How to mark the center line and divide a pipe circumference
How to mark the center line and divide a pipe circumference

CBSE Compartment Exam Date Sheet 2023

June 1, 2023

39 Insightful Publications

Section Formula in 3D: Points are the most basic characteristics of geometry, and it has no dimensions. Two points can be connected using exactly one straight line. A point on this straight line divides the line into two parts. In two-dimensional geometry, the coordinates of the point that divide a line segment internally or externally in a particular ratio can be calculated using the coordinates of the endpoints.

A similar formula can be extended into three-dimensional space. It has a lot of applications in three-dimensional geometry, such as the ratio in which a point in 3D space divides a line segment and to find the collinearity of points.

Then, the section formula in $$2D$$ for internal division $$\to B\left( {x,\,y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$$

Then, the section formula in $$2D$$ for external division $$\to B\left( {x,\,y} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\,\frac{{m{y_2} – n{y_1}}}{{m – n}}} \right)$$.

Step 1: Draw perpendiculars to the $$XY$$ plane from the points $$P,\,Q$$ and $$R$$ to intersect the $$XY$$ plane at the points $$L,\,M,$$ and $$N$$, respectively, such that $$PL\parallel RN\parallel QM$$.

Step 2: Join the points $$L,\,N,$$ and $$M$$. Draw a line parallel to $$LM$$ through $$R$$. Let this line intersect $$QM$$ at $$T$$. Extend the line $$PL$$ to intersect the parallel line at $$S$$.

Since $$ST\parallel LM$$ and $$PL\parallel RN\parallel QM$$, the quadrilaterals $$LNRS$$ and $$NMTR$$ are parallelograms.
Also, by $$AA$$ Similarity Theorem, $$\Delta PSR \sim \Delta QTR$$.
Corresponding sides of similar triangles are proportional.
Also, you already have $$\frac{{PR}}{{QR}} = \frac{m}{n}$$
Thus, $$\frac{{PR}}{{QR}} = \frac{{RS}}{{RT}} = \frac{{SP}}{{QT}} = \frac{m}{n}$$
By the construction of the line segments,
$$SP = SL – PL$$
$$= RN – PL$$
$$= z – z_1$$
and
$$QT = QM – TM$$
$$= QM – RM$$
$$= z_2 – z$$
Using these measures:
$$\frac{{SP}}{{QT}} = \frac{m}{n} = \frac{{z – {z_1}}}{{{z_2} – z}} \to nz – n{z_1} = m{z_2} – mz$$
This can be simplified as $$z = \frac{{m{z_2} + n{z_1}}}{{m + n}}$$.
Now, if we start with the perpendiculars to the $$XZ$$ plane from the point $$P,\,Q,\,R$$ where $$R$$ divides the line segment $$\overline {PQ}$$ in the ratio $$m : n$$,
and following the same arguments, we get the $$y-$$coordinate of $$R$$ as $$y = \frac{{m{y_2} + n{y_1}}}{{m + n}}$$
We can draw perpendiculars $$PL,\,RN,$$ and $$QM$$ to the $$YZ-$$plane to get the $$x-$$coordinate of the point $$R$$ as $$x = \frac{{m{x_2} + n{x_1}}}{{m + n}}$$.
Therefore, we have the coordinates of the point $$R$$ as:
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$
Since the point $$R$$ internally divides the line segment $$\overline {PQ}$$ in the ratio $$m : n$$, this is called the section formula for internal division.

Consider a line segment $$\overline {PQ}$$ and the point $$R$$ externally divides the line segment in the ratio $$m : n$$.

Then the coordinates of the point $$R$$ can be calculated by replacing $$n$$ with $$-n$$.
That is, the coordinates of $$R$$ can be written as:
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)$$

There are some special cases of the formula for specific values of $$m$$ and $$n$$.

This formula is used to find the coordinates of the points that trisect a line segment. Consider the line segment $$\overline {PQ}$$ that joins the points $$P\left( {x_1,\,y_1,\,z_1}\right)$$ and $$Q\left( {x_2,\,y_2,\,z_2}\right)$$ and the points $$R$$ and $$S$$ that divide $$\overline {PQ}$$ into three equal parts.

