Three Dimensional Geometry — Class 12 Maths Solution

Here, we use the formula $\overrightarrow {\rm{r}} = \overrightarrow {\rm{b}} + \lambda \overrightarrow {\rm{a}}$,

where $\overrightarrow {\rm{r}}$ is

the equation of the line which passes through $\overrightarrow {\rm{b}}$

and parallel to $\overrightarrow {\rm{a}}$.

Let $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = \widehat {\rm{i}} - 2\widehat {\rm{j}} + 3\widehat {\rm{k}}$

So, vector equation of the line, which is parallel to the vector

$\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}}$ and passes through the vector $\overrightarrow {\rm{b}}$

$= \widehat {\rm{i}} - 2\widehat {\rm{j}} + 3\widehat {\rm{k}}$

is $\overrightarrow {\rm{r}} = \overrightarrow {\rm{b}} + \lambda \overrightarrow {\rm{a}}$.

$\therefore$ $\overrightarrow {\rm{r}} = (\hat i - 2\hat j + 3\hat k) + \lambda (3\hat i - 2\hat j + 6\hat k)$

$\Rightarrow$ $(x\hat i + y\hat j + z\hat k) - (\hat i - 2\hat j + 3\hat k) = \lambda (3\hat i - 2\hat j + 6\hat k)$

$\Rightarrow$ $(x - 1)\widehat {\rm{i}} + (y + 2)\widehat {\rm{j}} + (z - 3)\widehat {\rm{k}} = \lambda (\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}})$