The Midpoint and Distance Formulas in 3D
The Midpoint and Distance Formulas in 3D

In two dimension geometry, the concept of section formula is implemented to find the coordinates of a point dividing a line segment internally in a specific ratio. In order to locate the position of a point in space, we require a coordinate system. After choosing a fixed coordinate system in three dimensions, the coordinates of any point P in that system can be given. In case of a rectangular coordinate system, it is given by an ordered 3-tuple (x, y, z). Also, if the coordinates (x, y, z) is already known, then we can easily locate the point P in space. The concept of section formula can be extended to three-dimension geometry as well as to determine the coordinates of a point dividing a line in a certain ratio.

Section Formula

Let us consider two points A (x1, y1, z1) and B(x2, y2, z2). Consider a point P(x, y, z) dividing AB in the ratio m:n as shown in the figure given below.

To determine the coordinates of the point P, the following steps are followed:

  • Draw AL, PN, and BM perpendicular to XY plane such that AL || PN || BM as shown above.
  • The points L, M and N lie on the straight line formed due to the intersection of a plane containing AL, PN and BM and XY- plane.
  • From point P, a line segment ST is drawn such that it is parallel to LM.
  • ST intersects AL externally at S, and it intersects BM at T internally.

Since ST is parallel to LM and AL || PN || BM, therefore, the quadrilaterals LNPS and NMTP qualify as parallelograms.

Also, ∆ASP ~∆BTP therefore,

Rearranging the above equation we get,

The above procedure can be repeated by drawing perpendiculars to XZ and YZ- planes to get the x and y coordinates of the point P that divides the line segment AB in the ratio m:n internally.

Sectional Formula (Internally)

Thus, the coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:

Sectional Formula (Externally)

If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

This represents the section formula for three dimension geometry.

If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be

What if the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB?

In that case, if P is the midpoint, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1.

Coordinates of point P will be given as:

Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,

Section Formula in 3D Examples

Go through the below examples to understand the concept of section formula in a three-dimensional plane.

Example 1:

Determine the coordinate points which divide the line segment that joins the point (1, -2, 3) and (3, 4, -5) in the ratio of 2:3 internally and externally.

Solution:

Assume that the point P(x, y, z) divides the line segment that joins A(1, -2, 3) and B(3, 4, -5) internally in the ratio of 2:3.

Here, m=2, and n=3

Now, substitute the values in the section formula,

⇒ [(mx2+nx1)/m+n, (my2+ny1)/m+n, (mz2+nz1)/m+n ]

⇒ [(2(3)+3(1))/2+3,(2(4)+3(-2))/2+3, (2(-5)+3(3))/2+3 ]

⇒[(9/5), (⅖), (-⅕)]

Hence, the required point is [(9/5), (⅖), (-⅕)]

Let the point P(x, y, z) divides the line segment that joins A(1, -2, 3) and B(3, 4, -5) externally in the ratio of 2:3, then

⇒ [(mx2-nx1)/m-n, (my2-ny1)/m-n, (mz2-nz1)/m-n ]

⇒ [(2(3)-3(1))/2-3,(2(4)-3(-2))/2-3, (2(-5)-3(3))/2-3 ]

⇒[-3, -14, 19]

Hence, the required point is (-3, -14, 19).

Example 2:

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

Solution:

Assume that the three points are A(– 4, 6, 10), B(2, 4, 6), C(14, 0, –2). Now, the point P divides the line segment AB in the ratio of k:1, then the coordinates of p are:

⇒[(2k-4)/k+1, (4k+6)/k+1, (6k+10)/k+1]

Now, let’s check whether some value of k, the point P coincides with the point C.

By equating (2k-4)/k+1 = 14, we get the value of k as -3/2.

If k=-3/2, then

⇒(4k+6)/k+1

⇒(4(-3/2)+6)/(-3/2)+1

⇒ 0

Similarly,

⇒ (6k+10)/k+1

⇒(6(-3/2)+10)/(-3/2)+1

⇒-2

Hence, the point C(14, 0, -2) which divides the line segment externally in the ratio of 3:2, which is the same as the point P.

Therefore, the points A, B and C are collinear.

Practice Problems

Solve the problems given below:

  1. Show that the points P(2, –3, 4), Q(–1, 2, 1), R(0, ⅓, 2) are collinear using section formula.
  2. Calculate the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
  3. Find the coordinate point that divides the line segment that joins the points (– 2, 3, 5) and (1, – 4, 6) in the ratio of 2:3 externally.

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