Section Formula in Vector Algebra
Section formula for Internal Division of a Line Segment
The section formula in vector algebra is a powerful tool used to determine the position vector of a point that lies on a line segment and divides it into two smaller segments with a specific ratio of lengths. This is the vector equivalent of the section formula used in coordinate geometry.
Problem Statement
Consider two points P and Q in space. Let their position vectors relative to a fixed origin O be $\vec{a}$ and $\vec{b}$ respectively. Let R be a point that lies on the line segment PQ and divides it internally in the ratio $m:n$. This means that the ratio of the length of the segment PR to the length of the segment RQ is $m:n$. Our goal is to find the position vector $\vec{r}$ of the point R with respect to the same origin O.
$\frac{\text{Length}(PR)}{\text{Length}(RQ)} = \frac{m}{n}$
where $m$ and $n$ are positive scalar quantities representing the ratio.
Derivation of the Internal Section Formula
Let O be the origin. The position vectors of P, Q, and R are $\vec{OP} = \vec{a}$, $\vec{OQ} = \vec{b}$, and $\vec{OR} = \vec{r}$, respectively.
Since R divides the line segment PQ internally in the ratio $m:n$, the point R lies between P and Q. The vectors $\vec{PR}$ and $\vec{RQ}$ lie along the same straight line PQ and point in the same direction (from P to Q). Their magnitudes are in the ratio $m:n$.
$PR : RQ = m : n$
This implies $n \times \text{Length}(PR) = m \times \text{Length}(RQ)$.
Since $\vec{PR}$ and $\vec{RQ}$ are vectors in the same direction along the line PQ, and their magnitudes are in the ratio $m:n$, we can write the vector equation:
$n \vec{PR} = m \vec{RQ}$
Now, express the vectors $\vec{PR}$ and $\vec{RQ}$ in terms of the position vectors relative to the origin O:
$\vec{PR} = \text{Position Vector of R} - \text{Position Vector of P} = \vec{OR} - \vec{OP} = \vec{r} - \vec{a}$
$\vec{RQ} = \text{Position Vector of Q} - \text{Position Vector of R} = \vec{OQ} - \vec{OR} = \vec{b} - \vec{r}$
Substitute these expressions back into the vector equation $n \vec{PR} = m \vec{RQ}$:
$n (\vec{r} - \vec{a}) = m (\vec{b} - \vec{r})$
Expand using the distributive property of scalar multiplication over vector subtraction:
$n\vec{r} - n\vec{a} = m\vec{b} - m\vec{r}$
Rearrange the terms to group the $\vec{r}$ terms on one side and the other terms on the other side:
$n\vec{r} + m\vec{r} = m\vec{b} + n\vec{a}$
Factor out the vector $\vec{r}$ on the left side:
$(m + n)\vec{r} = m\vec{b} + n\vec{a}$
Since $m$ and $n$ are positive for internal division, $m+n$ is non-zero. Divide both sides by the scalar $(m+n)$ to solve for $\vec{r}$:
$$ \mathbf{\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}} $$
This is the section formula for the position vector of a point R that divides the line segment PQ, with position vectors $\vec{a}$ and $\vec{b}$, internally in the ratio $m:n$.
Interpretation: The formula shows that the position vector of the dividing point is a weighted average of the position vectors of the endpoints. The weight applied to the position vector of the terminal point ($\vec{b}$) is the ratio segment $m$ associated with the initial point side, and the weight applied to the position vector of the initial point ($\vec{a}$) is the ratio segment $n$ associated with the terminal point side. The sum of the weights is the denominator.
Example 1. Find the position vector of point R which divides the line joining two points P and Q, whose position vectors are $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ respectively, internally in the ratio $2:1$.
Answer:
Given the position vector of point P is $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$.
Given the position vector of point Q is $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.
The point R divides PQ internally in the ratio $m:n = 2:1$. Thus, $m=2$ and $n=1$.
We use the section formula for internal division:
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$
Substitute the given values:
$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1}$
$\vec{r} = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - \hat{k})}{3}$
Combine the corresponding components in the numerator:
$\vec{r} = \frac{(-2 + 1)\hat{i} + (2 + 2)\hat{j} + (2 - 1)\hat{k}}{3}$
$\vec{r} = \frac{-1\hat{i} + 4\hat{j} + 1\hat{k}}{3}$
Write the components separately:
$\vec{r} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$
The position vector of the point R is $-\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$.