Point $$R$$ divides the line segment $$\overline {PQ}$$ in the ratio $$1 : 2$$, and point $$S$$ divides the line segment $$\overline {PQ}$$ in the ratio $$2 ∶ 1$$.
Applying the section formula for $$m = 1$$ and $$n = 2$$, the coordinates of $$R$$ is calculated as,
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{\left( 1 \right){x_2} + \left( 2 \right){x_1}}}{{1 + 2}},\,\frac{{\left( 1 \right){y_2} + \left( 2 \right){y_1}}}{{1 + 2}},\,\frac{{\left( 1 \right){z_2} + \left( 2 \right){z_1}}}{{1 + 2}}} \right)$$ $$= \left( {\frac{{{x_2} + 2{x_1}}}{3},\,\frac{{{y_2} + 2{y_1}}}{3},\,\frac{{{z_2} + 2{z_1}}}{3}} \right)$$

Similarly, the coordinates of the point $$S$$ can be calculated by employing the section formula for $$m = 2$$ and $$n = 1$$ as,
$$S\left( {x’,\,y’,\,z’} \right) = \left( {\frac{{\left( 2 \right){x_2} + \left( 1 \right){x_1}}}{{2 + 1}},\,\frac{{\left( 2 \right){y_2} + \left( 1 \right){y_1}}}{{2 + 1}},\,\frac{{\left( 2 \right){z_2} + \left( 1 \right){z_1}}}{{2 + 1}}} \right)$$
$$= \left( {\frac{{2{x_2} + {x_1}}}{3},\,\frac{{2{y_2} + {y_1}}}{3},\,\frac{{2{z_2} + {z_1}}}{3}} \right)$$
In general, if a point $$R(x,\,y,\,z)$$ divides the line joining the points $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ internally in the ratio $$1 : k$$, then the coordinates of the point $$R$$ are written as,
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{\left( 1 \right){x_2} + \left( k \right){x_1}}}{{1 + k}},\frac{{\left( 1 \right){y_2} + \left( k \right){y_1}}}{{1 + k}},\frac{{\left( 1 \right){z_2} + \left( k \right){z_1}}}{{1 + k}}} \right)$$
$$= \left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\,\frac{{{y_2} + k{y_1}}}{{k + 1}},\,\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right)$$

The section formula in $$3D$$ can be applied to derive many other useful results in three-dimensional geometry. For example, the coordinates of the centroid of a triangle can be derived from the section formula.

The centroid of a triangle is the point of intersection of medians. Centroid divides each median in the ratio of $$2 : 1$$.
Consider $$\Delta ABC$$ with the coordinates $$A(x_1,\,y_1,\,z_1),\;B(x_2,\,y_2,\,z_2),$$ and $$C(x_3,\,y_3,\,z_3)$$ with the point $$G(x,\,y,\,z)$$ as the centroid.

$$D$$ is the midpoint of the side $$AB$$, and the coordinates of $$D$$ can be written using the midpoint formula as,
$$D\left( {\frac{{{x_1} + {x_2}}}{2},\;\frac{{{y_1} + {y_2}}}{2},\;\frac{{{z_1} + {z_2}}}{2}} \right)$$
Now, point $$G$$ divides the line segment $$CD$$ in the ratio $$2 : 1$$. So, the coordinates of the point $$G$$ are given by the section formula as,
$$\left( {\frac{{{x_3} + 2\left( {\frac{{{x_1} + {x_2}}}{2}} \right)}}{3},\frac{{{y_3} + 2\left( {\frac{{{y_1} + {y_2}}}{2}} \right)}}{3},\frac{{{z_3} + 2\left( {\frac{{{z_1} + {z_2}}}{2}} \right)}}{3}} \right) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3},\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$$
Therefore, the coordinates of the centroid of a triangle with vertices
$$A\left( {{x_1},\,{y_1},\,{z_1}} \right),\,B\left( {{x_2},\,{y_2},\,{z_2}} \right),$$ and $$C\left( {{x_3},\,{y_3},\,{z_3}} \right)$$ are given by: $$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$$.

Three points are collinear if they lie on the same line. For three points, $$L(x_1,\,y_1,\,z_1),\,M(x_2,\,y_2,\,z_2)$$, and $$N(x_3,\,y_3,\,z_3)$$, it is enough to show that one of the points divide the line segment in a particular ratio, say $$1 : k$$.
If a point $$R(x,\,y,\,z)$$ divides the line joining the points $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ internally in the ratio $$1 : k$$, then the coordinates of the point $$R$$ are written as, $$R\left( {x,\;y,\;z} \right) = \;\left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\,\frac{{{y_2} + k{y_1}}}{{k + 1}},\,\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right)$$.
Now, if we can find a value of $$k$$ such that $$N\left( {{x_3},\;{y_3},\;{z_3}} \right) = \;\left( {\frac{{{x_2} + k{x_1}}}{{k + 1}},\;\frac{{{y_2} + k{y_1}}}{{k + 1}},\;\frac{{{z_2} + k{z_1}}}{{k + 1}}} \right)$$, then the points $$L,\,M,$$ and $$N$$ are collinear.

