Question

The point of intersection of the lines $\overrightarrow{r}=(-\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})+t(-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$ and $\overrightarrow{r}=(2\overrightarrow{i}+3\overrightarrow{j}+5\overrightarrow{k})+s(\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})$ is:

• A. $(1,1,2)$
• B. $(2,1,1)$
• C. $(1,1,1)$
• D. $(1,2,1)$

Answer 1

The correct option is A $(1,1,2)$

Given two lines $\overrightarrow{r}=(-\hat{i}+2\hat{j}+3\hat{k})+t(-2\hat{i}+\hat{j}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}+3\hat{j}+5\hat{k})+t(\hat{i}+2\hat{j}+3\hat{k})$

$\Rightarrow \overrightarrow{r}=(-1-2t)\hat{i}+(2+t)\hat{j}+(3+t)\hat{k}$

$\Rightarrow \overrightarrow{r}=(2+s)\hat{i}+(3+2s)\hat{j}+(5+3s)\hat{k}$

Equating the coefficients of the above vector equations, we get

$-1-2t=2+s$ or $s+2t=-3$ .........$\left(1\right)$

$2+t=3+2s$ or $2s-t=-1$ .........$\left(2\right)$

$3+t=5+3s$ or $3s-t=-2$ .........$\left(3\right)$

Eqn.$\left(2\right)$-Eqn.$\left(3\right)=2s-t-3s+t=-1+2$

$\Rightarrow -s=1$ or $s=-1$

Substituting the value of $s=-1$ in $\left(1\right)$ we get

$s+2t=-3\Rightarrow 2t=-3-s=-3-(-1)=-3+1=-2$

$\Rightarrow t=-1$

$\therefore \overrightarrow{r}=(-1-2t)\hat{i}+(2+t)\hat{j}+(3+t)\hat{k}$ becomes

$\overrightarrow{r}=(-1-2\times -1)\hat{i}+(2-1)\hat{j}+(3-1)\hat{k}=\hat{i}+\hat{j}+2\hat{k}$

Thus, the point intersects at $(1,1,2)$