The section formula and the midpoint

What you will be able to do

• write the point that divides a segment internally or externally in a given ratio;
• find the midpoint and the ratio in which a given point divides a segment;
• find where a coordinate plane cuts a segment.

A clamp on a cable

A clamp holds a cable at a point $P$ that is twice as far from $A$ as from $B$. Where is the clamp? "In the ratio $2 : 1$" means $AP : PB = 2 : 1$, so $P$ lies between $A$ and $B$, nearer to $B$.

In the chapter on Straight Lines you met this for two coordinates. The rule is the same in space, one coordinate at a time.

Internal division

If $P$ divides $AB$ internally in the ratio $m : n$, then $P$ is the fraction $\dfrac{m}{m + n}$ of the way from $A$ to $B$. Each coordinate moves that fraction of its change:

and in the same way for $y$ and $z$. The weight $m$ goes with the far point $B$ and $n$ goes with $A$.

For $m = n = 1$ we get the midpoint

External division

A point $Q$ divides $AB$ externally in the ratio $m : n$ if $Q$ lies on the line $AB$, outside the segment, and $AQ : QB = m : n$. Replace $n$ by $-n$:

The ratio $m : n$ cannot be $1 : 1$ here. For $m > n$ the point $Q$ lies beyond $B$, and for $m < n$ it lies beyond $A$.

A worked example

Let $A(2, -3, 1)$ and $B(7, 2, -4)$. Find the points that divide $AB$ in the ratio $2 : 3$ internally and externally.

1. Internal. $x = \dfrac{2 \times 7 + 3 \times 2}{5} = 4$, $y = \dfrac{2 \times 2 + 3 \times (-3)}{5} = -1$, $z = \dfrac{2 \times (-4) + 3 \times 1}{5} = -1$. So $P(4, -1, -1)$.
2. External. $x = \dfrac{2 \times 7 - 3 \times 2}{2 - 3} = -8$, $y = \dfrac{2 \times 2 - 3 \times (-3)}{-1} = -13$, $z = \dfrac{2 \times (-4) - 3 \times 1}{-1} = 11$. So $Q(-8, -13, 11)$.
3. Check $Q$. $AQ$ has differences $(-10, -10, 10)$ and $QB$ has $(15, 15, -15)$. Their lengths are $10\sqrt{3}$ and $15\sqrt{3}$, in the ratio $2 : 3$.
4. Midpoint. It is $\left(\dfrac{9}{2}, -\dfrac{1}{2}, -\dfrac{3}{2}\right)$.

Finding the ratio

To find the ratio in which a point $P$ divides $AB$, let the ratio be $\lambda : 1$. Then $x = \dfrac{\lambda x_2 + x_1}{\lambda + 1}$, so $\lambda = \dfrac{x - x_1}{x_2 - x}$. The same $\lambda$ must come from $y$ and from $z$. If it does not, $P$ is not on the line $AB$. A positive $\lambda$ means internal division, and a negative $\lambda$ means external.

Test $P(3, 0, -1)$ on the line through $A(1, 2, -3)$ and $B(7, -4, 3)$. From $x$: $\lambda = \dfrac{3 - 1}{7 - 3} = \dfrac{1}{2}$. From $y$: $\lambda = \dfrac{0 - 2}{-4 - 0} = \dfrac{1}{2}$. From $z$: $\lambda = \dfrac{-1 + 3}{3 + 1} = \dfrac{1}{2}$. All agree, so $P$ divides $AB$ in the ratio $1 : 2$.

Where a coordinate plane cuts a segment

The $yz$-plane is $x = 0$. Put $x = 0$ in $\lambda = \dfrac{x - x_1}{x_2 - x}$ to get $\lambda = -\dfrac{x_1}{x_2}$. So the $yz$-plane divides $AB$ in the ratio $-x_1 : x_2$. In the same way the $zx$-plane gives $-y_1 : y_2$ and the $xy$-plane gives $-z_1 : z_2$.

If the two ends are on opposite sides of the plane, the ratio is positive and the division is internal. If both ends are on one side, the ratio is negative and the plane cuts the line outside the segment.

Practise with new numbers

Try it in Python

The function divide takes a negative n for an external division. The program checks the lengths $AP$ and $PB$ in both cases.

Common mistakes

• Swapping the weights. The weight $m$ multiplies the coordinates of $B$, the far end. The ratio is $AP : PB$.
• Using $m + n$ for an external division. For an external division use $m - n$ in the denominator and $-n$ in the numerator.
• Trusting one coordinate. A point is on the line $AB$ only if all three coordinates give the same ratio.
• Forgetting the sign of the ratio. A negative ratio is an external division.

Remember these

• Internal: $P = \dfrac{mB + nA}{m + n}$, taken coordinate by coordinate. Midpoint: $m = n = 1$.
• External: $Q = \dfrac{mB - nA}{m - n}$, with $m \ne n$.
• The ratio from a coordinate: $\lambda = \dfrac{x - x_1}{x_2 - x}$. All three coordinates must agree.
• A coordinate plane such as $x = 0$ divides $AB$ in the ratio $-x_1 : x_2$.

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 $(4, -2, 7)$ and $(-6, 8, 3)$.

2. Find the point that divides $A(1, 3, -2)$ and $B(6, -2, 3)$ internally in the ratio $2 : 3$.

3. Find the point that divides $A(2, 1, 0)$ and $B(4, 3, -2)$ externally in the ratio $3 : 1$.

4. In what ratio does the $xy$-plane divide the segment from $(2, 1, -4)$ to $(3, -1, 6)$? Find the point.

5. In what ratio does $P(4, 0, 2)$ divide the join of $A(0, 2, -4)$ and $B(6, -1, 5)$?

References

• NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry, section on the section formula.
• OpenStax, Calculus Volume 3, Chapter 2, section 2.2 on vectors in three dimensions.