Q.1. What is the midpoint of the line joining the points $$J(-3,\,4,\,7)$$ and $$K(9,\,0,\,3)$$?
Ans: The coordinates of the midpoint of the line joining the points $$\left( {{x_1},\;{y_1},\;{z_1}} \right)$$ and $$\left( {{x_2},\;{y_2},\;{z_2}} \right)$$ is given by $$\left( {\frac{{{x_1} + {x_2}}}{2},\;\,\frac{{{y_1} + {y_2}}}{2},\,\frac{{{z_1} + {z_2}}}{2}} \right)$$.
So, the coordinates of the line joining the points $$J(-3,\,4,\,7)$$ and $$K(9,\,0,\,3)$$ are:
$$\left( {\frac{{ – 3 + 9}}{2},\frac{{4 + 0}}{2},\frac{{7 + 3}}{2}\;} \right) = \left( {3,\,2,\,5} \right)$$.

Q.2. Find the coordinates of the points that divide the line segment joining the points $$X(2,\,3\,-2)$$ and $$Y(6,\,3\,2)$$ internally in the ratio $$3 : 1$$?
Ans: If point $$R \left( {{x},\;{y},\;{z}} \right)$$ divides the line segment $$\overline {PQ}$$ joining the points $$P \left( {{x_1},\;{y_1},\;{z_1}} \right)$$ and $$Q\left( {{x_2},\;{y_2},\;{z_2}} \right)$$ internally in the ratio $$m : n$$, then the coordinates of $$R$$ are given by
$$R\left( {x,\,y,\,z} \right) = \;\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$
Here, $$\left( {{x_1},{y_1},\;{z_1}} \right) = \left( {2,\;3,\; – 2} \right)$$ and $$\left( {{x_2},{y_2},\;{z_2}} \right) = \;\left( {6,\;3,\;2} \right)$$.
The coordinates of the point that divides the line segment internally in the ratio $$3 : 1$$ are:
$$\left( {\frac{{3\left( 6 \right) + 1\left( 2 \right)}}{{3 + 1}},\frac{{3\left( 3 \right) + 1\left( 3 \right)}}{{3 + 1}},\frac{{3\left( 2 \right) + 1\left( { – 2} \right)}}{{3 + 1}}} \right) = \left( {5,\;3,\;1} \right)$$.

Q.3. Find the ratio in which the point $$(9,\,-11,\,1)$$ externally divides the line segment joining the points $$D(1,\,5,\,-3)$$ and $$E(3,\,1,\,-2)$$.
Ans: If point $$R(x,\,y,\,z)$$ divides the line segment $$PQ$$ joining the points $$P(x_1,\,y_1,\,z_1)$$and $$Q(x_2,\,y_2,\,z_2)$$externally in the ratio $$m : n$$, then the coordinates of $$R$$ are given by
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)$$
Here, $$(x_1,\,y_1,\,z_1 ) = (1,\,5,\,-3),\;(x_2,\,y_2,\,z_2 ) = (3,\,1,\,-2),$$ and $$(x,\,y,\,z) = (9,\,-11,\,1)$$. We need to find the ratio $$m ∶ n$$.
The coordinates of the point that divides the line segment externally in the ratio $$m : n$$ are
$$\left( {\frac{{m\left( 3 \right) – n\left( 1 \right)}}{{m – n}},\frac{{m\left( 1 \right) – n\left( 5 \right)}}{{m – n}},\frac{{m\left( { – 2} \right) – n\left( { – 3} \right)}}{{m – n}}} \right) = \left( {9,\; – 11,\;1} \right)$$
Equating the coordinates:
$$\frac{{3m – n}}{{m – n}} = 9$$
$$\frac{{m – 5n}}{{m – n}} = – 11$$
$$\frac{{-2m + 3n}}{{m – n}} = 1$$
These equations can be simplified as:
$$-3m + 4n = 0$$
That is, $$3m = 4n$$
or
$$\frac{m}{n} = \frac{4}{3}$$
Therefore the ratio in which point $$(9,\,-11,\,1)$$ externally divides $$\overline {DE}$$ is $$4 : 3$$.

