10.2 Collinearity of points

2.1 Concept of Unit Vector

3.1 Parallelogram Law of Vector Addition
3.2 Properties of Vector Addition

4.1 Resolution of a Vector:

8.1 Scalar (or dot) products of two vectors
8.2 Vector or Cross Product of two Vectors

10.1 Collinearity of Vectors
10.2 Collinearity of points

Let $A$, $B$ and $C$ be three collinear points. Then each pair of the vector $\overrightarrow {AB} $, $\overrightarrow {BC} $; $\overrightarrow {AB} $, $\overrightarrow {AC} $; $\overrightarrow {BC} $, $\overrightarrow {AC} $ is a pair of collinear vector. In order to check the collinearity of three points, we can check the collinearity of any two vectors obtained with the help of given three points. Therefore, we can conclude

Three vectors are said to be collinear if and only if there exists some scalars $x$, $y$ and $z$ not all zero simultaneously such that $$x\overrightarrow a + y\overrightarrow b + z\overrightarrow c = 0$$ together with $x + y + z = 0$

Proof: Let us assume three points $A$, $B$ and $C$ are collinear. We can say that $\overrightarrow {AB} $ and $\overrightarrow {AC} $ are also collinear. Using the collinearity of vectors, we can write $$\overrightarrow {AB} = \lambda \overrightarrow {AC} ...(1)$$ On the other hand, we can write $$\overrightarrow {AB} = \overrightarrow b - \overrightarrow a {\text{ and }}\overrightarrow {AC} = \overrightarrow c - \overrightarrow a $$ Put these values in equation $(1)$, we get $$\begin{equation} \begin{aligned} \overrightarrow b - \overrightarrow a = \lambda \left( {\overrightarrow c - \overrightarrow a } \right) \\ (\lambda + 1)\overrightarrow a + (1)\overrightarrow b - (\lambda )\overrightarrow c = 0 \\ \Rightarrow x\overrightarrow a + y\overrightarrow b + z\overrightarrow c = 0 \\\end{aligned} \end{equation} $$ where $x = \lambda + 1$, $y=1$ and $z=-\lambda$.

Question 18. Show that the points $A$, $B$ and $C$ with position vectors $ - 2\overrightarrow a + 3\overrightarrow b + 5\overrightarrow c $ , $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c $ and $7\overrightarrow a - \overrightarrow c $ are collinear.

Solution: Vector $\overrightarrow {AB} $ = Position vector of $B$-Position vector of $A$$$(\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ) - ( - 2\overrightarrow a + 3\overrightarrow b + 5\overrightarrow c )$$$$3\overrightarrow a - 2\overrightarrow b - 2\overrightarrow c $$

$\overrightarrow {BC} $ = Position vector of $C$-Position vector of $B$$$ = (7\overrightarrow a - \overrightarrow c) -(\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c) $$$$6\overrightarrow a - 2\overrightarrow b - 4\overrightarrow c $$$$2\overrightarrow {AB} = \overrightarrow {BC} $$

So these points are collinear.