Section formula for External Division of a Line Segment
The section formula can also be used to find the position vector of a point that lies on the line extending the segment PQ, but outside the segment itself. This is called external division.
Problem Statement
Consider two points P and Q in space with position vectors $\vec{a}$ and $\vec{b}$ respectively, relative to an origin O. Let R be a point on the line passing through P and Q, but lying outside the line segment PQ. R divides the line segment PQ externally in the ratio $m:n$. This means that the ratio of the distance from P to R to the distance from Q to R is $m:n$.
$\frac{\text{Length}(PR)}{\text{Length}(QR)} = \frac{m}{n}$
where $m$ and $n$ are positive scalar quantities. We assume $m \neq n$, because if $m=n$, the ratio PR/QR = 1, implying PR = QR, which is only possible if R is infinitely far from P and Q (or P and Q coincide, in which case the line is not defined). If $m > n$, R lies on the line on the side of Q. If $m < n$, R lies on the line on the side of P.
Derivation of the External Section Formula
Let O be the origin, with position vectors $\vec{OP} = \vec{a}$, $\vec{OQ} = \vec{b}$, and $\vec{OR} = \vec{r}$.
Let's consider the case where R lies on the line PQ such that Q is between P and R (i.e., $m > n$). The points lie in the order P-Q-R. The vectors $\vec{PR}$ and $\vec{QR}$ lie along the same line and point in the same direction (from P towards R, and from Q towards R). Their magnitudes are in the ratio $m:n$.
$PR : QR = m : n$
This implies $n \times \text{Length}(PR) = m \times \text{Length}(QR)$.
Since $\vec{PR}$ and $\vec{QR}$ are vectors in the same direction, we can write the vector equation:
$n \vec{PR} = m \vec{QR}$
(Note: If R lies on the side of P such that P is between R and Q, i.e., R-P-Q ($m < n$), the vectors $\vec{RP}$ and $\vec{RQ}$ would be used, pointing in the same direction along the line. The derivation would yield the same final formula.)
Express the vectors $\vec{PR}$ and $\vec{QR}$ in terms of position vectors relative to the origin O:
$\vec{PR} = \text{Position Vector of R} - \text{Position Vector of P} = \vec{OR} - \vec{OP} = \vec{r} - \vec{a}$
$\vec{QR} = \text{Position Vector of R} - \text{Position Vector of Q} = \vec{OR} - \vec{OQ} = \vec{r} - \vec{b}$
Substitute these expressions back into the vector equation $n \vec{PR} = m \vec{QR}$:
$n (\vec{r} - \vec{a}) = m (\vec{r} - \vec{b})$
Expand using the distributive property:
$n\vec{r} - n\vec{a} = m\vec{r} - m\vec{b}$
Rearrange terms to isolate $\vec{r}$ on one side:
$m\vec{b} - n\vec{a} = m\vec{r} - n\vec{r}$
Factor out $\vec{r}$ on the right side:
$m\vec{b} - n\vec{a} = (m - n)\vec{r}$
Since $m \neq n$, $m-n$ is a non-zero scalar. Divide by $(m-n)$ to solve for $\vec{r}$:
$$ \mathbf{\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}} $$
This is the section formula for the position vector of a point R that divides the line segment PQ, with position vectors $\vec{a}$ and $\vec{b}$, externally in the ratio $m:n$.
Comparison with Internal Division: The external division formula is very similar to the internal division formula ($\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$), but with the plus signs in the numerator and denominator replaced by minus signs for the term involving $n\vec{a}$. This can be seen as dividing internally in the ratio $m:(-n)$ or $(-m):n$. However, using the formula with positive $m$ and $n$ and applying the minus signs as shown is generally more straightforward.
Example 1. Find the position vector of point R which divides the line joining the points P with coordinates $(2, -1, 4)$ and Q with coordinates $(1, 3, -5)$ externally in the ratio $3:2$.
Answer:
First, write the position vectors for points P and Q relative to the origin O.
The position vector of P is $\vec{a} = \vec{OP} = 2\hat{i} - \hat{j} + 4\hat{k}$.
The position vector of Q is $\vec{b} = \vec{OQ} = \hat{i} + 3\hat{j} - 5\hat{k}$.