Q.4. Find the centroid of $$\Delta SRT$$ whose vertices have the coordinates $$S(-4,\,1,\,0),\;R(5,\,3,\,-3)$$ and $$T(2,\,-10,\,3)$$.
Ans: The coordinates of the centroid of a triangle with vertices $$A(x_1,\,y_1,\,z_1),\;B(x_2,\,y_2,\,z_2)$$, and $$C(x_3,\,y_3,\,z_3)$$ are $$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$$.
The coordinates of the centroid of the triangle with vertices $$S(-4,\,1,\,0),\;R(5,\,3,\,-3)$$ and $$T(2,\,-10,\,3)$$ are given by $$\left( {\frac{{ – 4 + 5 + 2}}{3},\frac{{1 + 3 + \left( { – 10} \right)}}{3},\frac{{0 + \left( { – 3} \right) + 3}}{3}} \right) = \left( {1,\; – 2,\;0} \right)$$.

Q.5. Use the section formula to show that the points $$A(2,\,-3,\,4),\;B(-1,\,2,\,1)$$ and $$C(0,\,\frac{1}{3},\,2)$$ are collinear.
Ans: Let point $$P$$ divides the segment $$\overline {AE}$$ in the ratio $$k : 1$$.
Then, using the section formula, the coordinates of $$P$$ are:
$$\left( {\frac{{k\left( { – 1} \right) + 1\left( 2 \right)}}{{k + 1}},\,\frac{{k\left( 2 \right) + 1\left( { – 3} \right)}}{{k + 1}},\,\frac{{k\left( 1 \right) + 1\left( 4 \right)}}{{k + 1}}} \right) = \left( {\frac{{ – k + 2}}{{k + 1}},\,\frac{{2k – 3}}{{k + 1}},\,\frac{{k + 4}}{{k + 1}}} \right)$$.
Now, if we can find a value of $$k$$ for which these coordinates coincide with $$C$$, we can say that $$C$$ divides the line segment $$\overline {AB}$$ internally or externally in the ratio $$k : 1$$, and therefore the points $$A,\,B,$$ and $$C$$ are collinear.
The $$x-$$coordinate of point $$C$$ is zero.
$$\frac{{ – k + 2}}{{k + 1}} = 0$$ only when $$k = 2$$
When $$k = 2$$:
$$\frac{{2k – 3}}{{k + 1}} = \frac{1}{3}$$ and $$\frac{{k + 4}}{{k + 1}} = 2$$
That is, when $$k = 2$$, the point $$P$$ coincides with the point $$C$$.
In other words, point $$C$$ divides the line segment $$\overline {AB}$$ internally in the ratio of $$2 : 1$$.
Therefore, the three points are collinear.

The article provides a foundation of the concept of section formula for two dimensions before moving into three dimensions. The derivation of the section formula for internal division is explained well, whereas the same for the external division is procured by applying the internal division formula. The special cases of the formula like points of trisection and midpoint formula are also dealt with clarity.

The article also discusses a couple of applications of section formula such as, finding the coordinates of the centroid of a triangle and checking the collinearity of three points. Further, the article concluded with a few solved examples to reinforce the concepts and calculations learnt.

Q.1. What is the section formula in 3D?
Ans: If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ internally in the ratio $$m : n$$, then the section formula for internal division is given by, $$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$
If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ externally in the ratio m:n, then the section formula for external division is given by, $$R\left( {x,\,y,\,z} \right) = \;\left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)$$.

Q.2. What is the section formula for external division?
Ans: If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ externally in the ratio $$m:n,$$ then the section formula for external division is given by $$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)$$.

Q.3. What is the section formula and distance formula?
Ans: If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ internally in the ratio $$m : n$$, then the section formula for internal division is given by $$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$.
The distance formula for finding the distance $$d$$ between the points $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ is given by $$d = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}}$$.

Q.4. How do you find the ratio in which a point divides a line?
Ans: If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ internally in the ratio $$m : n$$, then the section formula for internal division is given by $$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\;\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$
If a point $$R(x,\,y,\,z)$$ divides $$\overline {PQ}$$ where $$P(x_1,\,y_1,\,z_1)$$ and $$Q(x_2,\,y_2,\,z_2)$$ externally in the ratio $$m : n$$, then the section formula for external division is given by,
$$R\left( {x,\,y,\,z} \right) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\;\frac{{m{y_2} – n{y_1}}}{{m – n}},\frac{{m{z_2} – n{z_1}}}{{m – n}}} \right)$$.

Q.5. Where is the section formula used?
Ans: In coordinate geometry, the section formula is used to find the ratio in which a point divides a line segment internally or externally. There are other applications like finding the coordinates of the centroid, incentre, etc. and checking the collinearity of three points.

Practice 3D Questions with Hints & Solutions