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

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

The correct answer is Option (1) → $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}); \frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$ ##

Vector equation of line is

$\vec{r} = \vec{a} + \lambda\vec{b}$

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

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, put $\vec{r} = (x\hat{i} + y\hat{j} + z\hat{k})$ in eq. (i), we get

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

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

On comparing both sides, we get

$x = 5 + 3\lambda, \quad y = 2 + 2\lambda, \quad z = -(4 + 8\lambda)$

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

Thus, Cartesian equation of line is

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