Section Formula in Three Dimensions

Section Formula for Internal Division in 3D

The section formula in three-dimensional geometry is a direct extension of the 2D section formula. It provides a method to find the coordinates of a point that divides the line segment connecting two given points in space, either internally (between the endpoints) or externally (on the extension of the segment), in a specific ratio.

This subheading focuses on the case of internal division in 3D space.

Internal Division

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two distinct points in three-dimensional space. A point $R(x, y, z)$ is said to divide the line segment $PQ$ internally in the ratio $m:n$ if $R$ lies on the line segment $PQ$ (between $P$ and $Q$) such that the ratio of the length of the segment $PR$ to the length of the segment $RQ$ is $m:n$.

Here, $m$ and $n$ are positive real numbers representing the ratio.

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Formula

The coordinates of the point $R(x, y, z)$ that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the following formulas for each coordinate:

$\mathbf{x = \frac{mx_2 + nx_1}{m + n}}$

$\mathbf{y = \frac{my_2 + ny_1}{m + n}}$

$\mathbf{z = \frac{mz_2 + nz_1}{m + n}}$

These can be compactly written as a single coordinate triplet:

$\mathbf{R(x, y, z) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n} \right)}$

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Derivation Outline

The derivation of the 3D internal section formula is analogous to the 2D case and can be understood by considering the projections of the points onto the coordinate planes or axes, or more elegantly using vectors.

Method 1: Using Projections (Conceptual)

Imagine projecting the points $P(x_1, y_1, z_1)$, $Q(x_2, y_2, z_2)$, and $R(x, y, z)$ onto the XY-plane. Let the projected points be $P'(x_1, y_1, 0)$, $Q'(x_2, y_2, 0)$, and $R'(x, y, 0)$ respectively. Since P, Q, R are collinear and R divides PQ in the ratio m:n, their projections $P', Q', R'$ onto the XY-plane are also collinear, and $R'$ divides the segment $P'Q'$ in the same ratio $m:n$.

We can apply the 2D section formula to the points $P'(x_1, y_1)$ and $Q'(x_2, y_2)$ in the XY-plane, with the point $R'(x, y)$ dividing the segment in the ratio $m:n$. This gives us the formulas for the x and y coordinates of R:

Similarly, if we projected the points onto the XZ-plane ($y=0$), the projected points would be $(x_1, 0, z_1)$, $(x_2, 0, z_2)$, and $(x, 0, z)$. Applying the 2D section formula (for x and z coordinates this time) would confirm the formula for x and yield the formula for z:

The same results would be obtained by projecting onto the YZ-plane ($x=0$).

Method 2: Using Vectors (More Formal)

Let $\vec{p}$, $\vec{q}$, and $\vec{r}$ be the position vectors of points P$(x_1, y_1, z_1)$, Q$(x_2, y_2, z_2)$, and R$(x, y, z)$ respectively, with respect to the origin O. The position vector of a point is the vector from the origin to that point, e.g., $\vec{p} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}$.

The point R divides the segment PQ internally in the ratio m:n. This means $\vec{PR} = \frac{m}{n} \vec{RQ}$ (as vectors pointing in the same direction). However, it's more standard to use the fact that R is on the segment PQ and $\frac{|\vec{PR}|}{|\vec{RQ}|} = \frac{m}{n}$. As vectors with direction, $n\vec{PR} = m\vec{RQ}$.

In terms of position vectors: $n(\vec{r} - \vec{p}) = m(\vec{q} - \vec{r})$

$n\vec{r} - n\vec{p} = m\vec{q} - m\vec{r}$

Gather terms with $\vec{r}$ on one side:

$n\vec{r} + m\vec{r} = m\vec{q} + n\vec{p}$

$(m + n)\vec{r} = n\vec{p} + m\vec{q}$

Solve for $\vec{r}$:

Now, substitute the position vectors in terms of coordinates:

$x\hat{i} + y\hat{j} + z\hat{k} = \frac{n(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + m(x_2\hat{i} + y_2\hat{j} + z_2\hat{k})}{m + n}$

$x\hat{i} + y\hat{j} + z\hat{k} = \frac{(nx_1 + mx_2)\hat{i} + (ny_1 + my_2)\hat{j} + (nz_1 + mz_2)\hat{k}}{m + n}$

$x\hat{i} + y\hat{j} + z\hat{k} = \left(\frac{mx_2 + nx_1}{m + n}\right)\hat{i} + \left(\frac{my_2 + ny_1}{m + n}\right)\hat{j} + \left(\frac{mz_2 + nz_1}{m + n}\right)\hat{k}$

Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides, we obtain the coordinates of R:

These are the same formulas obtained using projections.

