Question

Show that the points whose position vectors are $-2\hat{i}+3\hat{j},\hat{i}+2\hat{j}+3\hat{k}\mathrm{and}7\hat{i}-\hat{k}$ are collinear.

Answer 1

Let the given points be P, Q and R and let their position vectors be $\overrightarrow{a},\overrightarrow{b}\mathrm{and}\overrightarrow{c},\mathrm{respectively}.$

$$\begin{gathered} \overrightarrow{a}=-2\hat{i}+3\hat{j} \\ \overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{c}=7\hat{i}+9\hat{k} \end{gathered}$$

Vector equation of line passing through P and Q is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a}+\lambda \left(\overrightarrow{b}-\overrightarrow{a}\right) \\ \Rightarrow \overrightarrow{r}=\left(-2\hat{i}+3\hat{j}\right)+\lambda \left\{\left(\hat{i}+2\hat{j}+3\hat{k}\right)-\left(-2\hat{i}+3\hat{j}\right)\right\} \\ \Rightarrow \overrightarrow{r}=\left(-2\hat{i}+3\hat{j}\right)+\lambda \left(3\hat{i}-\hat{j}+3\hat{k}\right)...\left(1\right) \end{gathered}$$

If points P, Q and R are collinear, then point R must satisfy (1).

$\mathrm{Replacing}\overrightarrow{r}\mathrm{by}\overrightarrow{c}=7\hat{i}+9\hat{k}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}$
$7\hat{i}+9\hat{k}=\left(-2\hat{i}+3\hat{j}\right)+\lambda \left(3\hat{i}-\hat{j}+3\hat{k}\right)$

Comparing the coefficients of $\hat{i},\hat{j}\mathrm{and}\hat{k}$, we get
$7=-2+3\lambda ,0=3-\lambda ,9=3\lambda$
∴ $\lambda$=3

These three equations are consistent, i.e. they give the same value of $\lambda$.
Hence, the given three points are collinear.

Disclaimer: The question given in the book has a minor error. The third vectors should be $7\hat{i}+9\hat{k}$. The solution here is created accordingly.