Find the vector and the Cartesian equations of the line through the point $(5, 2, -4)$ and which is parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$.

Vector: $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$; Cartesian: $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$

The correct answer is Option (2) → Vector: $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$; Cartesian: $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$ ##

We have

$\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k} \quad \text{and} \quad \vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$

Therefore, the vector equation of the line is

$\vec{r} = 5\hat{i} + 2\hat{j} - 4\hat{k} + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$

Now, $\vec{r}$ is the position vector of any point $P(x, y, z)$ on the line. Therefore,

$x\hat{i} + y\hat{j} + z\hat{k} = 5\hat{i} + 2\hat{j} - 4\hat{k} + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$

$= (5 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (-4 - 8\lambda)\hat{k}$

Eliminating $\lambda$, we get

$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$

which is the equation of the line in Cartesian form.