Discussion Forum

The distance of the point  (1,0,2)  from the point of intersection of the line  $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$   and the plane x-y+z=16 is

Answer:

Explanation:

 Given equation of the line is

                                  $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ =   $\lambda$              (say)......(i)

  and equation of plane is

  x-y+z=16        ...........(ii)

 Any point on the  line (i) is 

$(3\lambda+2,4\lambda-1,12\lambda+2)$

 Let this point be point of intersection of line and plane 

$\therefore$      $(3\lambda+2)-(4\lambda-1)+(12\lambda+2)=16$

 $\Rightarrow$        $11\lambda+5=16$

 $\Rightarrow$          $11\lambda=11$

$\Rightarrow$                       $\lambda=1$

  $\therefore$          Point  of intersection is (5,3,14).

 Now , distance  between  the points (1,0,2) and (5,3,14)

             $= \sqrt{(5-1)^{2}+(3-0)^{2}+(14-2)^{2}}$          

                  $=\sqrt{16+9+144}=\sqrt{169}=13$