Question

Using vector method, prove that the following points are collinear:
(i) A (6, −7, −1), B (2, −3, 1) and C (4, −5, 0)
(ii) A (2, −1, 3), B (4, 3, 1) and C (3, 1, 2)
(iii) A (1, 2, 7), B (2, 6, 3) and C (3, 10, −1)
(iv) A (−3, −2, −5), B (1, 2, 3) and C (3, 4, 7)

Answer 1

(i) Given the points $A\left(6,-7,-1\right),B\left(2,-3,1\right)$ and $C\left(4,-5,0\right)$. Then,
$\overrightarrow{AB}=$Position vector of B $-$ Position vector of A
$$\begin{gathered} =2\hat{i}-3\hat{j}+\hat{k}-6\hat{i}+7\hat{j}+\hat{k} \\ =-4\hat{i}+4\hat{j}+2\hat{k} \\ =-2\left(2\hat{i}-2\hat{j}-\hat{k}\right) \end{gathered}$$

$\overrightarrow{BC}=$Position vector of C $-$ Position vector of B
$$\begin{gathered} =4\hat{i}-5\hat{j}-2\hat{i}+3\hat{j}-\hat{k} \\ =2\hat{i}-2\hat{j}-\hat{k} \end{gathered}$$

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

(ii) Given the points $A\left(2,-1,3\right),B\left(4,3,1\right)$ and $C\left(3,1,2\right)$. Then,
$\overrightarrow{AB}=$ Position vector of B $-$ Position vector of A
$$\begin{gathered} =4\hat{i}+3\hat{j}+\hat{k}-2\hat{i}+\hat{j}-3\hat{k} \\ =2\hat{i}+4\hat{j}-2\hat{k} \\ =-2\left(-\hat{i}-2\hat{j}+\hat{k}\right) \end{gathered}$$

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

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

(iii) Given the points $A\left(1,2,7\right),B\left(2,6,3\right)$ and $C\left(3,10,-1\right)$. Then,
$\overrightarrow{AB}=$ Position vector of B $-$ Position vector of A
$$\begin{gathered} =2\hat{i}+6\hat{j}+3\hat{k}-\hat{i}-2\hat{j}-7\hat{k} \\ =\hat{i}+4\hat{j}-4\hat{k} \end{gathered}$$

$\overrightarrow{BC}=$Position vector of C $-$ Position vector of B.
$$\begin{gathered} =3\hat{i}+10\hat{j}-\hat{k}-2\hat{i}-6\hat{j}-3\hat{k} \\ =\hat{i}+4\hat{j}-4\hat{k} \end{gathered}$$

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

(iv) Given the points $A\left(-3,-2,-5\right),B\left(1,2,3\right)$ and $C\left(3,4,7\right)$. Then,
$\overrightarrow{AB}=$Position vector of B $-$ Position vector of A
$$\begin{gathered} =\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}+2\hat{j}+5\hat{k} \\ =4\hat{i}+4\hat{j}+8\hat{k} \\ =2\left(2\hat{i}+2\hat{j}+4\hat{k}\right) \end{gathered}$$

$\overrightarrow{BC}=$ Position vector of C $-$ Position vector of B
$$\begin{gathered} =3\hat{i}+4\hat{j}+7\hat{k}-\hat{i}-2\hat{j}-3\hat{k} \\ =2\hat{i}+2\hat{j}+4\hat{k} \end{gathered}$$

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