Question

Equation of a line passing through the point $\hat{i}+\hat{j}-\hat{k}$ and parallel to the vector $2\hat{i}+\hat{j}+2\hat{k}$ is

• A. $\overrightarrow{r}=(1+2t)\hat{i}+(1+t)\hat{j}+(1+2t)\hat{k}$
• B. $\overrightarrow{r}=(1+2t)\hat{i}+(1+t)\hat{j}+(2t-1)\hat{k}$
• C. $\overrightarrow{r}=(2t-1)\hat{i}+(1+t)\hat{j}+(2t+1)\hat{k}$
• D. $\overrightarrow{r}=(1-2t)\hat{i}+(1+t)\hat{i}+(2t+1)\hat{k}$

Answer 1

Answer: (B)

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

The required equation of the line in Cartesian format will be
$\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{2}=t$
Hence, any point on the line can be written as
$x=2t+1$, $y=t+1$ and $z=2t-1$
Hence in vector format,
$\overrightarrow{r}=xi+yj+zk$
$=(2t+1)i+(t+1)j+(2t-1)k$