The point R divides PQ externally in the ratio $m:n = 3:2$. Thus, $m=3$ and $n=2$. Note that $m > n$, so R will lie on the side of Q.
We use the section formula for external division:
$\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$
Substitute the given values:
$\vec{r} = \frac{3(\hat{i} + 3\hat{j} - 5\hat{k}) - 2(2\hat{i} - \hat{j} + 4\hat{k})}{3 - 2}$
$\vec{r} = \frac{(3\hat{i} + 9\hat{j} - 15\hat{k}) - (4\hat{i} - 2\hat{j} + 8\hat{k})}{1}$
Distribute the scalar 2 into the second parenthesis and the negative sign:
$\vec{r} = 3\hat{i} + 9\hat{j} - 15\hat{k} - 4\hat{i} + 2\hat{j} - 8\hat{k}$
Combine the corresponding components:
$\vec{r} = (3 - 4)\hat{i} + (9 + 2)\hat{j} + (-15 - 8)\hat{k}$
$\vec{r} = -1\hat{i} + 11\hat{j} - 23\hat{k}$
$\vec{r} = -\hat{i} + 11\hat{j} - 23\hat{k}$
The position vector of the point R is $-\hat{i} + 11\hat{j} - 23\hat{k}$.
Midpoint Formula in Vector Form
A specific case of the section formula arises when the point dividing the line segment is exactly in the middle. This point is known as the midpoint, and there's a simplified vector formula to find its position vector.
Problem Statement
Let P and Q be two points in space, with position vectors $\vec{a}$ and $\vec{b}$ respectively, relative to a fixed origin O. Let M be the midpoint of the line segment PQ. We want to determine the position vector $\vec{m}$ of the point M with respect to the origin O.
Derivation of the Midpoint Formula
The midpoint M of the line segment PQ is the point that divides the segment internally into two equal parts. Therefore, M divides PQ in the ratio $1:1$.
We can use the section formula for internal division with the ratio $m:n = 1:1$. The position vector $\vec{m}$ of the point R dividing PQ internally in the ratio $m:n$ is given by:
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$
(Section formula for internal division)
In the case of the midpoint M, the ratio is $m=1$ and $n=1$. Substituting these values into the formula, replacing $\vec{r}$ with $\vec{m}$:
$\vec{m} = \frac{1\cdot\vec{b} + 1\cdot\vec{a}}{1 + 1}$
$\vec{m} = \frac{\vec{b} + \vec{a}}{2}$
Rearranging the terms in the numerator (since vector addition is commutative):
$$ \mathbf{\vec{m} = \frac{\vec{a} + \vec{b}}{2}} $$
This is the midpoint formula in vector form. It states that the position vector of the midpoint of a line segment is the average of the position vectors of its two endpoints.
Example 1. Find the position vector of the midpoint of the line segment joining the points P$(2, 3, 4)$ and Q$(4, 1, -2)$.
Answer:
First, write the position vectors for points P and Q relative to the origin O.
Position vector of P: $\vec{a} = \vec{OP} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Position vector of Q: $\vec{b} = \vec{OQ} = 4\hat{i} + \hat{j} - 2\hat{k}$.
Let $\vec{m}$ be the position vector of the midpoint M of PQ.
Using the midpoint formula in vector form:
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$
Substitute the component forms of $\vec{a}$ and $\vec{b}$:
$\vec{m} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} - 2\hat{k})}{2}$
Add the vectors in the numerator by adding their corresponding components:
$\vec{m} = \frac{(2+4)\hat{i} + (3+1)\hat{j} + (4+(-2))\hat{k}}{2}$
$\vec{m} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}$
Now, divide each component by the scalar 2:
$\vec{m} = \frac{6}{2}\hat{i} + \frac{4}{2}\hat{j} + \frac{2}{2}\hat{k}$
$\vec{m} = 3\hat{i} + 2\hat{j} + \hat{k}$
The position vector of the midpoint of the line segment joining P and Q is $3\hat{i} + 2\hat{j} + \hat{k}$.
Centroid of a Triangle/Tetrahedron in Vector Form
The concept of averaging position vectors to find a central point, as seen with the midpoint, extends to other geometric figures like triangles and tetrahedrons. The centroid of these figures can also be expressed concisely using vectors.
