Discussion Forum

Perpendicular are drawn  from points on the line   $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the  plane x+y+z=3 . The feet of perpendicular lie on the line

Answer:

Explanation:

Concept involved

  To find the foot of perpendicular and find its locus

 Formula used

 Foot  of perpendicular form  $(x_{1},y_{1},z_{1})$   to ax+by+cz+d=0 be    $(x_{2},y_{2},z_{2})$

  then

$\frac{x_{2}-x_{1}}{a}=\frac{y_{2}-y_{1}}{b}=\frac{z_{2}-z_{1}}{a}$

                               = $\frac{-(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}$

      Any point on,     $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$

                     $\Rightarrow x=2\lambda-2,y=\lambda-1, z=3\lambda$

Let focus of perpendicular from   $(2\lambda-2,\lambda-1, 3\lambda)$

 to x+y+z=3  be ($(x_{2},y_{2},z_{2})$)

 $\therefore$       $\frac{x_{2}-(2\lambda-2)}{1}=\frac{y_{2}-(-\lambda-1)}{1}=\frac{z_{2}-(3\lambda)}{1}$

   =  $-\frac{(2\lambda-2-\lambda-1+3\lambda-3)}{1+1+1}$

               =  $x_{2}-2\lambda+2=y_{2}+\lambda+1$

   =  $z_{2}-3\lambda=2-\frac{4\lambda}{3}$

 $\therefore$        $x_{2}=\frac{2\lambda}{3}$     ,  $y_{2}=1-\frac{7\lambda}{3}$,  $z_{2}=2+\frac{5\lambda}{3}$

 $\Rightarrow \lambda=\frac{x_{2}-0}{2/3}=\frac{y_{2}-1}{-7/3}=\frac{z_{2}-2}{5/3}$

$\therefore$  foot  of perpendicular lie on

                 $\frac{x}{2/3}=\frac{y_{}-1}{-7/3}=\frac{z_{}-2}{5/3}$

   $\Rightarrow\frac{x}{2}=\frac{y_{}-1}{-7}=\frac{z_{}-2}{5}$