Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1, -2, 3)$.
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
Here,the position vector of the point $(1, -2, 3)$ is $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$.
The line is parallel to the vector $\vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$.
Substituting these values into the formula,we get:
$\vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + \lambda (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$,where $\lambda$ is a scalar.
Here,the position vector of the point $(1, -2, 3)$ is $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$.
The line is parallel to the vector $\vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$.
Substituting these values into the formula,we get:
$\vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + \lambda (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$,where $\lambda$ is a scalar.
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