Centroid of a Triangle
The centroid of a triangle is a point of concurrency where the three medians of the triangle intersect. A median is a line segment joining a vertex to the midpoint of the opposite side. A key property of the centroid is that it divides each median in the ratio $2:1$, with the longer segment being from the vertex to the centroid.
Let the vertices of a triangle be A, B, and C, and let their position vectors relative to an origin O be $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively. Let G be the centroid of the triangle. The position vector of the centroid G is denoted by $\vec{g}$.
The position vector $\vec{g}$ of the centroid G is given by the average of the position vectors of its vertices:
$$ \mathbf{\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}} $$
Derivation (Implicit basis on Section Formula)
Consider the median AD, where D is the midpoint of the side BC. Using the midpoint formula, the position vector of D is:
$\vec{d} = \frac{\vec{b} + \vec{c}}{2}$
The centroid G lies on the median AD and divides it in the ratio $2:1$ (AG : GD = 2 : 1). We can use the internal section formula to find the position vector of G, dividing the line segment AD with position vectors $\vec{a}$ (for A) and $\vec{d}$ (for D) in the ratio $m:n = 2:1$.
$\vec{g} = \frac{m\vec{d} + n\vec{a}}{m + n}$
Substitute $m=2, n=1$ and $\vec{d} = \frac{\vec{b} + \vec{c}}{2}$:
$\vec{g} = \frac{2\left(\frac{\vec{b} + \vec{c}}{2}\right) + 1\cdot\vec{a}}{2 + 1}$
$\vec{g} = \frac{(\vec{b} + \vec{c}) + \vec{a}}{3}$
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
This confirms the formula for the centroid of a triangle.
Example 1. Find the position vector of the centroid of a triangle whose vertices are A$(1,1,1)$, B$(2,3,4)$, and C$(6,2,1)$.
Answer:
Let the position vectors of the vertices A, B, and C relative to the origin O be $\vec{a}, \vec{b}, \vec{c}$ respectively.
$\vec{a} = \vec{OA} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \vec{OB} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
$\vec{c} = \vec{OC} = 6\hat{i} + 2\hat{j} + \hat{k}$
Let $\vec{g}$ be the position vector of the centroid G.
Using the formula for the centroid of a triangle:
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
Substitute the component forms of the position vectors:
$\vec{g} = \frac{(\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + 3\hat{j} + 4\hat{k}) + (6\hat{i} + 2\hat{j} + \hat{k})}{3}$
Add the vectors in the numerator by combining corresponding components:
$\vec{g} = \frac{(1+2+6)\hat{i} + (1+3+2)\hat{j} + (1+4+1)\hat{k}}{3}$
$\vec{g} = \frac{9\hat{i} + 6\hat{j} + 6\hat{k}}{3}$
Divide each component by the scalar 3:
$\vec{g} = \frac{9}{3}\hat{i} + \frac{6}{3}\hat{j} + \frac{6}{3}\hat{k}$
$\vec{g} = 3\hat{i} + 2\hat{j} + 2\hat{k}$
The position vector of the centroid of the triangle is $3\hat{i} + 2\hat{j} + 2\hat{k}$.
Centroid of a Tetrahedron
A tetrahedron is a polyhedron with four vertices and four triangular faces. The centroid of a tetrahedron is the point where the four lines joining each vertex to the centroid of the opposite triangular face intersect. This centroid divides each such line segment in the ratio $3:1$, with the longer segment being from the vertex to the centroid of the tetrahedron.
Let the vertices of a tetrahedron be A, B, C, and D, and let their position vectors relative to an origin O be $\vec{a}$, $\vec{b}$, $\vec{c}$, and $\vec{d}$ respectively. Let G be the centroid of the tetrahedron. The position vector of the centroid G is denoted by $\vec{g}$.
The position vector $\vec{g}$ of the centroid G is given by the average of the position vectors of its four vertices:
$$ \mathbf{\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}} $$
Derivation (Implicit basis on Section Formula and Triangle Centroid)
Consider the line segment joining vertex A to the centroid of the opposite face BCD. Let $G_{bcd}$ be the centroid of the triangular face BCD. Using the formula for the centroid of a triangle, the position vector of $G_{bcd}$ is:
$\vec{g}_{bcd} = \frac{\vec{b} + \vec{c} + \vec{d}}{3}$
The centroid G of the tetrahedron lies on the line segment $AG_{bcd}$ and divides it in the ratio $3:1$ (AG : $G_{bcd}$G = 3 : 1). We use the internal section formula to find the position vector of G, dividing the line segment $AG_{bcd}$ with position vectors $\vec{a}$ (for A) and $\vec{g}_{bcd}$ (for $G_{bcd}$) in the ratio $m:n = 3:1$.
