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# The equation of the line parallel to $\large\frac{x-3}{1}=\frac{y+3}{5}=\frac{2z-5}{3}$ and passing through the point $(1 , 2 ,5 )$ in vector form is

(1)$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{5j}+\overrightarrow{3k})+ t(\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{5k})$(2)$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{5k})+ t(\overrightarrow{i}+\overrightarrow{5j}+\overrightarrow{3k})$ (3)$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{5j}+\frac{3}{2}\overrightarrow{k})+ t(\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{5k})$(4)$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{5k})+ t(\overrightarrow{i}+\overrightarrow{5j}+\frac{3}{2}\overrightarrow{k})$

The equation of the given line is $\large\frac{x-3}{1}=\frac{y+3}{5}=\frac{2z-5}{3}$
The vector equation of the line passing through the point whose position vector is
$\overrightarrow{a}=\overrightarrow{i}+3 \overrightarrow{j}+5 \overrightarrow{k}$
and parallel to the line
ie parallel to the vector $\overrightarrow{u}=\overrightarrow{i}+5 \overrightarrow{j}+\large\frac{3}{2} \overrightarrow{k}$ is
$\overrightarrow{r}=\overrightarrow{a}+t \overrightarrow{u}$
$\overrightarrow{r}=(\overrightarrow{i}+3 \overrightarrow{j}+5 \overrightarrow{k})+ t(\overrightarrow{i}+5 \overrightarrow{j}+\large\frac{3}{2} \overrightarrow{k})$
Hence 4 is the correct answer.