Three Dimensional Geometry — Class 12 Maths Solution

We have, $\vec a = 2\hat i - \hat j + 4\hat k,\vec b = \hat i + 2\hat j - \hat k$

Vector equation of the line is $\vec r = \vec a + \lambda \vec b$

$\Rightarrow$ $\vec r = (2\hat i - \hat j + 4\hat k) + \lambda (\hat i + 2\hat j - \hat k)$

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

$\therefore$ $x\hat i + y\hat j + z\hat k = (2\hat i - \hat j + 4\hat k) + \lambda (\hat i + 2\hat j - \hat k)$

$\Rightarrow$ $x\hat i + y\hat j + z\hat k = (2 + \lambda )\hat i + ( - 1 + 2\lambda )\hat j + (4 - \lambda )\hat k$

Eliminating $\lambda$,

we get
$\cfrac{{x - 2}}{1} = \cfrac{{y + 1}}{2} = \cfrac{{z - 4}}{{ - 1}}$ is the cartesian equation of line.