Vectors: Section Formula and Remaining Questions of Exercise 10.2

This lesson completes Exercise 10.2 on vectors. It applies the section formula to divide a line in a given ratio (internally, externally, and at the midpoint), then proves a right-angled triangle, checks collinearity, and finds direction cosines.

The section formula

Let $P$ and $Q$ have position vectors $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ divides $PQ$ in the ratio $m : n$, then:

Internal division

$\vec{OR} = \frac{m\,\vec{b} + n\,\vec{a}}{m + n}$

External division

$\vec{OR} = \frac{m\,\vec{b} - n\,\vec{a}}{m - n}$

Midpoint (the case $m : n = 1 : 1$)

$\vec{OR} = \frac{\vec{a} + \vec{b}}{2}$

Dividing a line in the ratio 2 to 1

Given $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$, with $m : n = 2 : 1$.

Internally

$\vec{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1}$

$\vec{OR} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\tfrac{1}{3}\hat{i} + \tfrac{4}{3}\hat{j} + \tfrac{1}{3}\hat{k}$

Externally

$\vec{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1}$

$\vec{OR} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 3\hat{k}$

Midpoint of two points

For $P(2, 3, 4)$ and $Q(4, 1, -2)$, the position vectors are $\vec{OP} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{OQ} = 4\hat{i} + \hat{j} - 2\hat{k}$.

$\text{Midpoint} = \frac{\vec{OP} + \vec{OQ}}{2} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$

Proving a right-angled triangle

The vertices have position vectors $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$, and $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$. Form the side vectors and their lengths.

Side AB

$\vec{AB} = \vec{b} - \vec{a} = -\hat{i} + 3\hat{j} + 5\hat{k}, \qquad |\vec{AB}|^2 = 1 + 9 + 25 = 35$

Side BC

$\vec{BC} = \vec{c} - \vec{b} = -\hat{i} - 2\hat{j} - 6\hat{k}, \qquad |\vec{BC}|^2 = 1 + 4 + 36 = 41$

Side CA

$\vec{CA} = \vec{a} - \vec{c} = 2\hat{i} - \hat{j} + \hat{k}, \qquad |\vec{CA}|^2 = 4 + 1 + 1 = 6$

The longest side is $BC$. Since

$|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 = |\vec{BC}|^2,$

the converse of Pythagoras' theorem holds, so the triangle is right-angled at $A$.

Testing collinearity

Given $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$, factor out $-2$:

$\vec{b} = -2(2\hat{i} - 3\hat{j} + 4\hat{k}) = -2\,\vec{a}$

Since $\vec{b} = k\,\vec{a}$ for a constant $k = -2$, the two vectors are collinear.

Direction cosines

For the vector $\hat{i} + 2\hat{j} + 3\hat{k}$, the magnitude is

$\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}.$

The direction cosines are each component divided by this magnitude:

$l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}.$

Key takeaways

• The section formula divides a line in ratio $m : n$ internally as $\tfrac{m\vec{b} + n\vec{a}}{m + n}$ and externally as $\tfrac{m\vec{b} - n\vec{a}}{m - n}$; the midpoint is $\tfrac{\vec{a} + \vec{b}}{2}$.
• Three points form a right-angled triangle when the squared lengths of the two shorter sides add to the squared length of the longest side.
• Two vectors are collinear when one is a scalar multiple of the other, and the direction cosines of a vector are its components divided by its magnitude.