Question

The two straight lines $\overrightarrow{r}={\overrightarrow{r}}_1+t{\overrightarrow{a}}_1$ and $\overrightarrow{r}={\overrightarrow{r}}_2+t{\overrightarrow{a}}_2$, where ${\overrightarrow{r}}_1=j,{\overrightarrow{a}}_1=i+2j-k$ and ${\overrightarrow{r}}_2=i+j+k,{\overrightarrow{a}}_2=-2i-2j$ have a common point. Determine the coordinates of that point.

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

Answer 1

The correct option is C $(-1,-1,1)$

given lines

$\overrightarrow{r}=\overrightarrow{r_1}+t\overrightarrow{a_1}$

$\overrightarrow{r_1}=\hat{j},\overrightarrow{a_1}=\hat{i}+2\hat{j}-\hat{k}$

putting $r_1,a_1$

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

in cartesian form

$\frac{x}{1}=\frac{y-1}{2}=\frac{z}{-1}=t$

$x=t,y=2t+1,z=-t$

$\overrightarrow{r}=\overrightarrow{r_2}+t\overrightarrow{a_2}$

$\overrightarrow{r_2}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{a_2}=-2\hat{i}-2\hat{j}$

putting $r_2,a_2$

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

in cartesian form

$\frac{x-1}{-2}=\frac{y-1}{-2}=\frac{z-1}{0}$

putting value of $x,y,z$

$\frac{t-1}{-2}=\frac{2t+1-1}{-2}=\frac{-t-1}{0}$

$\frac{t-1}{-2}=\frac{2t}{-2}=\frac{-t-1}{0}$

$\frac{t-1}{-2}=\frac{2t}{-2}$

$t-1=2t$

$t=-1$

common point

$x=t=-1$

$y=2t+1=-2+1=-1$

$z=-t=1$

Point$(-1,-1,1)$