Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

(N/A) The line passes through the point with position vector $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$.
The direction of the line is given by the vector $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.
The vector equation of a line passing through a point $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\lambda$ is a scalar.
Substituting the values,we get the vector equation:
$\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$.
For the Cartesian form,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Then $x\hat{i} + y\hat{j} + z\hat{k} = (2 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (4 - \lambda)\hat{k}$.
Comparing the components,we have:
$x = 2 + \lambda \Rightarrow \lambda = x - 2$
$y = -1 + 2\lambda \Rightarrow \lambda = \frac{y + 1}{2}$
$z = 4 - \lambda \Rightarrow \lambda = 4 - z = \frac{z - 4}{-1}$
Equating the values of $\lambda$,we get the Cartesian equation:
$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$.