Mathematics Section Formula
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Class 11 Chapter 12 - INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Section Formula

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`star` Section Formula

Section Formula

`\color{red} ✍️` In two dimensional geometry, we have learnt how to find the coordinates of a point dividing a line segment in a given ratio internally. Now, we extend this to three dimensional geometry as follows:

`\color{red} ✍️` Let the two given points be `P(x_1, y_1, z_1)` and `Q (x_2, y_2, z_2).`

`\color{red} ✍️` Let the point `R (x, y, z)` `PL, QM` and `RN` perpendicular to the `XY`-plane.

Obviously `PL ∥ RN ∥ QM` and feet of these perpendiculars lie in a `XY`-plane.

The points `L, M` and `N` will lie on a line which is the intersection of the plane containing `PL, RN` and `QM` with the `XY`-plane.

Through the point R draw a line `ST` parallel to the line `LM`. Line `ST` will intersect the line `LP` externally at the point `S` and the line `MQ` at `T`, as shown in Fig 12.5.

`\color{red} ✍️` Also note that quadrilaterals `LNRS` and `NMTR` are parallelograms.

`\color{red} ✍️` The triangles `PSR` and `QTR` are similar. Therefore,

`m/n = (PR)/(QR) = (SP)/(QT) = (SL - PL)/(QM - TM) = (NR - PL)/(QM - NR) = (z-z_1)/(z_2-z)`

This implies `color{fuchsia}(z = ( m z_2+ n z_1)/(m+n))`

`\color{red} ✍️` Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get

`color{fuchsia}(y = ( my_2+n y_1)/(m+n))` and `color{fuchsia}(x = (m x_2+n x_1)/(m+n))`

Hence, the coordinates of the point `R` which divides the line segment joining two points `P (x_1, y_1, z_1)` and `Q (x_2, y_2, z_2)` internally in the ratio `m : n` are

`color{red}((( mx_2- nx_1)/(m-n) , (my_2-ny_1)/(m-n) , (mz_2-nz_1)/(m-n)))`

`color(blue)("Case 1 :")` Coordinates of the mid-point: In case `R` is `color(blue)("the mid-point of PQ,")` then

`color{fuchsia}(m : n = 1 : 1 )` so that `color{fuchsia}(x = (x_1+x_2)/2 , y = (y_1+y_2)/2)` and `color{fuchsia}(z = (z_1+z_2)/2)`

These are the coordinates of the mid point of the segment joining `P( x_1 , y_1 , z_1)` and `Q ( x_2 , y_2 , z_2)`

`color(blue)("Case 2 :")` The coordinates of the point `R` which divides `PQ` in the ratio `k : 1` are obtained by taking `k = m/n` which are as given below:

`color{fuchsia}((( kx_2+x_1)/(1+k) , (ky_2+y_1)/(1+k) , (kz_2+z_1)/(1+k)))`

Generally, this result is used in solving problems involving a general point on the line passing through two given points.


Q 3089034817

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

Solution:

(i) Let P (x, y, z) be the point which divides line segment joining A(1, – 2, 3) and B (3, 4, –5) internally in the ratio 2 : 3. Therefore

`x = (2(3) +3(1))/(2+3) = 9/5 , y = (2(4)+3(-2))/(2+3) = 2/5 , z = (2(-5)+3(3))/(2+3) = (-1)/5`

Thus, the required point is ` ( 9/5 , 2/5 , (-1)/5 )`

(ii) Let `P (x, y, z)` be the point which divides segment joining `A (1, –2, 3)` and `B (3, 4, –5)` externally in the ratio `2 : 3`. Then

`x = (2(3)+(-3)(1))/(2+(-3)) = -3 , y = (2(4)+(-3)(-2))/(2+(-3)) = -14 , z = (2 (-5) +(-3)(3))/(2+(-3)) = 19`

Therefore, the required point is` (–3, –14, 19).`
Q 3039134912

Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear.

Solution:

Let `A (– 4, 6, 10), B (2, 4, 6)` and `C(14, 0, – 2)` be the given points. Let the point P divides AB in the ratio `k : 1`. Then coordinates of the point P are

`((2k-4)/(k+1) , (4k+6)/(k+1) , (6k+10)/(k+1))`

Let us examine whether for some value of k, the point P coincides with point C.

On putting `(2k-4)/(k+1) = 14` we get `k = -3/2`

when `k = -3/2 `, then `(4k+6)/(k+1) = (4 (-3/2) +6)/(-3/2+1) = 0`

and `(6k+10)/(k+1) = (6(-3/2) +10)/(-3/2+1) = -2`

Therefore, `C (14, 0, –2)` is a point which divides AB externally in the ratio `3 : 2` and is same as P.Hence A, B, C are collinear.
Q 3059245114

Find the coordinates of the centroid of the triangle whose vertices are `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)`

Solution:

Let ABC be the triangle. Let the coordinates of the vertices A, B,C be `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)`, respectively. Let D be the mid-point of BC. Hence coordinates of D are

`( (x_2+x_3)/2 , (y_2+y_3)/2 , (z_2+z_3)/2)`

Let G be the centroid of the triangle. Therefore, it divides the median AD in the ratio `2 : 1`. Hence, the coordinates of G are

`( {(2 (x_1+x_2)/2)+x_1 }/(2+1) , {2((y_1+y_3)/2 )+y_1}/(2+1) , {2 ((z_2+z_3)/2)+z_1}/(2+1))`

or `(( x_1+x_2+x_3)/3 , (y_1+y_2+y_3)/3 , (z_1+z_2+z_3)/3)`
Q 3039345212

Find the ratio in which the line segment joining the points `(4, 8, 10)` and `(6, 10, – 8)` is divided by the YZ-plane.

Solution:

Let YZ-plane divides the line segment joining `A (4, 8, 10)` and `B (6, 10, – 8)` at `P (x, y, z) ` in the ratio `k : 1`. Then the coordinates of P are

`((4+6k)/(k+1) , (8+10k)/(k+1) , (10-8k)/(k+1))`

Since P lies on the YZ-plane, its x-coordinate is zero, i.e., `(4+6k)/(k+1) = 0`

or `k = -2/3`

Therefore, YZ-plane divides AB externally in the ratio `2 : 3.`
 
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