Question
Show that the points whose position vectors are $-2\hat{\text{i}}+3\hat{\text{j}},\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $7\hat{\text{i}}-\hat{\text{k}}$ are collinear.
✓
Answer
Let the given points be P, Q and R and let their position vectors be $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}},$ respectively.
$\vec{\text{a}}=-2\hat{\text{i}}+3\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{c}}=7\hat{\text{i}}+9\hat{\text{k}}$
Vector equation of line passing through P and Q is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\Rightarrow\vec{\text{r}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big\{\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)\big\}$
$\Rightarrow\vec{\text{r}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)\dots(1)$
If points P, Qand R are collinear, then point R must satisfy (1).
Replacing $\vec{\text{r}}$ by $\vec{\text{c}}=7\hat{\text{i}}+9\hat{\text{k}}$ in (1), we get
$7\hat{\text{i}}+9\hat{\text{k}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)$
Comparing the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$7=-2+3\lambda,0=3-\lambda,9=3\lambda$
$\therefore\lambda=3$
These three equations are consistent, i.e. they give the same value of $\lambda.$
Hence, the given three points are colinear.
Disclaimar: The question given in the book has a minor error. The third vectors should be $7\hat{\text{i}}+9\hat{\text{k}}.$
The solution here is created accordingly
$\vec{\text{a}}=-2\hat{\text{i}}+3\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{c}}=7\hat{\text{i}}+9\hat{\text{k}}$
Vector equation of line passing through P and Q is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\Rightarrow\vec{\text{r}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big\{\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)\big\}$
$\Rightarrow\vec{\text{r}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)\dots(1)$
If points P, Qand R are collinear, then point R must satisfy (1).
Replacing $\vec{\text{r}}$ by $\vec{\text{c}}=7\hat{\text{i}}+9\hat{\text{k}}$ in (1), we get
$7\hat{\text{i}}+9\hat{\text{k}}=\big(-2\hat{\text{i}}+3\hat{\text{j}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)$
Comparing the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$7=-2+3\lambda,0=3-\lambda,9=3\lambda$
$\therefore\lambda=3$
These three equations are consistent, i.e. they give the same value of $\lambda.$
Hence, the given three points are colinear.
Disclaimar: The question given in the book has a minor error. The third vectors should be $7\hat{\text{i}}+9\hat{\text{k}}.$
The solution here is created accordingly
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