vector equation of line joining the points having position vectors \(\vec a_1=\hat i-\hat j+4\hat k\) and \(\vec a_2=2\hat i+\hat j+2\hat k\) is

\(\vec b = \vec a_2-\vec a_1=(2\hat i+\hat j+2\hat k)-(\hat i-\hat j+4\hat k)\) 

 = \(\hat i+2\hat j-2\hat k\) 

Given that required line is passing through the point having position vector \(\vec a=\hat i-2\hat j-3\hat k\) 

\(\therefore\) Equation of required line is

\((\vec r-\vec a)+\lambda \vec b=0\)

(\(\because\) Required line is passing through \(\vec a\) and parallel to line \(\vec b\))

⇒ (\(\vec r-(\hat i-2\hat j-3\hat k)\)) + \(\lambda(\hat i+2\hat j-2\hat k)=0\)

which is vector equation of required line

Let \(\vec r = x\hat i+y\hat j+z\hat k\) 

Then (x\(\hat i\) + y\(\hat j\) + z\(\hat k\)) - (\(\hat i-2\hat j-3\hat k\)) = \(-\lambda(\hat i+2\hat j-2\hat k)\)

⇒ (x - 1)\(\hat i+(y+2)\hat j+(z+3)\hat k=-\lambda\hat i-2\lambda \hat j+2\lambda\hat k\)

⇒ x - 1 = -\(\lambda\), y + 2 = -2\(\lambda\) and z + 3 = 2\(\lambda\)

(On comparing coefficient of \(\hat i, \hat j\&\hat k\))

⇒ \(\lambda\) = \(\frac{x-1}{-1}, \lambda=\frac{y+2}{-2},\lambda=\frac{z+3}{2}\)

⇒ \(\frac{x-1}{-1}=\frac{y+2}{-2}=\frac{z+3}2\) which is cartesian equation of required line

Alternate method:

Position vector of point is \(\hat i-2\hat j-3\hat k\) 

\(\therefore\) Point is (1, -2, -3)

Hence, required line is passing through point (1, -2, -3)

Similarly, required line is parallel to the line joining points (1, -1, 4) and (2, 1, 2).

\(\therefore\) Direction ratios of line joining points (1, -1, 4) & (2, 1, 2) are \(\pm(2-1,\pm(1-(-1))),\pm(2,-4)\)

or 1, 2, -2 or -1, -2, 2.

\(\therefore\) Direction ratios of required line are 1,2,-2 or-1, -2, 2 and it is passing through point (1, -2, -3)

 \(\therefore\) Cartesian equation of required line \(\frac{x-1}1=\frac{y-(-2)}2=\frac{z-(-3)}{-2}\)

or \(\frac{x-1}{-1}=\frac{y-(-2)}{-2}=\frac{2-(-3)}2\) 

⇒ \(\frac{x-1}1=\frac{y+2}2=\frac{z+3}{-2}\)

or \(\frac{x-1}{-1}=\frac{y+2}{-2}=\frac{z+3}2\)