Mathematics - Vector Question with Solution | TestHub

A variable point on line$\frac{x + 1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$is$\left(-3t-1, 2t+3,-t+2\right)$

$\therefore DR$of variable point &$\left(-4, 3, 1\right)$is

$\left(-3t+3,2t,-t+1\right)$

Since line is parallel to$x+2y-z-5=0$

$\therefore -3t+3+4t+t-1=0$

$\Rightarrow t=-1$

$\therefore DR$of line is$\left(6,-2, 2\right)$or$\left(3, -1, 1\right)$

$\therefore$Equation of line is$\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$