.e., `2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z = 2k^2 – 109.`