Question

Prove that the points having position vectors $\hat{i}+2\hat{j}+3\hat{k},3\hat{i}+4\hat{j}+7\hat{k},-3\hat{i}-2\hat{i}-5\hat{k}$ are collinear.

Answer 1

Let $A,B,C$ be the points with position vectors $\hat{i}+2\hat{j}+3\hat{k},3\hat{i}+4\hat{j}+7\hat{k},-3\hat{i}-2\hat{j}-5\hat{k}.$ Then,
$\overrightarrow{AB}=$ Position vector of B $-$ Position vector of A
$$\begin{gathered} =3\hat{i}+4\hat{j}+7\hat{k}-\hat{i}-2\hat{j}-3\hat{k} \\ =2\hat{i}+2\hat{j}+4\hat{k} \end{gathered}$$

$\overrightarrow{BC}=$Position vector of C $-$ Position vector of B
$$\begin{gathered} =-3\hat{i}-2\hat{j}-5\hat{k}-3\hat{i}-4\hat{j}-7\hat{k} \\ =-6\hat{i}-6\hat{j}-12\hat{k} \\ =-3\left(2\hat{i}+2\hat{j}+4\hat{k}\right) \end{gathered}$$
∴ $\overrightarrow{BC}=-3\overrightarrow{AB}$
$\mathrm{So},\overrightarrow{AB}\mathrm{and}\overrightarrow{BC}$ are parallel vectors.
But $B$ is a point common to them.
So, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear.
Hence, $A,B,C$ are collinear.