Position vectors and the section formula

What you will be able to do

• use the section formula for internal and external division;
• find the midpoint of a segment and the centroid of a triangle;
• test whether three points are collinear, and find the ratio in which a plane divides a segment.

The hanger on a bridge cable

A cable runs from the top of one tower to the top of another. A hanger must be fixed one third of the way along it. Where exactly is that point? If we know the position vectors of the two tower tops, vectors give the answer in one line.

The section formula

Let $A$ and $B$ have position vectors $\vec a$ and $\vec b$. A point $P$ on the segment $AB$ divides it internally in the ratio $m : n$ when $AP : PB = m : n$. Then $\overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}$, so

Look at the formula as a recipe. The ratio number $m$ goes with the far end $B$. When $m$ is small, $P$ is near $A$, and $B$ has a small weight. In coordinates, the same formula works for each coordinate: $x = \dfrac{n x_1 + m x_2}{m + n}$, and likewise for $y$ and $z$.

When $P$ lies on the line $AB$ but outside the segment, it divides $AB$ externally in the ratio $m : n$. Replace $n$ by $-n$:

Midpoint, centroid and the parallelogram

• Midpoint. Put $m = n = 1$: $\dfrac{\vec a + \vec b}{2}$.
• Centroid. The medians of a triangle meet at the centroid, which has the position vector $\dfrac{\vec a + \vec b + \vec c}{3}$.
• Parallelogram. The diagonals of a parallelogram bisect each other. So for the parallelogram $ABCD$, the midpoints of $AC$ and $BD$ are the same point. This gives $\vec d = \vec a + \vec c - \vec b$.

Collinear points and the ratio from a plane

Three points $A$, $B$, $C$ are collinear when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel, that is, when one is a number times the other. The number tells you the ratio.

To find where a coordinate plane cuts a segment, put the right coordinate of the section formula equal to zero. This gives $m : n$. A negative ratio means that the plane cuts the line outside the segment.

A worked example

The points are $A(2, -1, 3)$ and $B(8, 5, -3)$.

1. Internal, $1 : 2$. $\vec p = \dfrac{2\vec a + \vec b}{3} = \dfrac{(4 + 8,\ -2 + 5,\ 6 - 3)}{3} = (4, 1, 1)$. Check: $\overrightarrow{AP} = (2, 2, -2)$ and $\overrightarrow{PB} = (4, 4, -4)$, so $AP : PB = 1 : 2$.
2. External, $2 : 1$. $\vec q = \dfrac{2\vec b - \vec a}{2 - 1} = (16 - 2,\ 10 + 1,\ -6 - 3) = (14, 11, -9)$. Here $\overrightarrow{AQ} = (12, 12, -12)$ and $\overrightarrow{QB} = (-6, -6, 6)$, so $AQ$ is twice $QB$ and $B$ lies between $A$ and $Q$.
3. Midpoint. $(5, 2, 0)$.

A plane example. Let $A(4, -2, 1)$ and $B(0, 6, 9)$. The $xz$-plane is $y = 0$. Put the $y$-coordinate of the section formula equal to zero: $n(-2) + m(6) = 0$, so $m : n = 1 : 3$. The point is $\dfrac{3\vec a + \vec b}{4} = (3, 0, 3)$.

A triangle with $A(0, 6, 3)$, $B(9, 0, -3)$ and $C(3, -3, 0)$ has the centroid $\left(\dfrac{12}{3}, \dfrac{3}{3}, 0\right) = (4, 1, 0)$.

A parallelogram $ABCD$ has $A(1, 0, 2)$, $B(4, 3, -1)$ and $C(6, 7, 3)$. Then $D = A + C - B = (3, 4, 6)$, and the diagonals meet at $\left(\dfrac{7}{2}, \dfrac{7}{2}, \dfrac{5}{2}\right)$.

For collinear points, take $A(0, 1, -1)$, $B(2, 5, 3)$ and $C(5, 11, 9)$. Then $\overrightarrow{AB} = (2, 4, 4)$ and $\overrightarrow{AC} = (5, 10, 10) = \dfrac{5}{2}\overrightarrow{AB}$. So the points are collinear, and $B$ divides $AC$ in the ratio $AB : BC = 2 : 3$.

Practise with new numbers

Try it in Python

The program uses exact fractions. It divides $AB$ internally and externally and finds the midpoint, for the points of the worked example.

Common mistakes

• Swapping $m$ and $n$. In $AP : PB = m : n$ the number $m$ goes with $B$.
• Forgetting to divide by $m + n$.
• A wrong sign in the external division. It is $\dfrac{m\vec b - n\vec a}{m - n}$, with a minus in both places.
• Reading $AP : PB$ as $AP : AB$. The second number of the ratio is the other part of the segment, not the whole of it.
• Taking $\overrightarrow{AB}$ as $\vec a - \vec b$.

Remember these

• Internal division, $AP : PB = m : n$: $\ \vec p = \dfrac{n\vec a + m\vec b}{m + n}$.
• External division: $\ \vec p = \dfrac{m\vec b - n\vec a}{m - n}$.
• Midpoint $\dfrac{\vec a + \vec b}{2}$, centroid $\dfrac{\vec a + \vec b + \vec c}{3}$, and $\vec d = \vec a + \vec c - \vec b$ in a parallelogram $ABCD$.
• Collinear points: $\overrightarrow{AC} = \lambda\overrightarrow{AB}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the midpoint of $P(4, -2, 6)$ and $Q(-2, 8, 0)$.

2. Find the point that divides $A(1, 2, 3)$ and $B(7, -4, 9)$ internally in the ratio $2 : 1$.

3. Find the point that divides $A(1, 2, -1)$ and $B(3, 4, -5)$ externally in the ratio $3 : 2$.

4. In what ratio does the plane $x = 4$ divide the segment joining $A(2, 1, 5)$ and $B(8, 7, -1)$? Find the point.

5. Find the centroid of the triangle with vertices $(3, 1, 5)$, $(-2, 4, 0)$ and $(5, -2, -2)$.

References

• NCERT, Mathematics Part II for Class XII, Chapter 10, section on the section formula.
• OpenStax, Calculus Volume 3, Chapter 2, section on vectors in three dimensions.