Question

Show that the points whose position vectors are as given below are collinear:
(i) $2\hat{i}+\hat{j}-\hat{k},3\hat{i}-2\hat{j}+\hat{k}\mathrm{and}\hat{i}+4\hat{j}-3\hat{k}$

(ii) $3\hat{i}-2\hat{j}+4\hat{k},\hat{i}+\hat{j}+\hat{k}\mathrm{and}-\hat{i}+4\hat{j}-2\hat{k}$

Answer 1

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

$\overrightarrow{BC}=$Position vector of C $-$ Position vector of B
$$\begin{gathered} =\hat{i}+4\hat{j}-3\hat{k}-3\hat{i}+2\hat{j}-\hat{k} \\ =-2\hat{i}+6\hat{j}-4\hat{k} \\ =-2\left(\hat{i}-3\hat{j}+2\hat{k}\right) \end{gathered}$$

$\therefore \overrightarrow{AB}=-2\overrightarrow{BC}$.
$\mathrm{So},\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel vectors. But B is a point common to them.
Hence, $A,B$ and C are collinear.

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

$\overrightarrow{BC}=$Position vector of C $-$ Position vector of B
$$\begin{gathered} =-\hat{i}+4\hat{j}-2\hat{k}-\hat{i}-\hat{j}-\hat{k} \\ =-2\hat{i}+3\hat{j}-3\hat{k} \end{gathered}$$

$\therefore \overrightarrow{AB}=\overrightarrow{BC}$
$\mathrm{So},\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel vectors.But $B$ is a point common to them.
Hence, $A,B$ and $C$ are collinear.