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Mid-point Formula in 3D

The midpoint of a line segment in three dimensions is the point that divides the segment into two segments of equal length. It is located exactly halfway between the two endpoints. In terms of the section formula, the midpoint is the point that divides the segment internally in the ratio $1:1$.

The 3D midpoint formula is a special case of the 3D section formula for internal division, where the ratio $m:n$ is $1:1$.

Derivation of the Mid-point Formula in 3D

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be the coordinates of the two endpoints of a line segment in 3D space.

Let $M(x, y, z)$ be the coordinates of the midpoint of the segment PQ.

The midpoint M divides the segment PQ internally in the ratio $1:1$. This means we can use the internal section formula with $m=1$ and $n=1$.

The section formula for internal division is:

$x = \frac{mx_2 + nx_1}{m + n}$

$y = \frac{my_2 + ny_1}{m + n}$

$z = \frac{mz_2 + nz_1}{m + n}$

Substitute $m=1$ and $n=1$ into these formulas:

Calculating the x-coordinate of M:

Simplify equation (i):

Calculating the y-coordinate of M:

Simplify equation (iii):

Calculating the z-coordinate of M:

Simplify equation (v):

Mid-point Formula in 3D

Combining the results from equations (ii), (iv), and (vi), the coordinates of the midpoint $M(x, y, z)$ of the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are given by:

$\mathbf{M(x, y, z) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)}$

In essence, the coordinates of the midpoint are found by taking the average of the corresponding x, y, and z coordinates of the two endpoints.

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Section Formula for External Division in 3D

In contrast to internal division, which locates a point between two endpoints, external division concerns a point that lies on the line containing the segment, but outside the segment itself. The section formula can also be used to find the coordinates of such a point in three dimensions.

External Division

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two distinct points in 3D space. A point $R(x, y, z)$ is said to divide the line segment $PQ$ externally in the ratio $m:n$ if R lies on the line passing through P and Q, but not on the segment PQ, such that the ratio of the distance from P to R ($PR$) to the distance from Q to R ($QR$) is equal to the given ratio $m:n$.

Here, $m$ and $n$ are positive real numbers representing the ratio of lengths. The location of R relative to P and Q depends on the comparison of $m$ and $n$:

• If $m > n$, the point R is closer to Q than to P relative to the segment PQ, and it lies on the extension of the segment beyond Q. The order of points on the line is P - Q - R.
• If $m < n$, the point R is closer to P than to Q relative to the segment PQ, and it lies on the extension of the segment beyond P. The order of points on the line is R - P - Q.

The case $m=n$ is not possible for external division of distinct points P and Q, as it would imply $PR=QR$, which means R is the midpoint of PQ (internal division) or R is on the perpendicular bisector plane of PQ (not on the line PQ unless it's the midpoint, which is internal).

Formula

The coordinates of the point $R(x, y, z)$ that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ externally in the ratio $m:n$ ($m \neq n$) are given by:

$\mathbf{x = \frac{mx_2 - nx_1}{m - n}}$

$\mathbf{y = \frac{my_2 - ny_1}{m - n}}$

$\mathbf{z = \frac{mz_2 - nz_1}{m - n}}$

These can be written as a single coordinate triplet:

$\mathbf{R(x, y, z) = \left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n} \right)}$

Relation to Internal Division Formula:

Notice that the formula for external division is very similar to the formula for internal division. The only difference is that the plus signs in the internal division formula are replaced by minus signs in the external division formula. This is consistent with the interpretation that external division in the ratio $m:n$ is equivalent to internal division in the ratio $m:(-n)$ or $(-m):n$ in some contexts (related to directed ratios).