$\vec{g} = \frac{m\vec{g}_{bcd} + n\vec{a}}{m + n}$
Substitute $m=3, n=1$ and $\vec{g}_{bcd} = \frac{\vec{b} + \vec{c} + \vec{d}}{3}$:
$\vec{g} = \frac{3\left(\frac{\vec{b} + \vec{c} + \vec{d}}{3}\right) + 1\cdot\vec{a}}{3 + 1}$
$\vec{g} = \frac{(\vec{b} + \vec{c} + \vec{d}) + \vec{a}}{4}$
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$
This confirms the formula for the centroid of a tetrahedron.
Example 2. Find the position vector of the centroid of a tetrahedron whose vertices have position vectors $\hat{i}$, $2\hat{j}$, $3\hat{k}$, and $\hat{i}+\hat{j}+\hat{k}$.
Answer:
Let the position vectors of the four vertices of the tetrahedron be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.
$\vec{a} = \hat{i}$
$\vec{b} = 2\hat{j}$
$\vec{c} = 3\hat{k}$
$\vec{d} = \hat{i} + \hat{j} + \hat{k}$
Let $\vec{g}$ be the position vector of the centroid G of the tetrahedron.
Using the formula for the centroid of a tetrahedron:
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$
Substitute the given position vectors:
$\vec{g} = \frac{(\hat{i}) + (2\hat{j}) + (3\hat{k}) + (\hat{i} + \hat{j} + \hat{k})}{4}$
Add the vectors in the numerator by combining corresponding components:
$\vec{g} = \frac{(1+1)\hat{i} + (2+1)\hat{j} + (3+1)\hat{k}}{4}$
$\vec{g} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{4}$
Divide each component by the scalar 4:
$\vec{g} = \frac{2}{4}\hat{i} + \frac{3}{4}\hat{j} + \frac{4}{4}\hat{k}$
$\vec{g} = \frac{1}{2}\hat{i} + \frac{3}{4}\hat{j} + \hat{k}$
The position vector of the centroid of the tetrahedron is $\frac{1}{2}\hat{i} + \frac{3}{4}\hat{j} + \hat{k}$.
Summary for Competitive Exams
Let P and Q be points with position vectors $\vec{a}$ and $\vec{b}$ respectively, relative to an origin O. Let R be a point dividing the line segment PQ in a ratio $m:n$. Let $\vec{r}$ be the position vector of R.
Section Formulas:
- Internal Division: If R divides PQ internally in the ratio $m:n$ ($m, n > 0$), then $$ \vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} $$
- External Division: If R divides PQ externally in the ratio $m:n$ ($m, n > 0, m \neq n$), then
$$ \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} $$
(If $m>n$, R is beyond Q; if $m
- Midpoint Formula: The midpoint M of PQ divides it internally in the ratio $1:1$. Its position vector $\vec{m}$ is $$ \vec{m} = \frac{1\vec{b} + 1\vec{a}}{1 + 1} = \frac{\vec{a} + \vec{b}}{2} $$
Centroid Formulas: The centroid of a geometric figure is the average of the position vectors of its vertices.
- Centroid of a Triangle: For a triangle with vertices A, B, C having position vectors $\vec{a}, \vec{b}, \vec{c}$, the position vector $\vec{g}$ of the centroid is $$ \vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} $$ (This follows from the section formula on a median, which the centroid divides in a 2:1 ratio).
- Centroid of a Tetrahedron: For a tetrahedron with vertices A, B, C, D having position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$, the position vector $\vec{g}$ of the centroid is $$ \vec{g} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4} $$ (This follows from the section formula on a line joining a vertex to the centroid of the opposite face, which the tetrahedron centroid divides in a 3:1 ratio).
Key Takeaway: The position vectors of special points like midpoints and centroids can be calculated efficiently by averaging the position vectors of the defining points (endpoints for midpoint, vertices for centroid), weighted by the respective ratios in the general section formula.