Three Dimensional Geometry — Class 12 Maths Solution

.: Let $\vec a$ and $\vec b$ be the position vectors of points

$A(3, - 2, - 5)$ and $B(3, - 2,6)$

Let $\vec a = 3\hat i - 2\hat j - 5\hat k$

and Let $\vec b = 3\hat i - 2\hat j + 6\hat k$
$\Rightarrow$ $\vec b - \vec a = 0\hat i + 0\hat j + 11\hat k$

Let $\vec r$ be the position vector of any point on the line. Then, the vector equation of line is $\vec r = \vec a + \lambda (\vec b - \vec a)$

$\Rightarrow$ $\vec r = 3\hat i - 2\hat j - 5\hat j + \lambda (11\hat k)$

Now, $\vec r = x\hat i + y\hat j + z\hat k = 3\hat i - 2\hat j + ( - 5 + 11\lambda )\hat k$

Eliminating $\lambda$,

we get $\cfrac{{x - 3}}{0} = \cfrac{{y + 2}}{0} = \cfrac{{z + 5}}{{11}}$

is the equation of line in cartesian form.