If you replace $n$ with $-n$ in the internal section formula: $x = \frac{mx_2 + (-n)x_1}{m + (-n)} = \frac{mx_2 - nx_1}{m - n}$ $y = \frac{my_2 + (-n)y_1}{m + (-n)} = \frac{my_2 - ny_1}{m - n}$ $z = \frac{mz_2 + (-n)z_1}{m + (-n)} = \frac{mz_2 - nz_1}{m - n}$

This provides an easy way to remember the external division formula if you know the internal one.

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Centroid of a Triangle/Tetrahedron in 3D

The concept of the centroid extends naturally to three-dimensional figures. For both triangles (which can be considered as 2D figures embedded in 3D space) and tetrahedrons, the centroid represents the geometric center or the center of mass of a uniform solid. The calculation of the centroid's coordinates in 3D relies implicitly on the section formula.

Centroid of a Triangle in 3D

The centroid of a triangle in 3D space is defined in the same way as in 2D: it is the point of intersection of the triangle's medians. A median connects a vertex to the midpoint of the opposite side.

Let the vertices of the triangle in 3D space be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.

To find the centroid $G$, we can use the property that the centroid divides each median in a 2:1 ratio. Consider the median from vertex A to the midpoint of side BC. Let D be the midpoint of BC.

Using the midpoint formula in 3D, the coordinates of D are:

The centroid $G$ divides the median AD internally in the ratio $AG : GD = 2 : 1$. So, G divides the segment joining A$(x_1, y_1, z_1)$ and D$\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right)$ internally in the ratio $m:n = 2:1$.

Using the section formula for internal division in 3D for point G$(x, y, z)$ dividing AD:

Calculating the x-coordinate of G:

Simplify equation (ii):

$G_x = \frac{\frac{2(x_2 + x_3)}{2} + x_1}{3} = \frac{x_2 + x_3 + x_1}{3}$

Calculating the y-coordinate of G:

Simplify equation (iv):

$G_y = \frac{\frac{2(y_2 + y_3)}{2} + y_1}{3} = \frac{y_2 + y_3 + y_1}{3}$

Calculating the z-coordinate of G:

Simplify equation (vi):

$G_z = \frac{\frac{2(z_2 + z_3)}{2} + z_1}{3} = \frac{z_2 + z_3 + z_1}{3}$

Centroid Formula for a Triangle in 3D:

The coordinates of the centroid $G(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are the average of the corresponding coordinates of the vertices:

$\mathbf{G \equiv \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)}$

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Centroid of a Tetrahedron

A tetrahedron is the three-dimensional analogue of a triangle; it is a polyhedron with four triangular faces, four vertices, and six edges. The centroid of a tetrahedron is its geometric center.

The centroid of a tetrahedron is the point of intersection of the four lines joining each vertex to the centroid of the opposite triangular face.

Let the vertices of a tetrahedron be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$, and $D(x_4, y_4, z_4)$.

Consider the line segment joining vertex D to the centroid of the opposite face, $\triangle ABC$. Let $G_{ABC}$ be the centroid of $\triangle ABC$. Using the centroid formula for a triangle:

The centroid $G$ of the tetrahedron lies on the segment $D G_{ABC}$ and divides it internally in the ratio $3:1$ (vertex to opposite face centroid). So, G divides the segment joining $D(x_4, y_4, z_4)$ and $G_{ABC}(x_{G_{ABC}}, y_{G_{ABC}}, z_{G_{ABC}})$ internally in the ratio $m:n = 3:1$.

Using the section formula for internal division in 3D for point G$(x, y, z)$ dividing $D G_{ABC}$:

Calculating the x-coordinate of G:

Simplify equation (ix):

$G_x = \frac{\frac{3(x_1 + x_2 + x_3)}{3} + x_4}{4} = \frac{x_1 + x_2 + x_3 + x_4}{4}$

Similarly, for the y and z coordinates:

Centroid Formula for a Tetrahedron:

The coordinates of the centroid $G(x, y, z)$ of a tetrahedron with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$, and $(x_4, y_4, z_4)$ are the average of the corresponding coordinates of the vertices:

$\mathbf{G \equiv \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4} \right)}$

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